Calculus Appendix
Implicit Differentiation, p. 564
1. The chain rule states that if y is a composite
dy
dy du
function, then dx 5 du dx . To differentiate an
equation implicitly, first differentiate both sides of
the equation with respect to x, using the chain rule
dy
for terms involving y, then solve for dx.
d
d
2. a. (x 2 1 y 2 ) 5 (36)
dx
dx
d 2
d 2
d
(x ) 1
(y ) 5
(36)
dx
dx
dx
dy 2
dy
2x 1
3
50
dy
dx
dy
50
2x 1 2y
dx
dy
x
52
y
dx
d
d
b.
(15y 2 ) 5
(2x 3 )
dx
dx
dy 2
dy
d
15
3
5
(2x 3 )
dy
dx
dx
dy
30y
5 6x 2
dx
dy
x2
5
dx
5y
3xy 2 1 y 3 5 8
c.
dy
dy
1 3y 2
50
3y 2 1 3x(2y)
dx
dx
dy
(2xy 1 y 2 ) 5 2y 2
dx
2y 2
dy
5
dx
2xy 1 y 2
d. 9x 2 2 16y 2 5 2144
dy
18x 2 32y
50
dx
9x
dy
5
dx
16y
x2
3 2
1 y 51
e.
16
13
6 dy
2x
1 y
50
16
13 dx
Calculus and Vectors Solutions Manual
dy
50
dx
dy
13x
52
dx
48y
d 2
d
(x 1 y 2 1 5y) 5
(10)
f.
dx
dx
d 2
d 2
d
d
(x ) 1
(y ) 1
(5y) 5
(10)
dx
dx
dx
dx
dy 2
dy
dy
dy
2x 1
3
15
3
50
dy
dx
dy
dx
dy
2x 1
(2y 1 5) 5 0
dx
dy
2x
52
dx
2y 1 5
3. a. x 2 1 y 2 5 13
dy
2x 1 2y
50
dx
At (2, 23),
dy
2(2) 1 2(23)
50
dx
dy
2
5 .
dx
3
The equation of the tangent at (2, 23) is
2
y 5 x 1 b.
3
At (2, 23),
2
23 5 (2) 1 b
3
29 5 4 1 3b
213 5 3b
13
2 5 b.
3
Therefore, the equation of the tangent to
2
13
x 2 1 y 2 5 13 is y 5 x 2 .
3
3
d 2
d
(x 1 4y 2 ) 5
(100)
b.
dx
dx
d
d
d 2
(x ) 1 4 (y 2 ) 5
(100)
dx
dx
dx
dy 2
dy
2x 1 4
3
50
dy
dx
26x 1 96y
1
dy
50
dx
x
dy
52
dx
4y
At the point (28, 3), the slope of the tangent is
2x 1 8y
dy
2
therefore 5 2 428
The equation of the
(3) 5 3 .
dx
tangent line is therefore
2
y 5 (x 1 8) 1 3
3
x2
y2
2
5 21
c.
25
36
2x
2y dy
2
50
25
36 dx
dy
36x 2 25y
50
dx
At (5 !3, 212),
dy
180!3 1 300
50
dx
dy
3 !3
52
.
dx
5
The equation of the tangent is y 5 mx 1 b.
At (5 !3, 212) and with m 5 2 3 !3
5 ,
3!3
212 5 2
(5 !3) 1 b
5
212 5 29 1 b
23 5 b
Therefore, the equation of the tangent is
3"3
y52
x 2 3.
5
d x2
5y 2
d
d.
a 2
b5
(1)
dx 81
162
dx
1 d
5 d 2
d
a b (x 2 ) 2 a
b (y ) 5
(1)
81 dx
162 dx
dx
1
5 dy 2
dy
a b2x 2 a
b
3
50
81
162 dy
dx
2
5 dy
x2 y
50
81
81 dx
dy
2x
5
dx
5y
At the point (211, 24), the slope of the tangent is
therefore
dy
5
2 (211)
5 (24)
The equation of the
5 11
10 .
tangent line is therefore
11
y 5 (x 1 11) 2 4
10
2
dx
4. x 1 y 2 5 1
The line x 1 2y 5 0 has slope of 2 12.
dy
50
1 1 2y
dx
Since the tangent line is parallel to x 1 2y 5 0, then
dy
5 2 12.
dx
1
1 1 2ya2 b 5 0
2
12y50
y51
Substituting,
x1151
x50
Therefore, the tangent line to the curve x 1 y 2 5 1
is parallel to the line x 1 2y 5 0 at (0, 1).
5. a. 5x 2 2 6xy 1 5y 2 5 16
dy
dy
10x 2 a6y 1
(6x)b 1 10y
5 0(1)
dx
dx
At (1, 21),
dy
dy
10 2 a26 1 6 b 2 10
50
dx
dx
dy
16 2 16
50
dx
dy
5 1.
dx
dy
b. When the tangent line is horizontal, dx 5 0.
Substituting,
10x 2 (6y 1 0) 1 0 5 0.
y 5 53 x at the point (x1, y1 ) of tangency:
substitute y1 5 53 x1 into 5x21 2 6x1 y1 1 5y21 5 16.
5
25
5x21 2 6x1 a x1 b 1 5a x21 b 5 16
3
9
45x21 2 90x21 1 125x21 5 144
80x21 5 144
5x21 5 9
3
3
or x1 5 2
x1 5
!5
!5
5
or y1 5 !5
y1 5
!5
5
or y1 5 2 !5
y1 5 2
!5
Therefore, the required points are !5, "5 and
3
2 !5, 2"5 .
(
)
(3
)
Calculus Appendix
6. 5x 2 1 y 2 5 21
dy
50
10x 1 2y
dx
dy
5x
52
y
dx
dy
210
52
At (22, 21),
or 210.
dx
21
The slope of the tangent line to the ellipse at
(22, 21) is 210.
x 3 1 y 3 2 3xy 5 17
7.
dy
dy
3x 2 1 3y 2 2 c 3y 1
(3x) d 5 0
dx
dx
At (2, 3),
dy
dy
12 1 27
2926
50
dx
dx
dy
21 5 23.
dx
dy
The slope of the tangent is 5 2 17.
dx
Therefore, the slope of the normal at (2, 3) is 7.
d
d
("x 1 y 2 2x) 5
(1)
dx
dx
d
d
d
1
( (x 1 y)2 ) 2
(2x) 5
(1)
dx
dx
dx
1
d
1
(x 1 y)22 3
(x 1 y) 2 2 5 0
2
dx
1
dy
1
(x 1 y)22 a1 1
b52
2
dx
dy
5 4"x 1 y 2 1
dx
10. 4x 2y 2 3y 5 x 3
dy
dy
a. 8xy 1
(4x 2 ) 2 3
5 3x 2
dx
dx
dy
(4x 2 2 3) 5 3x 2 2 8xy
dx
dy
3x 2 2 8xy
(1)
5
dx
4x 2 2 3
b. y(4x 2 2 3) 5 x 3
b.
x3
4x 2 3
dy
3x 2 (4x 2 2 3) 2 8x(x 3 )
5
dx
(4x 2 2 3)2
y5
y23
The equation of the normal at (2, 3) is x 2 2 5 7
y 2 3 5 7x 2 14 or 7x 2 y 2 11 5 0
d 2
d
x3
8.
(y ) 5
a
b
dx
dx 2 2 x
dy 2
dy
(2 2 x)(3x 2 ) 2 x 3 (21)
3
5
dy
dx
(2 2 x)2
dy
6x 2 2 2x 3
2y
5
dx
(2 2 x)2
dy
3x 2 2 x 3
5
dx
y(2 2 x)2
dy
3(1)2 2 (1)3
At (1, 21),
5
5 22
dx
(21)(2 2 1)2
The slope of the normal is 12. The equation of the
normal at (1, 21) is y 2 (21) 5 12 (x 2 1) or
y 5 12 x 2 32.
d
d
9. a.
(x 1 y)3 5
(12x)
dx
dx
d
3(x 1 y)2 3
(x 1 y) 5 12
dx
dy
3(x 1 y)2 a1 1
b 5 12
dx
4
dy
5
21
dx
(x 1 y)2
Calculus and Vectors Solutions Manual
5
2
12x 4 2 9x 2 2 8x 4
(4x 2 2 3)2
4x 4 2 9x 2
(4x 2 2 3)2
c. We must show that (1) is equivalent to (2).
5
dy
(2)
3x 2 2 8xy
From (1): dx 5 4x 2 2 3 and substituting,
y5
x3
4x 2 3
2
3x 2 2 8xa 4x 2 2 3 b
x3
dy
5
dx
4x 2 2 3
3x 2 2 (4x 2 2 3) 2 8x 4
5
(4x 2 2 3)2
12x 4 2 9x 2 2 8x 4
5
(4x 2 2 3)2
4x 4 2 9x 2
5
5 (2), as required.
(4x 2 2 3)2
3
11. a.
13.
y
P (4, 6)
(0, 2)
Q
There is one tangent when x 5 1.
b.
A
(4, 0)
(–4, 0)
x
(0, –2)
There is one tangent when x 5 1.
c.
There is one tangent when x 5 1.
d.
There are two tangents when x 5 1.
dy
y
y
x
12.
1
5 10, x 2 y 2 0,
5
x
dx
Äy Äx
dy
dy
1
1
1 y 22 dx x 2 y
1 x 22 1y 2 dxx
50
1 a b
a b
x2
2 x
2 y
y2
1
2
dy
1y 2 x
dx
1
2
dy
x2y
dx
y
x
1 1
50
1
2
2
y
x2
2x
2y2
Multiply by 2x 2y 2:
dy
dy
3 1
1 3
x2y2 ay 2 x b 1 x2y2 a x 2 yb 5 0
dx
dx
3 3
5 1 dy
3 3 dy
1 5
x 2y 2 2 x 2y 2
1 x 2y 2
2 x 2y 2 5 0
dx
dx
dy 32 32
3 1
1 3
3 3
Q x y 2 x2y2 R 5 x2y 2 2 x 2y2
dx
1 3
x 2y 2 (y 2 x)
dy
5 1 1
dx
x 2y 2 (y 2 x)
Let Q have coordinates
"16 2 q 2
(q, f(q)) 5 aq,
b, q , 0.
2
For x 2 1 4y 2 5 16
dy
2x 1 3y
50
dx
dy
x
52 .
dx
4y
q
dy
At Q,
52
dx
2 !16 2 q 2
y26
The line through P has equation x 2 4 5 m.
Since PQ is the slope of the tangent line to
x 2 1 4y 2 5 16, we conclude:
dy
at point Q.
m5
dx
"16 2 q 2
26
2
q24
52
q
2"16 2 q 2
9
"16 2 q 2 2 12
52
2(q 2 4)
2"16 2 q 2
16 2 q 2 2 12"16 2 q 2 5 2q(q 2 4)
16 2 q 2 2 12"16 2 q 2 5 2q 2 1 4q
4 2 q 5 3"16 2 q 2
16 2 8q 1 q 2 5 9(16 2 q 2 )
16 2 8q 1 q 2 5 144 2 9q 2
10q 2 2 8q 2 128 5 0
5q 2 2 4q 2 64 5 0
(5q 1 16)(q 2 4) 5 0
q 5 2 165 or q 5 4 (as expected; see graph)
f(q) 5 65 or f(q) 5 0
dy
y
5 , as required.
x
dx
4
Calculus Appendix
16
dy
5 56 or f(q) 5 0
dx
4Q5 R
2
3
y26
Equation of the tangent at Q is x 2 4 5 23 or
2x 2 3y 1 10 5 0 or equation of tangent at A is
x 5 4.
14.
y
xy = p
x2 – y2 = k
5
P(a, b)
0
x2 – y2 = k
x
xy = p
Let P(a, b) be the point of intersection where a 2 0
and b 2 0.
For x 2 2 y 2 5 k
dy
50
2x 2 2y
dx
dy
x
5
y
dx
At P(a, b),
dy
a
5 .
dx
b
For xy 5 P,
dy
x5P
1?y1
dx
y
dy
52
x
dx
At P(a, b),
dy
b
52 .
a
dx
At point P(a, b), the slope of the tangent line of
xy 5 P is the negative reciprocal of the slope of the
tangent line of x 2 2 y 2 5 k. Therefore, the tangent
lines intersect at right angles, and thus, the two
curves intersect orthogonally for all values of the
constants k and P.
15. !x 1 !y 5 !k
Differentiate with respect to x:
1 12
1 1 dy
x 1 y2
50
2
2 dx
dy
!y
52
dx
!x
Calculus and Vectors Solutions Manual
Let P(a, b) be the point of tangency.
!b
dy
2
dx
!a
Equation of tangent line l at P is
y2b
!b
.
52
x2a
!a
x-intercept is found when y 5 0.
2b
!b
52
x2a
!a
2b!a 5 2 !bx 1 a !b
a !b 1 b!a
x5
!b
a !b 1 b !a
.
!b
Therefore, the x-intercept is
For the y-intercept, let x 5 0,
y2b
!b
.
52
2a
!a
a !b
1 b.
y-intercept is
!a
The sum of the intercepts is
a !b 1 b!a
a!b 1 b!a
1
!b
!a
3
1
3
1
a 2 b2 1 2ab 1 b 2 a 2
5
1 1
a 2b 2
1
1
a 2 b 2 (a 1 2"a"b 1 b)
5
1
1
a2 b2
5 a 1 2"a"b 1 b
1
1
5 (a 2 1 b2 )2
Since P(a, b) is on the curve, then
!a 1 !b 5 !k, or a2 1 b2 5 k2.
Therefore, the sum of the intercepts is 5 k,
as required.
16.
y
1
1
1
y2 = 4x
(–2, 5)
(1, 2)
(4, –1)
x
x + y = –3
5
y 2 5 4x
dy
2t
54
dx
dy
At (1, 2),
5 1.
dx
Therefore, the slope of the tangent line at (1, 2) is 1
and the equation of the normal is
y22
5 21 or x 1 y 5 3.
x21
The centres of the two circles lie on the straight line
x 1 y 5 3. Let the coordinates of the centre of each
circle be (p, q) 5 (p, 3 2 p). The radius of each
circle is 3!2. Since (1, 2) is on the circumference
of the circles,
(p 2 1)2 1 (3 2 p 2 2)2 5 r 2
p 2 2 2p 1 1 1 1 2 2p 1 p 2 5 (3 !2)2
2p 2 2 4p 1 2 5 18
p 2 2 2p 2 8 5 0
(p 2 4)(p 1 2) 5 0
p 5 4 or p 5 22
q 5 21 or q 5 5.
Therefore, the centres of the circles are (22, 5) and
(4, 21). The equations of the circles are
(x 1 2)2 1 (y 2 5)2 5 18 and
(x 2 4)2 1 (y 1 1)2 5 18.
dT(x)
dx
5 2200(1 1 x 2 )22 2x
dt
dt
2400x
dx
5
? .
(1 1 x 2 )2 dt
At a specific time, when x 5 5,
dT(5)
2400(5)
5
(2)
dt
(26)2
24000
5
676
21000
5
169
dT(5)
8 25.9.
dt
Therefore, the temperature is decreasing at a rate of
5.9 °C>s.
b. To determine the distance at which the temperature
is changing fastest, graph the derivative of the
function on a graphing calculator. Since the
temperature decreases as the person moves away
from the fire, the temperature will be changing
fastest when the value of the derivative is at its
minimum value.
For Y1, enter nDeriv( from the MATH menu. Then
enter 200 4 (1+X2), X, X).
Related Rates, pp. 569–570
dA
5 4 m> s2
dt
dS
b.
5 23 m2> min.
dt
ds
5 70 km> h when t 5 0.25
c.
dt
dx
dy
5
d.
dt
dt
du
p
5
e.
rad> s
dt
10
200
2. T(x) 5
1 1 x2
dx
5 2 m> s
a.
dt
1. a.
dT(x)
dt
when x 5 5 m:
200
T(x) 5
1 1 x2
5 200(1 1 x 2 )21
Find
6
The temperature is changing fastest at about 0.58 m.
c. Solve Ts (x) 5 0.
2400x
Tr(x) 5
(1 1 x 2 )2
2400(1 1 x 2 )2 2 2(1 1 x 2 )(2x)(2400x)
Tr(x) 5
(1 1 x 2 )4
Let Ts (x) 5 0,
2400(1 1 x 2 )2 1 1600x 2 (1 1 x 2 ) 5 0.
Divide,
400(1 1 x 2 ) 2 (1 1 x 2 ) 1 4x 2 5 0
3x 2 5 1
1
x2 5
!3
1
x5
!3
x . 0 or x 8 0.58.
Calculus Appendix
y 2 5 4x
dy
2t
54
dx
dy
At (1, 2),
5 1.
dx
Therefore, the slope of the tangent line at (1, 2) is 1
and the equation of the normal is
y22
5 21 or x 1 y 5 3.
x21
The centres of the two circles lie on the straight line
x 1 y 5 3. Let the coordinates of the centre of each
circle be (p, q) 5 (p, 3 2 p). The radius of each
circle is 3!2. Since (1, 2) is on the circumference
of the circles,
(p 2 1)2 1 (3 2 p 2 2)2 5 r 2
p 2 2 2p 1 1 1 1 2 2p 1 p 2 5 (3 !2)2
2p 2 2 4p 1 2 5 18
p 2 2 2p 2 8 5 0
(p 2 4)(p 1 2) 5 0
p 5 4 or p 5 22
q 5 21 or q 5 5.
Therefore, the centres of the circles are (22, 5) and
(4, 21). The equations of the circles are
(x 1 2)2 1 (y 2 5)2 5 18 and
(x 2 4)2 1 (y 1 1)2 5 18.
dT(x)
dx
5 2200(1 1 x 2 )22 2x
dt
dt
2400x
dx
5
? .
(1 1 x 2 )2 dt
At a specific time, when x 5 5,
dT(5)
2400(5)
5
(2)
dt
(26)2
24000
5
676
21000
5
169
dT(5)
8 25.9.
dt
Therefore, the temperature is decreasing at a rate of
5.9 °C>s.
b. To determine the distance at which the temperature
is changing fastest, graph the derivative of the
function on a graphing calculator. Since the
temperature decreases as the person moves away
from the fire, the temperature will be changing
fastest when the value of the derivative is at its
minimum value.
For Y1, enter nDeriv( from the MATH menu. Then
enter 200 4 (1+X2), X, X).
Related Rates, pp. 569–570
dA
5 4 m> s2
dt
dS
b.
5 23 m2> min.
dt
ds
5 70 km> h when t 5 0.25
c.
dt
dx
dy
5
d.
dt
dt
du
p
5
e.
rad> s
dt
10
200
2. T(x) 5
1 1 x2
dx
5 2 m> s
a.
dt
1. a.
dT(x)
dt
when x 5 5 m:
200
T(x) 5
1 1 x2
5 200(1 1 x 2 )21
Find
6
The temperature is changing fastest at about 0.58 m.
c. Solve Ts (x) 5 0.
2400x
Tr(x) 5
(1 1 x 2 )2
2400(1 1 x 2 )2 2 2(1 1 x 2 )(2x)(2400x)
Tr(x) 5
(1 1 x 2 )4
Let Ts (x) 5 0,
2400(1 1 x 2 )2 1 1600x 2 (1 1 x 2 ) 5 0.
Divide,
400(1 1 x 2 ) 2 (1 1 x 2 ) 1 4x 2 5 0
3x 2 5 1
1
x2 5
!3
1
x5
!3
x . 0 or x 8 0.58.
Calculus Appendix
x dt 5 5 cm> s.
dx
3. Given square
x
dA
Find dt when x 5 10 cm.
Solution
Let the side of a square be x cm.
A 5 x2
dA
dx
5 2x
dt
dt
At a specific time, x 5 10 cm.
dA
5 2(10)(5)
dt
5 100
Therefore, the area is increasing at 100 cm2> s
when a side is 10 cm.
P 5 4x
dx
dP
54
dt
dt
dx
At any time, dt 5 5.
dP
5 20.
dt
Therefore, the perimeter is increasing at 20 cm> s.
4. Given cube with sides x cm,
dv
a. Find dt when x 5 5 cm:
dx
55
dt
cm> s.
V 5 x3
dV
dx
5 3x 2
dt
dt
At a specific time, x 5 5 cm.
dV
5 3(5)2 (4)
dt
5 300
Therefore, the volume is increasing at 300 cm3> s.
dS
b. Find dt when x 5 7 cm.
S 5 6x 2
dS
dx
5 12x
dt
dt
At a specific time, x 5 7 cm,
dS
5 12(7)(4)
dt
5 336.
Therefore, the surface area is increasing at a rate of
336 cm2> s.
Calculus and Vectors Solutions Manual
5. Given rectangle
dx
5 2 cm> s
dt
dy
5 23 cm> s
dt
y
x
dA
Find dt when x 5 20 cm and y 5 50 cm.
Solution
A 5 xy
dA
dx
dy
5
y1
x
dt
dt
dt
At s specific time, x 5 20, y 5 50,
dA
5 (2)(50) 1 (23)(20)
dt
5 100 2 50
5 40.
Therefore, the area is increasing at a rate of 40 cm2> s.
6. Given circle with radius r,
dA
5 25 m2> s.
dt
dr
a. Find dt when r 5 3 m.
A 5 pr 2
dA
dr
5 2pr
dt
dt
When r 5 3,
25 5 2p(3)
dr
dt
dr
25
5
dt
6p
5
Therefore, the radius is decreasing at a rate of 6p m> s
when r 5 3 m.
dD
b. Find dt when r 5 3.
dD
dr
52
dt
dt
25
5 2a
b
6p
25
5
3p
Therefore, the diameter is decreasing at a rate of
5
m> s.
3p
7
7. Given circle with radius r, dt 5 6 km2> h
dr
Find dt when A 5 9p km2.
A 5 pr 2
dA
dr
5 2pr
dt
dt
When A 5 9p,
9p 5 pr 2
r2 5 9
r53
r . 0.
When r 5 3,
dA
56
dt
dr
6 5 2p(3)
dt
dr
1
5 .
p
dt
1
Therefore, the radius is increasing at a rate of p km> h.
8. Let x represent the distance from the wall and y
the height of the ladder on the wall.
dA
y
r
9.
x
y
kite
r
Let the variables represent the distances as shown
on the diagram.
x2 1 y2 5 r2
dx
dy
dr
2x
1 2y
5 2r
dt
dt
dt
dx
dy
dr
x
1y
5r
dt
dt
dt
x 5 30, y 5 40
r 2 5 302 1 402
r 5 50
dy
dr
dx
5 ?,
5 10,
50
dt
dt
dt
dr
30(10) 1 40(0) 5 50a b
dt
dr
58
dt
Therefore, she must let out the line at a rate of
8 m> min.
10.
r
y
x
2
x 1 y2 5 r2
dx
dy
dr
2x
1 2y
5 2r
dt
dt
dt
dx
dy
dr
x
1y
5r
dt
dt
dt
When r 5 5, y 5 3,
x 2 5 25 2 9
5 16
x54
x 5 4, y 5 3, r 5 5
dx
1 dr
5 ,
5 0.
dt
3 dt
Substituting,
1
dy
4a b 1 3a b 5 5(0)
3
dt
dy
4
52 .
dt
9
Therefore, the top of the ladder is sliding down at
4 m> s.
8
915 m
Label diagram as shown.
r 2 5 y 2 1 9152
dr
dy
2r 5 2y
dt
dt
dy
dr
r 5y
dt
dt
dy
When y 5 1220, dt 5 268 m> s.
r 5 "12202 1 9152
5 1525
dr
1525a b 5 1220 3 268
dt
dr
5 214 m>s
dt
Therefore, the camera–to–rocket distance is
changing at 214 m> s.
Calculus Appendix
11.
4
12. Given sphere V 5 3pr 3
y
Starting
point
3
dV
5 8 cm3> s.
dt
r
dr
x
dx
5 15 km> h
dt
dy
5 20 km> h
dt
dr
Find
when t 5 2 h.
dt
Solution
Let x represent the distance cyclist 1 is from the starting point, x $ 0. Let y represent the distance cyclist
2 is from the starting point, y $ 0 and let r be the
distance the cyclists are apart. Using the cosine law,
p
r 2 5 x 2 1 y 2 2 2xy cos
3
1
5 x 2 1 y 2 2 2xya b
2
2
2
2
r 5 x 1 y 2 xy
dr
dx
dy
dx
dy
2r 5 2x
1 2y
2 c y 1 xd
dt
dt
dt
dt
dt
At t 5 2 h, x 5 30 km, y 5 40 km
p
and r 2 5 302 1 402 2 2(30)(40) cos 3
1
5 2500 2 2(1200)a b
2
5 1300
r 5 10"13, r . 0.
dr
6 2(10"13)
dt
5 2(30)(15) 1 2(40)(20) 2 315(40) 1 20(30)4
dr
20"13
5 900 1 1600 2 3600 2 6004
dt
5 1300
dr
130
5
dt
2"13
65
5
"13
65"13
5
13
5 5"13
Therefore, the distance between the cyclists is
increasing at a rate of 5"13 km> h after 2 h.
Calculus and Vectors Solutions Manual
a. Find dt when r 5 12 cm.
4
V 5 pr 3
3
dV
dr
5 4pr 2
dt
dt
At a specific time, when r 5 12 cm:
dr
8 5 4p(12)2
dt
dr
8 5 4p(144)
dt
1
dr
5 .
72p
dt
Therefore, the radius is increasing at a rate of
1
cm> s.
72p
dr
b. Find dt when V 5 1435 cm3.
4
V 5 pr 3
3
dV
dr
5 4pr 2 .
dt
dt
At a specific time, when V 5 1435 cm3:
V 5 1435
4 3
p 5 1435
3
r 3 8 342.581 015
8 6.997 148 6
57
dr
8 8 4p(7)2
dt
dr
8 5 196p
dt
2
dr
5
49p
dt
dr
0.01 5 .
dt
2
Therefore, the radius is increasing at 49p cm> s
(or about 0.01 cm> s).
dr
c. Find dt when t 5 33.5 s.
When t 5 33.5, v 5 8 3 33.5 cm3:
4 3
pr 5 268
3
r 3 8 63.980 287 12
9
r 8 3.999 589 273
8 4.
4
V 5 pr 3
3
dV
dr
5 4pr 2
dt
dt
At t 5 33.5 s,
dr
8 8 4p(4)2
dt
dr
8 5 64p
dt
1
dr
5 .
8p
dt
Therefore, the radius is increasing at a rate of
1
8p
cm> s (or 0.04 cm> s).
13. Given cylinder
15 m
Since dt 5 500 000 cm3>min, it takes 500 000 min,
5 30p min to fill
8 94.25 min.
Therefore, it will take 94.25 min, or just over 1.5 h
to fill the cylindrical tank.
14. There are many possible problems.
Samples:
a. The diameter of a right-circular cone is expanding
at a rate of 4 cm>min. Its height remains constant at
10 cm. Find its radius when the volume is increasing
at a rate of 80p cm3>min.
b. Water is being poured into a right-circular tank at
the rate of 12p m3>min. Its height is 4 m and its
radius is 1 m. At what rate is the water level rising?
c. The volume of a right-circular cone is expanding
because its radius is increasing at 12 cm>min and its
height is increasing at 6 cm>min. Find the rate at
which its volume is changing when its radius is
20 cm and its height is 40 cm.
15. Given cylinder
dV
15 000 000p
2m
V 5 pr 2h
dV
5 500 L> min
dt
5 500 000 cm> min
dh
Find .
dt
V 5 pr 2h
Since the diameter is constant at 2 m, the radius is
also constant at 1 m 5 100 cm.
V 5 10 000ph
dV
dh
5 10 000p
dt
dt
dh
500 000 5 10 000p
dt
50
dh
5
p
dt
Therefore, the fluid level is rising at a rate of
50
p
cm> min.
Find t, the time of fill the cylinder.
V 5 pr 2h
V 5 p(100)2 (1 500) cm3
V 5 150 000 00p cm3
10
d51m
h 5 15 m
dr
5 0.003 m>year
dt
dh
5 0.4 m>year
dt
dV
Find dt at the instant D 5 1
V 5 pr 2h
dV
dr
dh
5 a2pr b (h) 1 a b (pr 2 ).
dt
dt
dt
At a specific time, when D 5 1; i.e., r 5 0.5,
dV
5 2p(0.5)(0.003)(15) 1 0.4p(0.5)2
dt
5 0.045p 1 0.1p
5 0.145p
Therefore, the volume of the trunk is increasing at a
rate of 0.145p m3>year.
Calculus Appendix
16. Given cone
h2
5 cm
15 cm
r 5 5 cm
h 5 15 cm
dV
5 2 cm3> min
dt
dh
Find dt when h 5 3 cm,
1
V 5 pr 2h.
3
r
Using similar triangles, h 5 155 5 13
h
r5 .
3
Since the cross section is equilateral, V 5
3 l.
"3
h2
V5
3 5.
"3
dV
10 dh
5
h
dt
"3 dt
At a specific, time when h 5 0.1 5 101 ,
10 1 dh
0.25 5
"3 10 dt
dh
0.25"3 5
dt
"3
dh
5
4
dt
Therefore, the water level is rising at a rate of
"3
4 m> min.
18.
Substituting into V 5 13pr 2h,
1
h2
V 5 pa bh
3
9
1
5 ph 3
27
dV
1
dh
5 ph 2
dt
9
dt
At a specific time, when h 5 3 cm,
1
dh
22 5 p(3)2
9
dt
2
dh
2 5
.
p
dt
Therefore, the water level is being lowered at a
1.8
y
x
dy
5 120 m> min
dt
dx
Find
when t 5 5 s.
dt
Let x represent the length of the shadow. Let y
represent the distance the man is from the base of
the lamppost. Let h represent the height of the
lamppost. At a specific instant, we have
rate of p cm> min when height is 3 cm.
2
17. Given trough
h
1.8
1
25 cm
5m
dV
m3
5 0.25
dt
min
dh
Find dt when h 5 10 cm
5 0.1 m.
Calculus and Vectors Solutions Manual
x+y
1.2
Using similar triangles,
x1y
1.2
5
h
1.8
2.2
2
5
h
3
2h 5 6.6
h 5 3.3
Therefore, the lamppost is 3.3 m high.
11
At any time,
x1y
3.3
5
x
1.8
x1y
11
5
x
6
6x 1 6y 5 11x
6y 5 5x
dy
dx
6
55 .
dt
dt
At a specific time, when t 5 5 seconds
dy
5 120
dt
m> min,
6 3 120 5 5
dx
dt
dx
5 144.
dt
Therefore, the man’s shadow is lengthening at a rate
of 144 m> min after 5 s.
19. This question is similar to finding the rate of
change of the length of the diagonal of a rectangular
prism.
A
F
G
B
H
E
C
D
20
20 m 5
km
1000
1
km
5
50
Find
d(GH)
dt
at t 5 10 s,
1
h.
360
Let BG be the path of the train and CH be the path
of the boat:
Using the Pythagorean theorem,
GH 2 5 HD 2 1 DG 2 and HD 2 5 CD 2 1 CH 2
GH 2 5 CD 2 1 CH 2 1 DG 2
Since BG 5 CD and FE 5 GD 5 501 , it follows that
1
GH 2 5 BG 2 1 CH 2 1
.
2500
d(GH)
d(BG)
d(CH)
2(GH)
5 2(BG)
1 2(CH)
dt
dt
dt
At t 5 10 s,
d(GH)
1
1
GH
5 (60) 1 (20)
dt
6
18
"6331 d(GH)
100
5
450
dt
9
d(GH)
45000
5
dt
9"6331
8 62.8.
1 2
1 2
1 2
And GH 2 5 a b 1 a b 1 a b
6
18
50
1
1
1
5
1
1
36
324
2500
911 664
5
18
29 160 000
113 958
GH 2 5
1 18
364 500
6331
5
202 500
"6331
"13 3 487
GH 5
5
450
450
Therefore, they are separating at a rate of
approximately 62.8 km> h.
20. Given cone
r
5
d(CH)
d(BG)
5 60 km> h and
5 20 km> h.
dt
dt
At t 5
12
1
1
h, BG 5 60 3
360
360
1
5 km
6
1
and CH 5 20 3
360
1
km.
5
18
h
a. r 5 h
dV
5 200 2 20
dt
5 180 cm3> s
dh
Find dt when h 5 15 cm.
1
V 5 pr 2h and r 5 h
3
1
V 5 ph 3.
3
dV
dh
5 ph 2
dt
dt
Calculus Appendix
At a specific time, h 5 15 cm.
dh
180 5 p(15)2
dt
dh
4
5
dt
5p
Therefore, the height of the water in the funnel is
increasing at a rate of 5p cm> s.
y
k
P(x, y)
l–k
4
b. dt 5 200 cm3> s
dh
Find dt when h 5 25 cm.
0
dV
dV
dh
5 ph 2 .
dt
dt
At the time when the funnel is clogged, h 5 25 cm:
dh
200 5 p(25)2 .
dt
dh
8
5
.
dt
25p
Therefore, the height is increasing at 25p cm> s.
21.
y
8
x
Let P(x, y) be a general point on the ladder a
distance k from the top of the ladder. Let A(a, 0) be
the point of contact of the ladder with the ground.
a
x
xl
From similar triangles, l 5 k or a 5 k .
Using the Pythagorean Theorem:
xl
y 2 1 (a 2 x)2 5 (l 2 k)2, and substituting a 5 k ,
2
xl
y 2 1 a 2 xb 5 (l 2 k)2
k
y2 1 x2 a
l2k 2
b 5 (l 2 k)2
k
(l 2 k)2 2
x 1 y 2 5 (l 2 k)2
k2
x2
y2
1
5 1 is the required equation.
k2
(l 2 k)2
Therefore, the equation is the first quadrant portion
of an ellipse.
B(0, 2y)
M(x, y)
0
A(a, 0)
x
A(2x, 0)
Let the midpoint of the ladder be (x, y). From
similar triangles, it can be shown that the top of the
ladder and base of the ladder would have points
B(0, 2y) and A(2x, 0) respectively. Since the ladder
has length l,
(2x)2 1 (2y)2 5 l 2
4x 2 1 4y 2 5 l 2
l2
x2 1 y2 5
4
l 2
5 a b is the required equation.
2
Therefore, the equation of the path followed by the
midpoint of the ladder represents a quarter circle
l
with centre (0, 0) and radius 2, with x, y $ 0.
Calculus and Vectors Solutions Manual
The Natural Logarithm and its
Derivative, p. 575
1. A natural logarithm has base e; a common
logarithm has base 10.
1
1
2. Since e 5 lim (1 1 h) n , let h 5 n. Therefore,
hS0
n
e 5 lim a1 1 1 b ,
1 S0
n
n
But as
1
S 0, n S `.
n
1 n
Therefore, e 5 lim a1 1 b .
n
nS`
1 100
b
If n 5 100, e 8 a1 1
100
5 1.01100
8 2.70481.
Try n 5 100 000, etc.
13
At a specific time, h 5 15 cm.
dh
180 5 p(15)2
dt
dh
4
5
dt
5p
Therefore, the height of the water in the funnel is
increasing at a rate of 5p cm> s.
y
k
P(x, y)
l–k
4
b. dt 5 200 cm3> s
dh
Find dt when h 5 25 cm.
0
dV
dV
dh
5 ph 2 .
dt
dt
At the time when the funnel is clogged, h 5 25 cm:
dh
200 5 p(25)2 .
dt
dh
8
5
.
dt
25p
Therefore, the height is increasing at 25p cm> s.
21.
y
8
x
Let P(x, y) be a general point on the ladder a
distance k from the top of the ladder. Let A(a, 0) be
the point of contact of the ladder with the ground.
a
x
xl
From similar triangles, l 5 k or a 5 k .
Using the Pythagorean Theorem:
xl
y 2 1 (a 2 x)2 5 (l 2 k)2, and substituting a 5 k ,
2
xl
y 2 1 a 2 xb 5 (l 2 k)2
k
y2 1 x2 a
l2k 2
b 5 (l 2 k)2
k
(l 2 k)2 2
x 1 y 2 5 (l 2 k)2
k2
x2
y2
1
5 1 is the required equation.
k2
(l 2 k)2
Therefore, the equation is the first quadrant portion
of an ellipse.
B(0, 2y)
M(x, y)
0
A(a, 0)
x
A(2x, 0)
Let the midpoint of the ladder be (x, y). From
similar triangles, it can be shown that the top of the
ladder and base of the ladder would have points
B(0, 2y) and A(2x, 0) respectively. Since the ladder
has length l,
(2x)2 1 (2y)2 5 l 2
4x 2 1 4y 2 5 l 2
l2
x2 1 y2 5
4
l 2
5 a b is the required equation.
2
Therefore, the equation of the path followed by the
midpoint of the ladder represents a quarter circle
l
with centre (0, 0) and radius 2, with x, y $ 0.
Calculus and Vectors Solutions Manual
The Natural Logarithm and its
Derivative, p. 575
1. A natural logarithm has base e; a common
logarithm has base 10.
1
1
2. Since e 5 lim (1 1 h) n , let h 5 n. Therefore,
hS0
n
e 5 lim a1 1 1 b ,
1 S0
n
n
But as
1
S 0, n S `.
n
1 n
Therefore, e 5 lim a1 1 b .
n
nS`
1 100
b
If n 5 100, e 8 a1 1
100
5 1.01100
8 2.70481.
Try n 5 100 000, etc.
13
dy
d
5
(ln (5x 1 8))
dx
dx
1
d
5
(5x 1 8)
5x 1 8 dx
5
5
5x 1 8
dy
d
5
(ln (x 2 1 1))
b.
dx
dx
1
d 2
5 2
(x 1 1)
x 1 1 dx
2x
5 2
x 11
ds
d
5 (5 ln t 3 )
c.
dt
dt
d
5 (15 ln t)
dt
15
5
t
dy
d
5
(5 ln !x 1 1)
d.
dx
dx
d
1
5
Aln (x 1 1)2 B
dx
d 1
5
a ln (x 1 1)b
dx 2
1
5
2(x 1 1)
d
ds
5 (ln (t 3 2 2t 2 1 5))
e.
dt
dt
1
d 3
5 3
(t 2 2t 2 1 5)
2
t 2 2t 1 5 dx
3t 2 2 4t
5 3
t 2 2t 2 1 5
dw
d
5
(5 ln "z 2 1 3z)
f.
dz
dz
d
1
5
Aln (z 2 1 3z)2 B
dz
d 1
5
a ln (z 2 1 3z)b
dz 2
1
d 2
5
(z 1 3z)
2(z 2 1 3z) dz
2z 1 3
5
2(z 2 1 3z)
d
4. a. f r(x) 5 (x ln x)
dx
1
5 (1)ln x 1 xa b
x
5 ln x 1 1
3. a.
14
b. Since e x and ln x are inverse functions, the
composite function y 5 e ln x is equivalent to y 5 x.
dy
d
So dx 5 dx (x) 5 1.
dv
d
c.
5 (e t ln t)
dt
dt
1
5 (e t ) ln t 1 e t a b
t
t
e
5 e t ln t 1
t
2z
d. g(z) 5 ln (e 1 ze 2z )
1
gr(z) 5 2z
32e 2z 1 (e 2z 2 ze 2z )4
e 1 ze 2z
2ze 2z
5 2z
e 1 ze 2z
ds
d et
5 a
b
e.
dt
dt ln t
ln t(e t ) 2 e t a t b
1
5
(ln t)2
te ln t 2 e t
5
t(ln t)2
t
f. h(u) 5 e !a ln u2
1
5 e !a a ln ub
2
1
1
1 1
hr(u) 5 e !a a
b a ln ub 1 a be !a
2 !u 2
2 u
1
1
1
5 e !a a e 1!a ln u 1 b
u
2
2
2x21
5. a. g(x) 5 e
ln (2x 2 1)
1
gr(x) 5 e 2x21 (2) ln (2x 2 1) 1 a
b (2)e 2x21
2x 2 1
gr(1) 5 e 2 (2) ln (1) 1 1(2)e 1
5 2e
t21
b. f(t) 5 ln a
b
3t 1 5
3t 1 t 3t 1 5 2 3(t 2 1)
f r(t) 5 a
bc
d
t21
(3t 1 5)2
20 20 2 12
c
d
f r(5) 5
4
202
8
5
4 3 20
1
5
10
5 0.1
1
Calculus Appendix
c.
1
? x 2 ln x
1
f r(x) 5 £ x
§
3
x2
The value shown is approximately 2e, which matches
the calculation in part a.
This value matches the calculation in part b.
6. a. f(x) 5 ln (x 2 1 1)
1
f r(x) 5 a
b (2x)
1 1 x2
2x
5
1 1 x2
2
Since 1 1 x . 0 for all x, f r(x) 5 0 when 2x 5 0,
i.e., when x 5 0.
1
b. f(x) 5 (ln x 1 2x)3
1
1
2
f r(x) 5 (ln x 1 2x)23 a 1 2b
x
3
5
1
12
x
2
3(ln x 1 2x)3
f r(x) 5 0 if
1
1250
x
1
1250
x
2
and (ln x 1 2x)3 2 0.
when x 5 2 12.
Since f(x) is defined only for x . 0, there is no
solution to f r(x) 5 0.
c. f(x) 5 (x 2 1 1)21 ln (x 2 1 1)
f r(x) 5 2 (x 2 1 1)22 (2x) ln (x 2 1 1)
2x
1 (x 2 1 1)21 a 2
b
x 11
2x(1 2 ln (x 2 1 1))
5
(x 2 1 1)2
2
2
Since (x 1 1) $ 1 for all x, f r(x) 5 0, when
2x(1 2 ln (x 2 1 1)) 5 0.
Hence, the solution is
ln (x 2 1 1) 5 1
x50
or
x2 1 1 5 e
x 5 6 !e 2 1.
3
ln"
x
x
1
ln x
53
x
At the point (1, 0), the slope of the tangent line is
1 120
f r(1) 5 c
d
3
1
1
5 .
3
The equation of the tangent line is y 5 13 (x 2 1) or
x 2 3y 2 1 5 0.
b.
c. The equation on the calculator is in a different
form, but is equivalent to the equation in part a.
8. The line defined by 3x 2 6y 2 1 5 0 has slope 12.
dy
For y 5 ln x 2 1, the slope at any point is dx 5 12.
1
Therefore, at the point of tangency x 5 12, or x 5 2
and y 5 ln 2 2 1.
The equation of the tangent is
1
y 2 (ln 2 2 1) 5 (x 2 2) or
2
x 2 2y 1 (2 ln 2 2 4) 5 0
9. a. For a horizontal tangent line, the slope equals 0.
We solve:
1
f r(x) 5 2(x ln x)aln x 1 x ? b 5 0
x
or ln x 5 0 or
x50
ln x 5 21
1
No ln in the domain
x 5 e 21 5 .
x51
e
The points on the graph of f(x) at which there are
horizontal tangents are a e , e 2 b and (1, 0).
b.
1 1
7. a. f(x) 5
Calculus and Vectors Solutions Manual
c. The solution in part a. is more precise and
efficient.
15
d
dy
5
(ln (1 1 e 2x ))
dx
dx
1
d
5
(1 1 e 2x )
1 1 e 2x dx
2e 2x
5
1 1 e 2x
At x 5 0, y 5 ln (1 1 e 20 ) 5 ln 2 and
10.
dy
2e 20
5
5 2 12,
dx
1 1 e 20
so the equation of the tangent
line at this point is y 5 2 12x 1 ln 2
11. v(t) 5 90 2 30 ln (3t 1 1)
a. At t 5 0, v(0) 5 90 2 30 ln (1) 5 90 km> h.
230
290
b. a 5 vr(t) 5
?35
3t 1 1
3t 1 1
90
c. At t 5 2, a 5 2 8 212.8 km> h> s.
7
d. The car is at rest when v 5 0.
We solve:
v(t) 5 90 2 30 ln (3t 1 1) 5 0
ln (3t 1 1) 5 3
3t 1 1 5 e 3
e3 2 1
t5
5 6.36 s.
3
ln (x 1 h) 2 ln x
d
12. By definition, ln x 5 lim
dx
hS0
h
1
5 .
x
The derivative of ln x at x 5 2 is
ln (2 1 h) 2 ln 2
1
lim
5 .
hS0
h
2
d
13. a. f r(x) 5 (ln (ln x))
dx
1 d
(ln x)
5
ln x dx
1
5
x ln x
b. The natural logarithm is defined for x . 0, so
f(x) 5 ln (ln x) is defined for ln x . 0, which is
satisfied for x . 1. The function’s domain is
therefore 5xPR 0 x . 16.
1
The derivative f r(x) 5 x ln x is defined where ln x is
defined, which is for x . 0 and where x ln x 2 0,
which is for x 2 0 and x 2 1. The domain of the
derivative is therefore 5xPR0 x . 0 and x 2 16 .
16
The Derivatives of General
Logarithmic Functions, p. 578
1. a. y 5 log5x
dy
1
5
dx
x ln 5
b. y 5 log3x
dy
1
5
dx
x ln 3
c. y 5 2log4x
dy
2
5
dx
x ln 4
d. y 5 23log7x
dy
23
5
dx
x ln 7
e. y 5 2 (log x)
dy
21
5
dx
x ln 10
f. y 5 3log6x
dy
3
5
dx
x ln 6
2. a. y 5 log3 (x 1 2)
dy
f r(x)
5
dx
f(x) ln (a)
1
5
(x 1 2) ln 3
b. y 5 log8 (2x)
dy
f r(x)
5
dx
f(x) ln (a)
2
1
5
5
(2x) ln 8
x ln 8
c. y 5 23 log3 (2x 1 3)
dy
f r(x)
5
dx
f(x) ln (a)
26
5
(2x 1 3) ln 3
d. y 5 log10 (5 2 2x)
dy
f r(x)
5
dx
f(x) ln (a)
22
5
(5 2 2x) ln 10
e. y 5 log 8 (2x 1 6)
dy
f r(x)
5
dx
f(x) ln (a)
2
1
5
5
(2x 1 6) ln 8
(x 1 3) ln 8
Calculus Appendix
d
dy
5
(ln (1 1 e 2x ))
dx
dx
1
d
5
(1 1 e 2x )
1 1 e 2x dx
2e 2x
5
1 1 e 2x
At x 5 0, y 5 ln (1 1 e 20 ) 5 ln 2 and
10.
dy
2e 20
5
5 2 12,
dx
1 1 e 20
so the equation of the tangent
line at this point is y 5 2 12x 1 ln 2
11. v(t) 5 90 2 30 ln (3t 1 1)
a. At t 5 0, v(0) 5 90 2 30 ln (1) 5 90 km> h.
230
290
b. a 5 vr(t) 5
?35
3t 1 1
3t 1 1
90
c. At t 5 2, a 5 2 8 212.8 km> h> s.
7
d. The car is at rest when v 5 0.
We solve:
v(t) 5 90 2 30 ln (3t 1 1) 5 0
ln (3t 1 1) 5 3
3t 1 1 5 e 3
e3 2 1
t5
5 6.36 s.
3
ln (x 1 h) 2 ln x
d
12. By definition, ln x 5 lim
dx
hS0
h
1
5 .
x
The derivative of ln x at x 5 2 is
ln (2 1 h) 2 ln 2
1
lim
5 .
hS0
h
2
d
13. a. f r(x) 5 (ln (ln x))
dx
1 d
(ln x)
5
ln x dx
1
5
x ln x
b. The natural logarithm is defined for x . 0, so
f(x) 5 ln (ln x) is defined for ln x . 0, which is
satisfied for x . 1. The function’s domain is
therefore 5xPR 0 x . 16.
1
The derivative f r(x) 5 x ln x is defined where ln x is
defined, which is for x . 0 and where x ln x 2 0,
which is for x 2 0 and x 2 1. The domain of the
derivative is therefore 5xPR0 x . 0 and x 2 16 .
16
The Derivatives of General
Logarithmic Functions, p. 578
1. a. y 5 log5x
dy
1
5
dx
x ln 5
b. y 5 log3x
dy
1
5
dx
x ln 3
c. y 5 2log4x
dy
2
5
dx
x ln 4
d. y 5 23log7x
dy
23
5
dx
x ln 7
e. y 5 2 (log x)
dy
21
5
dx
x ln 10
f. y 5 3log6x
dy
3
5
dx
x ln 6
2. a. y 5 log3 (x 1 2)
dy
f r(x)
5
dx
f(x) ln (a)
1
5
(x 1 2) ln 3
b. y 5 log8 (2x)
dy
f r(x)
5
dx
f(x) ln (a)
2
1
5
5
(2x) ln 8
x ln 8
c. y 5 23 log3 (2x 1 3)
dy
f r(x)
5
dx
f(x) ln (a)
26
5
(2x 1 3) ln 3
d. y 5 log10 (5 2 2x)
dy
f r(x)
5
dx
f(x) ln (a)
22
5
(5 2 2x) ln 10
e. y 5 log 8 (2x 1 6)
dy
f r(x)
5
dx
f(x) ln (a)
2
1
5
5
(2x 1 6) ln 8
(x 1 3) ln 8
Calculus Appendix
f. y 5 log7 (x 2 1 x 1 1)
dy
f r(x)
5
dx
f(x)ln (a)
2x 1 1
5 2
(x 1 x 1 1)ln 7
t11
3. a. f(t) 5 log2
2t 1 7
f(t) 5 log2 (t 1 1) 2 log2 (2t 1 7)
1
2
f r(t) 5
2
(t 1 1)ln 2
(2t 1 7)ln 2
1
2
f r(3) 5
2
4 ln 2
13 ln 2
13
8
5
2
52 ln 2
52 ln 2
5
5
52 ln 2
b. h(t) 5 log3 (log 2 (t))
f r(t)
hr(t) 5
f(t) ln a
5
1
t ln 2
log2 (t) ln 3
1
5
t log2 (t)(ln 3)(ln 2)
1
hr(8) 5
8 log2 (8)(ln 3)(ln 2)
11x
4. a. y 5 log10
12x
y 5 log10 (1 1 x) 2 log10 (1 2 x)
dy
1
21
5
2
dx
(1 1 x)ln 10
(1 2 x)ln 10
1
1
5
1
(1 1 x)ln 10
(1 2 x)ln 10
12x
11x
5
1
2
(1 2 x )ln 10
(1 2 x 2 )ln 10
2
5
(1 2 x 2 )ln 10
b. y 5 log2"x 1 3x
1
y 5 log2 (x 2 1 3x)
2
dy
2x 1 3
5
2
dx
2(x 1 3x)ln (2)
c. y 5 2 log3 (5x ) 2 log3 (4x )
y 5 2x log3 5 2 x log3 4
dy
5 2 log3 5 2 log3 4
dx
2 ln 5
ln 4
5
2
ln 3
ln 3
2
Calculus and Vectors Solutions Manual
dy
2 ln 5 2 ln 4
5
dx
ln 3
d. y 5 3x log3 x
dy
1
5 ln 3(3x )(log3 x) 1 (3x )
dx
x ln 3
ln
x
1
5 ln 3(3x )
1 (3x )
ln 3
x ln 3
x ln 3(3x )(ln x) 1 3x
5
x ln 3
e. y 5 2x log4 (x)
dy
1
5 2 log4 (x) 1 (2x)
dx
x ln 4
ln x
2
52
1
ln 4
ln 4
2 ln x 1 2
5
ln 4
2 ln x 1 2
5
2 ln 2
ln x 1 1
5
ln 2
log5 (3x 2 )
f. y 5
!x 1 1
!x 1 1 2
2 log5 (3x 2 ) 2 !x 1 1
dy
3x ln 5
5
dx
x11
6x
2 !x 1 1
5
x ln 5
2
log (3x 2 )
5
2 !x 1 1
x11
4x 1 1
5
5
1
2x ln 5 !x 1 1
2
log (3x 2 )x ln 5
5
2 !x 1 1 x ln 5
x11
ln (3x 2 )
4x 1 1 2 ln 5 x ln 5
2x ln 5 !x 1 1
x11
4x 1 1 2 x ln (3x 2 )
3
2x ln 5(x 1 1)2
5. y 5 xlog x
dy
1
5 log x 1 (x)
dx
x ln 10
1
5 log x 1
ln 10
ln x
1
5
1
ln 10
ln 10
ln x 1 1
5
ln 10
5
17
To find the slope of the tangent line, find yr(10):
ln (10) 1 1
5 1.434
ln 10
We also need to find a point on this line to determine
its equation:
y 5 10 log (10)
y 5 10
So the point (10, 10) is on the tangent line.
y 2 10 5 1.434(x 2 10)
y 5 1.434x 2 4.343
20
15
10
5
–4 0
–5
y
x
4 8 12 16 20
6. y 5 loga kx
dy
f r(x)
5
dx
f(x) ln (a)
k
5
kx ln (a)
1
5
x ln (a)
1
5
x ln (a)
7. y 5 102x29 log10 (x 2 2 3x)
dy
5 2 ln 10(102x29 ) log10 (x 2 2 3x)
dx
2x 2 3
1 2
(102x29 )
(x 2 3x) ln 10
At x 5 5,
dy
5 2 ln 10(102(5)29 ) log10 ((10)2 2 3(10))
dx
2(10) 2 3
1
(102(10)29 )
2
((10) 2 3(10)) ln 10
dy
5 49.1
dx
The slope of the tangent line is 49.1 at the point
(5, 10), so the equation of the line is
y 2 10 5 49.1(x 2 5)
y 5 49.1x 2 235.5
8. s(t) 5 tlog 6 (t 1 1)
To find whether the function is increasing or
decreasing at t 5 15, we need to find the derivative
and evaluate it at t 5 15.
18
sr(t) 5 log6 (t 1 1) 1 (t)
1
(t 1 1) ln 6
t
(t 1 1) ln 6
15
sr(15) 5 log6 (16) 1
16 ln 6
8 2.14
Since the derivative is positive at t 5 15, the
distance is increasing at that point.
9. a. y 5 log3 x
dy
1
5
dx
x ln 3
To find the slope of the tangent line, let x 5 9
1
m5
9 ln 3
5 0.1
The equation of the tangent line is y 5 0.1x 1 1.1.
b.
y
8
6
4
2
x
0
–4 –2
2 4 6 8
–2
–4
5 log6 (t) 1
Vertical asymptote at x 5 0
c. The tangent line will intersect this asymptote
because it is defined for x 5 0.
10. f(x) 5 ln (x 2 2 4)
ln (x) is defined for x . 0, so ln (x 2 2 4) is defined
for x 2 2 4 . 0. Therefore, the domain is
5xPR0 x , 2 or x . 26
2x
f r(x) 5 2
x 24
Let f r(x) 5 0:
2x
50
2
x 24
Therefore, 2x 5 0
x50
There is a critical number at x 5 0
Since the derivative is undefined for x 5 6 2, there
are also critical numbers at x 5 2 and x 5 22.
However, since the function is undefined for
22 , x , 2, we do not need to include x 5 0 as a
critical number.
Calculus Appendix
x
x , 22
22
dy
dx
2
DNE
Graph
Dec
2
x.2
x
x,0
0
x.0
DNE
1
d 2y
dx 2
1
DNE
1
Inc
Graph
C. Up
22 , x , 2
Undefined
Therefore the function is decreasing for x , 22
and increasing for x . 2.
11. a. y 5 x ln x
To determine whether this function has points of
inflection, we need to look at the second derivative.
dy
1
5 ln x 1 (x)
x
dx
5 ln x 1 1
d 2y
1
2 5
x
dx
d 2y
Let 2 5 0:
dx
1
50
x
1
This has no solution, but x is undefined for x 5 0,
x 5 0 is a possible point of inflection.
x
x,0
0
x.0
DNE
1
2
d y
dx 2
Graph
C. Down
Point of
Inflection
C. Up
The graph switches from concave down to concave
up at x 5 0, so there is a point of inflection at
x 5 0.
b. y 5 3 2 2 log x
To determine whether this function has points of
inflection, we need to look at the second derivative.
dy
22
5
dx
x ln 10
d 2y
2
5 2
dx 2
x ln 10
d 2y
Let 2 5 0:
dx
2
50
x 2 ln 10
2
This has no solution, but x 2 ln 10 is undefined for
x 5 0, so x 5 0 is a possible point of inflection.
Calculus and Vectors Solutions Manual
C. Up
Since the graph is always concave up, there is no
point of inflection.
12. y 5 3x
dy
5 3x ln 3
dx
m 5 3(0) ln 3
5 ln 3
So the slope of y 5 3x at (0, 1) is ln3.
y 5 log 3 x
dy
1
5
dx
x ln 3
1
m5
(1)ln 3
1
5
ln 3
1
So the slope of y 5 log 3 x at (1, 0) is ln 3.
Since ln 3 . 1, the slope of y 5 3x at (0, 1) is
greater than the slope of y 5 log3 x at (1, 0)
8
6
4
2
–4 –2 0
–2
–4
y
x
2 4 6 8
Logarithmic Differentiation, p. 582
1. a. y 5 x "10 2 3
dy
5 "10x "1021
dx
b. f(x) 5 5x 3"2
f r(x) 5 15"2x 3"221
c. s 5 t p
ds
5 pt p21
dt
d. f(x) 5 x e 1 e x
f r(x) 5 ex e21 1 e x
19
x
x , 22
22
dy
dx
2
DNE
Graph
Dec
2
x.2
x
x,0
0
x.0
DNE
1
d 2y
dx 2
1
DNE
1
Inc
Graph
C. Up
22 , x , 2
Undefined
Therefore the function is decreasing for x , 22
and increasing for x . 2.
11. a. y 5 x ln x
To determine whether this function has points of
inflection, we need to look at the second derivative.
dy
1
5 ln x 1 (x)
x
dx
5 ln x 1 1
d 2y
1
2 5
x
dx
d 2y
Let 2 5 0:
dx
1
50
x
1
This has no solution, but x is undefined for x 5 0,
x 5 0 is a possible point of inflection.
x
x,0
0
x.0
DNE
1
2
d y
dx 2
Graph
C. Down
Point of
Inflection
C. Up
The graph switches from concave down to concave
up at x 5 0, so there is a point of inflection at
x 5 0.
b. y 5 3 2 2 log x
To determine whether this function has points of
inflection, we need to look at the second derivative.
dy
22
5
dx
x ln 10
d 2y
2
5 2
dx 2
x ln 10
d 2y
Let 2 5 0:
dx
2
50
x 2 ln 10
2
This has no solution, but x 2 ln 10 is undefined for
x 5 0, so x 5 0 is a possible point of inflection.
Calculus and Vectors Solutions Manual
C. Up
Since the graph is always concave up, there is no
point of inflection.
12. y 5 3x
dy
5 3x ln 3
dx
m 5 3(0) ln 3
5 ln 3
So the slope of y 5 3x at (0, 1) is ln3.
y 5 log 3 x
dy
1
5
dx
x ln 3
1
m5
(1)ln 3
1
5
ln 3
1
So the slope of y 5 log 3 x at (1, 0) is ln 3.
Since ln 3 . 1, the slope of y 5 3x at (0, 1) is
greater than the slope of y 5 log3 x at (1, 0)
8
6
4
2
–4 –2 0
–2
–4
y
x
2 4 6 8
Logarithmic Differentiation, p. 582
1. a. y 5 x "10 2 3
dy
5 "10x "1021
dx
b. f(x) 5 5x 3"2
f r(x) 5 15"2x 3"221
c. s 5 t p
ds
5 pt p21
dt
d. f(x) 5 x e 1 e x
f r(x) 5 ex e21 1 e x
19
2. a. y 5 x ln x
ln y 5 ln (x ln x )
ln y 5 ln x ln x
ln y 5 (ln x)2
1
1 dy
5 2 ln xa b
y dx
x
2y ln x
dy
5
x
dx
2x ln x ln x
5
x
(x 1 1)(x 2 3)2
b. y 5
(x 1 2)3
(x 1 1)(x 2 3)2
ln y 5 ln
(x 1 2)3
ln y 5 ln (x 1 1) 1 ln (x 2 3)2 2 ln (x 1 2)3
ln y 5 ln (x 1 1) 1 2 ln (x 2 3) 2 3 ln (x 1 2)
1 dy
1
2
3
5
1
2
y dx
x11
x23
x12
dy
1
2
3
5 ya
1
2
b
dx
x11
x23
x12
(x 1 1)(x 2 3)2
1
2
3
5
a
1
2
b
3
(x 1 2)
x11 x23 x12
c. y 5 x !x
!x ln x
1
1
ln x 1 !x
x
2 !x
ln x
1
1
2 !x
!x
ln x 1 2
2 !x
ln x 1 2
(y)
2 !x
ln x 1 2
(x "x )
2 !x
1 t
d. s 5 a b
t
1
ln s 5 t ln
t
1 ds
1
1 21
5 ln 1 ta 1 b a 2 b
s dt
t
t
t
ln y 5
1 dy
5
y dx
1 dy
5
y dx
1 dy
5
y dx
dy
5
dx
dy
5
dx
1 ds
1
21
5 ln 1 t(t)a 2 b
s dt
t
t
ds
1
5 saln 2 1b
dt
t
ds
1 t
1
5 a b aln 2 1b
dt
t
t
3. a. y 5 f(x) 5 x 2x
ln y 5 x ln x
1 dy
1
5 ln x 1 xa b
?
y dx
x
dy
5 x 2x (ln x 1 1)
dx
fr(e) 5 e e (ln e 1 1) 5 2e e
b. s 5 e t 1 t e
ds
5 e t 1 et e21
dt
ds
5 e 2 1 e ? 2e21
When t 5 2,
dt
3
(x 2 3)2 !
x11
c. y 5
(x 2 4)5
Let y 5 f(x)
1
ln y 5 2 ln (x 2 3) 1 ln (x 1 1) 2 5 ln (x 2 4)
3
1 dy
2
1
5
5
1
2
?
y dx
x23
3(x 1 1)
x24
dy
2
1
5
5 ya
1
2
b
dx
x23
3(x 1 1)
x24
2
1
5
fr(7) 5 f(7)a 1
2 b
4
24
3
32
27
4
5
a2 b 5 2
243
24
27
4. y 5 x(x 2 )
The point of contact is (2, 16). The slope of the
dy
tangent line at any point on the curve is given by dx.
We take the natural logarithm of both sides and
differentiate implicitly with respect to x.
2
y 5 x (x )
ln y 5 x 2 ln x
1 dy
5 2x ln x 1 x
?
y dx
dy
At the point (2, 16), dx 5 16(4 ln 2 1 2).
The equation of the tangent line at (2, 16) is
y 2 16 5 32(2 ln 2 1 1)(x 2 2) or
y 5 32(2 ln 2 1 1)x 2 128 ln 2 2 48.
1 ds
1
5 ln 2 1
s dt
t
20
Calculus Appendix
1
(x 1 1)(x 1 2)(x 1 3)
We take the natural logarithm of both sides and
dy
differentiate implicitly with respect to x to find dx,
the slope of the tangent line.
ln y 5 ln (1) 2 ln (x 1 1) 2 ln (x 1 2)
2 ln (x 1 3)
1 dy
1
1
1
52
2
2
?
y dx
x11
x12
x13
The point of contact is (0, 16 ).
5. y 5
1
At a0, b,
6
dy
1
1
1
1
11
11
5 a21 2 2 b 5 a2 b 5 2 .
dx
6
2
3
6
6
36
1
6. y 5 xx, x . 0
We take the natural logarithm of both sides and
dy
differentiate implicitly with respect to x to find dx.
the slope of the tangent.
1
y 5 xx
1
ln x
ln y 5 ln x 5
x
x
a x b (x) 2 (ln x)(1)
1 dy
5
?
y dx
x2
1
dy
xx (1 2 ln x)
5
dx
dy
We want the values of x so that dx 5 0.
1
1
xx (1 2 ln x)
50
Since x 2 0 and x 2 . 0, we have 1 2 ln x 5 0
ln x 5 1
x 5 e.
1
The slope of the tangent is 0 at (e, ee ).
7. We want to determine the points on the given
curve at which the tangent lines have slope 6. The
slope of the tangent at any point on the curve is
given by
4
dy
5 2x 1 .
x
dx
To find the required points, we solve:
4
2x 1 5 6
x
x 2 2 3x 1 2 5 0
(x 2 1)(x 2 2) 5 0
x 5 1 or x 5 2.
The tangents to the given curve at (1, 1) and
(2, 4 1 4 ln 2) have slope 6.
1
x
Calculus and Vectors Solutions Manual
8. We first must find the equation of the tangent at
A(4, 16). We need dx for y 5 x !x.
dy
ln y 5 !x ln x
1 dy
1 1
1
5 x 2x ln x 1 !x ?
?
y dx
x
2
ln x 1 2
5
2 !x
dy
(ln 4 1 2)
At (4, 16),
5 16
5 4 ln 4 1 8.
dx
4
The equation of the tangent is
y 2 16 5 (4 ln 4 1 8)(x 2 4).
The y-intercept is 216(ln 4 1 1).
24
4 ln 4 1 4
The x-intercept is
145
.
ln 4 1 2
ln 4 1 2
1 4 ln 4 1 4
The area of ^OBC is a
b (16)
2 ln 4 1 2
32(ln 4 1 1)2
(ln 4 1 1),which equals
.
ln 4 1 2
e x"x 2 1 1
(x 2 1 2)3
To find the slope of the tangent line at A0, 18 B, we
need to evaluate the derivative at x 5 0.
9. y 5
ln y 5 ln
e x"x 2 1 1
(x 2 1 2)3
ln y 5 ln e x 1 ln "x 2 1 1 2 ln (x 2 1 2)3
1
ln y 5 x ln e 1 ln (x 2 1 1) 2 3 ln (x 2 1 2)
2
1 dy
2x
3(2x)
511
2 2
y dx
2(x 2 1 1)
x 12
dy
1
6x
x
5 a1 1 2
2 2
b
y
dx
x 11
x 12
e x"x 2 1 1
6x
x
dy
5
2 2
b
2
3 a1 1 2
dx
(x 1 2)
x 11
x 12
e (0)"(0)2 1 1
(0)
6(0)
2
b
2
3 a1 1
2
( (0) 1 2)
(0) 1 1
(0)2 1 2
1
m 5 (1)
8
1
m5
8
x sin x 2
10. f(x) 5 a 2
b
x 21
x sin x 2
ln (f(x)) 5 ln a 2
b
x 21
x sin x
ln (f(x)) 5 2 ln 2
x 21
m5
21
ln (f(x)) 5 2 ln (x sin x) 2 2 ln (x 2 2 1)
1
2(sin x 1 x cos x)
2(2x)
f r(x) 5
2 2
f(x)
x sin x
x 21
2(sin x 1 x cos x)
4x
f r(x) 5 f(x)a
2 2
b
x sin x
x 21
5a
x sin x 2 2(sin x 1 x cos x)
4x
b a
2 2
b
x2 2 1
x sin x
x 21
11. y 5 x cos x
ln y 5 ln x cos x
ln y 5 cos x ln x
cos x
1 dy
5 sin x ln x 1
y dx
x
cos x
dy
5 yasin x ln x 1
b
x
dx
cos x
5 x cos x asin x ln x 1
b
x
12. y 5 x x
First use the derivative to find the slope of the
tangent line at the point (1, 1)
ln y 5 ln x x
ln y 5 x ln x
1 dy
1
5 ln x 1 (x)
y dx
x
1 dy
5 ln x 1 1
y dx
dy
5 y(ln x 1 1)
dx
5 x x (ln x 1 1)
m 5 (1)(1) (ln (1) 1 1)
5 1(0 1 1) 5 1
Since we already know a point on the curve, we can
use point slope form to write the equation of the
tangent line.
y 2 1 5 1(x 2 1)
y5x2111
y5x
1
13. s(t) 5 t t , t . 0
1
a. ln (s(t)) 5 t ln t
Differentiate with respect to t:
1
? sr(t) 5
s(t)
1
? t 2 ln t
t
t2
1 2 ln t
5
.
t2
1 2 ln t
1 2 ln t
1
Thus, v(t) 5 s(t) ? a
b 5 tt a
b.
2
t
t2
22
Now, a(t) 5 vr(t) 5 sr(t)a
1 s(t) a
2
1 2 ln t
b 1 s(t)
t2
1
? t 2 2 (1 2 ln t)(2t)
l
t4
Substituting for s(t) and sr(t) 5 v(t) gives
a(t) 5 t t a
1
b
1 2 ln t 2
2t ln t 2 3t
1
b 1 tt a
b
2
t
t4
1
5
tt
31 2 2 ln t 1 (ln t)2 1 2t ln t 2 3t4
t4
1
b. Since t t and t 2 are always positive, the velocity is
zero when
1 2 ln t 5 0 or when t 5 e.
1
ee
a(e) 5 4 31 2 2 1 1 1 2e 2 3e4
e
1
ee
52 3
e
1
5 2ee 23
14. If a b . b a, then
ln a b . ln b a
b ln a . a ln b Dividing both sides by ab yields
ln a
ln b
.
a
b
ln x
This indicates we should use the function y 5 x
to determine whether e p or pe is larger.
ln x
y5
x
dy
1 2 ln x
5
dx
x2
dy
Let
5 0:
dx
1 2 ln x
50
x2
1 2 ln x 5 0
x5e
Therefore e is a critical number.
x
x,e
e
x.e
dy
dx
1
0
Graph
Inc.
Local Max
Dec.
The graph attains a local maximum when x 5 e.
ln e
ln p
Therefore, e . p , and e p . pe.
Using a calculator, ep 8 23.14 and pe 8 22.46. This
verifies the work above.
Calculus Appendix
Cumulative Review of Calculus,
pp. 267–270
1. a. f(x) 5 3x2 1 4x 2 5
f(a 1 h) 2 f(a)
m 5 lim
hS0
h
f(2 1 h) 2 15
5 lim
hS0
h
3(2 1 h)2 1 4(2 1 h) 2 5 2 15
5 lim
hS0
h
12 1 12h 1 3h 2 1 8 1 4h 2 20
5 lim
hS0
h
3h 2 1 16h
5 lim
hS0
h
5 lim 3h 1 16
hS0
5 16
2
b. f(x) 5
x21
f(a 1 h) 2 f(a)
m 5 lim
hS0
h
f(2 1 h) 2 2
5 lim
hS0
h
5 lim
hS0
5 lim
2
22
21h21
h
2
2(1 1 h)
2 11h
11h
h
2 2 2(1 1 h)
5 lim
hS0
h(1 1 h)
22h
5 lim
hS0 h(1 1 h)
22
5 lim
hS0 1 1 h
5 22
c. f(x) 5 !x 1 3
f(a 1 h) 2 f(a)
m 5 lim
hS0
h
f(6 1 h) 2 3
5 lim
hS0
h
!h 1 9 2 3
5 lim
hS0
h
( !h 1 9 2 3)( !h 1 9 1 3)
5 lim
hS0
h( !h 1 9 1 3)
hS0
Calculus and Vectors Solutions Manual
h1929
hS0 h( !h 1 9 1 3)
h
5 lim
hS0 h( !h 1 9 1 3)
1
5 lim
hS0 ( !h 1 9 1 3)
1
5
6
d. f(x) 5 25x
f(a 1 h) 2 f(a)
m 5 lim
hS0
h
25(11h) 2 32
5 lim
hS0
h
25 ? 25h 2 32
5 lim
hS0
h
32(25h 2 1)
5 lim
hS0
h
5(25h 2 1)
5 32 lim
hS0
5h
5h
(2 2 1)
5 160 lim
hS0
5h
5 160 ln 2
change in distance
2. a. average velocity 5
change in time
s(t2 ) 2 s(t1 )
5
t2 2 t1
32(4)2 1 3(4) 1 14 2 3 (2(1)2 1 3(1) 1 1)4
5
421
45 2 6
5
3
5 13 m> s
b. instantaneous velocity 5 slope of the tangent
s(a 1 h) 2 s(a)
m 5 lim
hS0
h
s(3 1 h) 2 s(3)
5 lim
hS0
h
2(3 1 h)2 1 3(3 1 h) 1 1
5 lim c
hS0
h
(2(3)2 1 3(3) 1 1)
d
2
h
18 1 12h 1 2h 2 1 9 1 3h 1 1 2 28
5 lim
hS0
h
5 lim
5-1
5 lim 353.655 1 4.9h4
15h 1 2h 2
hS0
h
5 lim (15 1 2h)
5 lim
hS0
hS0
5 15 m> s
f(a 1 h) 2 f(a)
h
(4 1 h)3 2 64
f(4 1 h) 2 f(4)
lim
5 lim
hS0
h
hS0
h
(4 1 h)3 2 64 5 f(4 1 h) 2 f(4)
Therefore, f(x) 5 x 3.
4. a. Average rate of change in distance with respect
to time is average velocity, so
s(t2 ) 2 s(t1 )
average velocity 5
t2 2 t1
s(3) 2 s(1)
5
321
4.9(3)2 2 4.9(1)
5
321
5 19.6 m> s
b. Instantaneous rate of change in distance with
respect to time 5 slope of the tangent.
f(a 1 h) 2 f(a)
m 5 lim
hS0
h
f(2 1 h) 2 f(2)
5 lim
hS0
h
4.9(2 1 h)2 2 4.9(2)2
5 lim
hS0
h
19.6 1 19.6h 1 4.9h 2 2 19.6
5 lim
hS0
h
2
19.6h 1 4.9h
5 lim
hS0
h
5 lim 19.6 1 4.9h
m 5 lim
3.
hS0
hS0
5 19.6 m> s
c. First, we need to determine t for the given
distance:
146.9 5 4.9t 2
29.98 5 t 2
5.475 5 t
Now use the slope of the tangent to determine the
instantaneous velocity for t 5 5.475:
f(5.475 1 h) 2 f(5.475)
m 5 lim
hS0
h
4.9(5.475 1 h)2 2 4.9(5.475)2
5 lim
hS0
h
146.9 1 53.655h 1 4.9h 2 2 146.9
5 lim
hS0
h
53.655h 1 4.9h 2
5 lim
hS0
h
5-2
5 53.655 m> s
5. a. Average rate of population change
p(t2 ) 2 p(t1 )
5
t2 2 t1
2(8)2 1 3(8) 1 1 2 (2(0) 1 3(0) 1 1)
5
820
128 1 24 1 1 2 1
5
820
5 19 thousand fish> year
b. Instantaneous rate of population change
p(t 1 h) 2 p(t)
5 lim
hS0
h
p(5 1 h) 2 p(5)
5 lim
hS0
h
2(5 1 h)2 1 3(5 1 h) 1 1
5 lim c
hS0
h
(2(5)2 1 3(5) 1 1)
2
d
h
50 1 20h 1 2h 2 1 15 1 3h 1 1 2 66
5 lim
hS0
h
2h 2 1 23h
5 lim
hS0
h
5 lim 2h 1 23
5 23 thousand fish> year
6. a. i. f(2) 5 3
ii. lim2 f(x) 5 1
hS0
xS2
iii. lim1 f(x) 5 3
xS2
iv. lim f(x) 5 2
xS6
b. No, lim f(x) does not exist. In order for the limit
xS4
to exist, lim2 f(x) and lim1 f(x) must exist and they
xS4
xS4
must be equal. In this case, lim2 f(x) 5 `, but
xS4
lim1 f(x) 5 2 `, so lim f(x) does not exist.
xS4
xS4
7. f(x) is discontinuous at x 5 2. lim2 f(x) 5 5, but
xS2
lim1 f(x) 5 3.
xS2
2x 2 1 1
2(0)2 1 1
5
xS0 x 2 5
025
1
52
5
x23
b. lim
xS3 !x 1 6 2 3
(x 2 3)( !x 1 6 1 3)
5 lim
xS3 ( !x 1 6 2 3)( !x 1 6 1 3)
8. a. lim
Cumulative Review of Calculus
(x 2 3)( !x 1 6 1 3)
xS3
x1629
(x 2 3)( !x 1 6 1 3)
5 lim
xS3
x23
5 lim !x 1 6 1 3
5 lim
xS3
56
c. lim
xS23
5 lim
1
1
13
x
x13
x13
3x
x13
x13
5 lim
xS23 3x(x 1 3)
1
5 lim
xS23 3x
1
52
9
x2 2 4
d. lim 2
xS2 x 2 x 2 2
(x 1 2)(x 2 2)
5 lim
xS2 (x 1 1)(x 2 2)
x12
5 lim
xS2 x 1 1
4
5
3
x22
e. lim 3
xS2 x 2 8
x22
5 lim
2
xS2 (x 2 2)(x 1 2x 1 4)
1
5 lim 2
xS2 x 1 2x 1 4
1
5
12
!x 1 4 2 !4 2 x
f. lim
x
xS0
( !x 1 4 2 !4 2 x)( !x 1 4 1 !4 2 x)
5 lim
xS0
x( !x 1 4 1 !4 2 x)
x 1 4 2 (4 2 x)
5 lim
xS0 x( !x 1 4 1 !4 2 x)
2x
5 lim
xS0 x( !x 1 4 1 !4 2 x)
2
5 lim
xS0 ( !x 1 4 1 !4 2 x)
1
5
2
xS23
Calculus and Vectors Solutions Manual
9. a. f(x) 5 3x2 1 x 1 1
f(x 1 h) 2 f(x)
f r(x) 5 lim
hS0
h
3(x 1 h)2 1 (x 1 h) 1 1
5 lim c
hS0
h
2
(3x 1 x 1 1)
2
d
h
3x 2 1 6hx 1 6h 2 1 x 1 h
5 lim c
hS0
h
1 2 3x 2 2 x 2 1
1
d
h
6hx 1 6h 2 1 h
5 lim
hS0
h
5 lim 6x 1 6h 1 1
hS0
5 6x 1 1
1
b. f(x) 5
x
f(x 1 h) 2 f(x)
f r(x) 5 lim
hS0
h
1
1
2x
x
1
h
5 lim
h
x 2 (x 1 h)
5 lim
hS0 h(x)(x 1 h)
2h
5 lim
hS0 h(x)(x 1 h)
21
5 lim
hS0 x(x 1 h)
1
52 2
x
10. a. To determine the derivative, use the power rule:
y 5 x 3 2 4x 2 1 5x 1 2
dy
5 3x 2 2 8x 1 5
dx
b. To determine the derivative, use the chain rule:
y 5 "2x 3 1 1
dy
1
5
(6x 2 )
3
dx
2"2x 1 1
3x 2
5
"2x 3 1 1
c. To determine the derivative, use the quotient rule:
2x
y5
x13
dy
2(x 1 3) 2 2x
5
dx
(x 1 3)2
6
5
(x 1 3)2
hS0
5-3
d. To determine the derivative, use the product rule:
y 5 (x 2 1 3)2 (4x 5 1 5x 1 1)
dy
5 2(x 2 1 3)(2x)(4x 5 1 5x 1 1)
dx
1 (x 2 1 3)2 (20x 4 1 5)
5 4x(x 2 1 3)(4x 5 1 5x 1 1)
1 (x 2 1 3)2 (20x 4 1 5)
e. To determine the derivative, use the quotient rule:
(4x 2 1 1)5
y5
(3x 2 2)3
dy
5(4x 2 1 1)4 (8x)(3x 2 2)3
5
dx
(3x 2 2)6
3(3x 2 2)2 (3)(4x 2 1 1)5
2
(3x 2 2)6
5 (4x 2 1 1)4 (3x 2 2)2
40x(3x 2 2) 2 9(4x 2 1 1)
3
(3x 2 2)6
(4x 2 1 1)4 (120x 2 2 80x 2 36x 2 2 9)
5
(3x 2 2)4
(4x 2 1 1)4 (84x 2 2 80x 2 9)
5
(3x 2 2)4
2
f. y 5 3x 1 (2x 1 1)34 5
Use the chain rule
dy
5 53x 2 1 (2x 1 1)34 4 32x 1 6(2x 1 1)24
dx
11. To determine the equation of the tangent line,
we need to determine its slope at the point (1, 2).
To do this, determine the derivative of y and
evaluate for x 5 1:
18
y5
(x 1 2)2
5 18(x 1 2)22
dy
5 236(x 1 2)23
dx
236
5
(x 1 2)3
236
m5
(x 1 2)3
24
236
5
5
27
3
Since we have a given point and we know the slope,
use point-slope form to write the equation of the
tangent line:
24
y225
(x 2 1)
3
3y 2 6 5 24x 1 4
4x 1 3y 2 10 5 0
5-4
12. The intersection point of the two curves
occurs when
x 2 1 9x 1 9 5 3x
x 2 1 6x 1 9 5 0
(x 1 3)2 5 0
x 5 23.
At a point x, the slope of the line tangent to the
curve y 5 x 2 1 9x 1 9 is given by
dy
d 2
5
(x 1 9x 1 9)
dx
dx
5 2x 1 9.
At x 5 23, this slope is 2(23) 1 9 5 3.
d
13. a. pr(t) 5 (2t2 1 6t 1 1100)
dt
5 4t 1 6
b. 1990 is 10 years after 1980, so the rate of change
of population in 1990 corresponds to the value
pr(10) 5 4(10) 1 6
5 46 people per year.
c. The rate of change of the population will be 110
people per year when
4t 1 6 5 110
t 5 26.
This corresponds to 26 years after 1980, which is
the year 2006.
d
14. a. f r(x) 5 (x 5 2 5x 3 1 x 1 12)
dx
5 5x 4 2 15x 2 1 1
d
f s (x) 5
(5x 4 2 15x 2 1 1)
dx
5 20x 3 2 30x
b. f(x) can be rewritten as f(x) 5 22x 22
d
f r(x) 5
(22x 22 )
dx
5 4x 23
4
x3
d
f s (x) 5
(4x 23 )
dx
5 212x 24
12
52 4
x
1
c. f(x) can be rewritten as f(x) 5 4x 22
d
1
f r(x) 5
(4x 2 2 )
dx
3
5 22x22
2
52
"x 3
5
Cumulative Review of Calculus
d
3
(22x 22 )
dx
5
5 3x 22
3
5
"x 5
d. f(x) can be rewritten as f(x) 5 x 4 2 x 24
d 4
f r(x) 5
(x 2 x 24 )
dx
5 4x 3 1 4x 25
4
5 4x 3 1 5
x
d
f s (x) 5
(4x 3 1 4x 25 )
dx
5 12x 2 2 20x 26
20
5 12x 2 2 6
x
15. Extreme values of a function on an interval will
only occur at the endpoints of the interval or at a
critical point of the function.
d
a. f r(x) 5 (1 1 (x 1 3)2 )
dx
5 2(x 1 3)
The only place where f r(x) 5 0 is at x 5 23, but
that point is outside of the interval in question. The
extreme values therefore occur at the endpoints of
the interval:
f(22) 5 1 1 (22 1 3)2 5 2
f(6) 5 1 1 (6 1 3)2 5 82
The maximum value is 82, and the minimum
value is 6
1
b. f(x) can be rewritten as f(x) 5 x 1 x 22
d
1
f r(x) 5
(x 1 x 22 )
dx
1 3
5 1 1 2 x 22
2
1
512
2"x 3
On this interval, x $ 1, so the fraction on the right
is always less than or equal to 12. This means that
f r(x) . 0 on this interval and so the extreme values
occur at the endpoints.
1
52
f(1) 5 1 1
!1
1
1
f(9) 5 9 1
59
!9
3
The maximum value is 9 13, and the minimum
value is 2.
f s (x) 5
Calculus and Vectors Solutions Manual
c. f r(x) 5
d
ex
a
b
dx 1 1 e x
(1 1 e x )(e x ) 2 (e x )(e x )
(1 1 e x )2
ex
5
(1 1 e x )2
Since e x is never equal to zero, f r(x) is never zero,
and so the extreme values occur at the endpoints of
the interval.
e0
1
f(0) 5
0 5
11e
2
e4
f(4) 5
1 1 e4
e4
The maximum value is 1 1 e 4, and the minimum
value is 12.
d
d. f r(x) 5 (2 sin (4x) 1 3)
dx
5 8 cos (4x)
p
Cosine is 0 when its argument is a multiple of
5
2
3p
or 2 .
3p
p
1 2kp or 4x 5
1 2kp
2
2
3p
p
p
p
x5
x5 1 k
1 k
8
2
8
2
4x 5
p 3p 5p 7p
Since xP30, p4, x 5 8 , 8 , 8 , 8 .
Also test the function at the endpoints of the interval.
f(0) 5 2 sin 0 1 3 5 3
p
p
f a b 5 2 sin 1 3 5 5
8
2
3p
3p
1351
f a b 5 2 sin
8
2
5p
5p
1355
f a b 5 2 sin
8
2
7p
7p
f a b 5 2 sin
1351
8
2
f(p) 5 2 sin (4p) 1 3 5 3
The maximum value is 5, and the minimum
value is 1.
16. a. The velocity of the particle is given by
v(t) 5 sr(t)
d
5 (3t 3 2 40.5t 2 1 162t)
dt
5 9t 2 2 81t 1 162.
5-5
The acceleration is
a(t) 5 vr(t)
d
5 (9t 2 2 81t 1 162)
dt
5 18t 2 81
b. The object is stationary when v(t) 5 0:
9t 2 2 81t 1 162 5 0
9(t 2 6)(t 2 3) 5 0
t 5 6 or t 5 3
The object is advancing when v(t) . 0 and retreating
when v(t) , 0. Since v(t) is the product of two
linear factors, its sign can be determined using the
signs of the factors:
t-values
t23
t26
v(t)
Object
0,t,3
,0
,0
.0
Advancing
3,t,6
.0
,0
,0
Retreating
6,t,8
.0
.0
.0
Advancing
c. The velocity of the object is unchanging when the
acceleration is 0; that is, when
a(t) 5 18t 2 81 5 0
t 5 4.5
d. The object is decelerating when a(t) , 0, which
occurs when
18t 2 81 , 0
0 # t , 4.5
e. The object is accelerating when a(t) . 0, which
occurs when
18t 2 81 . 0
4.5 , t # 8
17.
w
l
Let the length and width of the field be l and w, as
shown. The total amount of fencing used is then
2l 1 5w. Since there is 750 m of fencing available,
this gives
2l 1 5w 5 750
5
l 5 375 2 w
2
The total area of the pens is
A 5 lw
5
5 375w 2 w 2
2
The maximum value of this area can be found by
expressing A as a function of w and examining its
derivative to determine critical points.
5-6
A(w) 5 375w 2 52w 2, which is defined for 0 # w
and 0 # l. Since l 5 375 2 52w, 0 # l gives the
restriction w # 150. The maximum area is therefore
the maximum value of the function A(w) on the
interval 0 # w # 150.
d
5
Ar(w) 5
a375w 2 w 2 b
dw
2
5 375 2 5w
Setting Ar(w) 5 0 shows that w 5 75 is the only
critical point of the function. The only values of
interest are therefore:
5
A(0) 5 375(0) 2 (0)2 5 0
2
5
A(75) 5 375(75) 2 (75)2 5 14 062.5
2
5
A(150) 5 375(150) 2 (150)2 5 0
2
The maximum area is 14 062.5 m2
18.
r
h
Let the height and radius of the can be h and r, as
shown. The total volume of the can is then pr 2h.
The volume of the can is also give at 500 mL, so
pr 2h 5 500
500
h5
pr 2
The total surface area of the can is
A 5 2prh 1 2pr 2
1000
5
1 2pr 2
r
The minimum value of this surface area can be
found by expressing A as a function of r and
examining its derivative to determine critical points.
1000
A(r) 5
1 2pr 2, which is defined for 0 , r and
r
500
0 , h. Since h 5 pr 2 , 0 , h gives no additional
restriction on r. The maximum area is therefore the
maximum value of the function A(r) on the interval
0 , r.
d 1000
Ar(r) 5 a
1 2pr 2 b
r
dr
1000
5 2 2 1 4pr
r
Cumulative Review of Calculus
The critical points of A(r) can be found by setting
Ar(r) 5 0:
2
1000
1 4pr 5 0
r2
4pr 3 5 1000
r 5 # 1000 8 4.3 cm
3
4p
So r 5 4.3 cm is the only critical point of the
function. This gives the value
500
h5
8 8.6 cm.
p(4.3)2
19.
r
h
Let the radius be r and the height h.
Minimize the cost:
C 5 2pr 2 (0.005) 1 2prh(0.0025)
V 5 pr2 h 5 4000
4000
h5
pr 2
4000
b (0.0025)
C(r) 5 2pr 2 (0.005) 1 2pr a
pr 2
20
5 0.01pr 2 1 , 1 # r # 36
r
20
C r(r) 5 0.02pr 2 2 .
r
For a maximum or minimum value, let C r(r) 5 0.
0.02pr 2 2
20
r2
50
20
0.02p
r 8 6.8
Using the max min algorithm:
C(1) 5 20.03, C(6.8) 5 4.39, C(36) 5 41.27.
The dimensions for the cheapest container are a
radius of 6.8 cm and a height of 27.5 cm.
20. a. Let the length, width, and depth be l, w, and
d, respectively. Then, the given information is that
l 5 x, w 5 x, and
l 1 w 1 d 5 140. Substituting gives
2x 1 d 5 140
d 5 140 2 2x
b. The volume of the box is V 5 lwh. Substituting
in the values from part a. gives
V 5 (x)(x)(140 2 2x)
5 140x 2 2 2x 3
r3 5
Calculus and Vectors Solutions Manual
In order for the dimensions of the box to make sense,
the inequalities l $ 0, w $ 0, and h $ 0 must be
satisfied. The first two give x $ 0, the third requires
x # 70. The maximum volume is therefore the
maximum value of V(x) 5 140x 2 2 2x 3 on the
interval 0 # x # 70, which can be found by
determining the critical points of the derivative Vr(x).
d
Vr(x) 5
(140x 2 2 2x 3 )
dx
5 280x 2 6x 2
5 2x(140 2 3x)
Setting Vr(x) 5 0 shows that x 5 0 and
140
x 5 3 8 46.7 are the critical points of the function.
The maximum value therefore occurs at one of these
points or at one of the endpoints of the interval:
V(0) 5 140(0)2 2 2(0)3 5 0
V(46.7) 5 140(46.7)2 2 2(46.7)3 5 101 629.5
V(0) 5 140(70)2 2 2(70)3 5 0
So the maximum volume is 101 629.5 cm3, from a
box with length and width 46.7 cm and depth
140 2 2(46.7) 5 46.6 cm.
21. The revenue function is
R(x) 5 x(50 2 x 2 )
5 50x 2 x 3. Its maximum for x $ 0 can be
found by examining its derivative to determine
critical points.
d
Rr(x) 5
(50x 2 x 3 )
dx
5 50 2 3x 2
The critical points can be found by setting Rr(x) 5 0:
50 2 3x 2 5 0
50
x5 6
8 64.1
Å3
Only the positive root is of interest since the number
of MP3 players sold must be positive. The number
must also be an integer, so both x 5 4 and x 5 5
must be tested to see which is larger.
R(4) 5 50(4) 2 43 5 136
R(4) 5 50(5) 2 53 5 125
So the maximum possible revenue is $136, coming
from a sale of 4 MP3 players.
22. Let x be the fare, and p(x) be the number of
passengers per year. The given information shows
that p is a linear function of x such that an increase
of 10 in x results in a decrease of 1000 in p. This
means that the slope of the line described by p(x) is
21000
10 5 2100. Using the initial point given,
p(x) 5 2100(x 2 50) 1 10 000
5 2100x 1 15 000
5-7
The revenue function can now be written:
R(x) 5 xp(x)
5 x(2100x 1 15 000)
5 15 000x 2 100x 2
Its maximum for x $ 0 can be found by examining
its derivative to determine critical points.
d
Rr(x) 5
(15 000x 2 100x 2 )
dx
5 15 000 2 200x
Setting Rr(x) 5 0 shows that x 5 75 is the only
critical point of the function. The problem states
that only $10 increases in fare are possible, however,
so the two nearest must be tried to determine the
maximum possible revenue:
R(70) 5 15 000(70) 2 100(70)2 5 560 000
R(80) 5 15 000(80) 2 100(80)2 5 560 000
So the maximum possible revenue is $560 000,
which can be achieved by a fare of either $70 or $80.
23. Let the number of $30 price reductions be n.
The resulting number of tourists will be 80 1 n
where 0 # n # 70. The price per tourist will be
5000 2 30n dollars. The revenue to the travel
agency will be (5000 2 30n)(80 1 n) dollars. The
cost to the agency will be 250 000 1 300(80 1 n)
dollars.
Profit 5 Revenue 2 Cost
P(n) 5 (5000 2 30n)(80 1 n)
2 250 000 2 300(80 1 n), 0 # n # 70
P r(n) 5 230(80 1 n) 1 (5000 2 30n)(1) 2 300
5 2300 2 60n
1
P r(n) 5 0 when n 5 38
3
Since n must be an integer, we now evaluate P(n)
for n 5 0, 38, 39, and 70. (Since P(n) is a quadratic
function whose graph opens downward with vertex
at 38 13 , we know P(38) . P(39).)
P(0) 5 126 000
P(38) 5 (3860)(118) 2 250 000 2 300(118)
5 170 080
P(39) 5 (3830)(119) 2 250 000 2 300(119)
5 170 070
P(70) 5 (2900)(150) 2 250 000 2 300(150)
5 140 000
The price per person should be lowered by $1140
(38 decrements of $30) to realize a maximum profit
of $170 080.
dy
d
24. a.
5
(25x 2 1 20x 1 2)
dx
dx
5 210x 1 20
5-8
dy
Setting dx 5 0 shows that x 5 2 is the only critical
number of the function.
b.
x
x,2
x52
x.2
y9
1
0
2
Graph
Inc.
Local Max
Dec.
d
dy
5
(6x 2 1 16x 2 40)
dx
dx
5 12x 1 16
dy
Setting dx 5 0 shows that x 5 2 43 is the only
critical number of the function.
x
x,2
4
3
x52
4
3
x.2
y9
2
0
1
Graph
Dec.
Local Min
Inc.
4
3
d
dy
5
(2x 3 2 24x)
dx
dx
5 6x 2 2 24
dy
The critical numbers are found by setting dx 5 0:
6x 2 2 24 5 0
6x 2 5 24
x 5 62
c.
x
x , 22
x 5 22
22 , x , 2
x52
x.2
y9
1
0
2
0
1
Graph
Inc.
Local Max
Dec.
Local Min
Inc.
dy
d
x
5
a
b
dx
dx x 2 2
(x 2 2)(1) 2 x(1)
5
(x 2 2)2
22
5
(x 2 2)2
This derivative is never equal to zero, so the
function has no critical numbers. Since the
numerator is always negative and the denominator
is never negative, the derivative is always negative.
This means that the function is decreasing
everywhere it is defined, that is, x 2 2.
25. a. This function is discontinuous when
x2 2 9 5 0
x 5 63. The numerator is non-zero at these
points, so these are the equations of the vertical
asymptotes.
d.
Cumulative Review of Calculus
To check for a horizontal asymptote:
8
8
lim 2
5 lim
9
xS` x 2 9
xS` 2
x a1 2 2 b
x
lim (8)
xS`
5
9
lim x 2 a1 2 2 b
xS`
x
lim (8)
xS`
5
9
lim (x)2 3 lim a1 2 2 b
xS`
xS`
x
1
8
5 lim 2 3
xS` x
120
50
8
Similarly, lim x 2 2 9 5 0, so y 5 0 is a horizontal
xS2`
asymptote of the function.
There is no oblique asymptote because the degree
of the numerator does not exceed the degree of the
denominator by 1.
Local extrema can be found by examining the
derivative to determine critical points:
(x 2 2 9)(0) 2 (8)(2x)
yr 5
(x 2 2 9)2
216x
5 2
(x 2 9)2
Setting yr 5 0 shows that x 5 0 is the only critical
point of the function.
x,0
x
x50
x.0
y9
1
0
1
Graph
Inc.
Local Max
Dec.
So (0, 2 89 ) is a local maximum.
b. This function is discontinuous when
x2 2 1 5 0
x 5 61. The numerator is non-zero at these
points, so these are the equations of the vertical
asymptotes.
To check for a horizontal asymptote:
4x 3
x 3 (4)
5 lim
lim 2
xS` x 2 1
xS` 2
1
x 1 2 x2
x(4)
5 lim
1
xS`
1 2 x2
(
)
lim (x(4))
5
xS`
(
1
lim 1 2 x 2
xS`
)
Calculus and Vectors Solutions Manual
lim (x) 3 lim (4)
5
xS`
xS`
(
lim 1 2
xS`
5 lim (x) 3
xS`
1
x2
)
4
120
5`
4x 3
Similarly, lim x 2 2 1 5 lim (x) 5 2`, so this
xS 2`
xS 2`
function has no horizontal asymptote.
To check for an oblique asymptote:
4x
x2 2 1q4x3 1 0x2 1 0x 1 0
4x3 1 0x2 2 4x
0 1 0 1 4x 1 0
So y can be written in the form
4x
y 5 4x 1 2
. Since
x 21
4x
x(4)
lim 2
5 lim
xS` x 2 1
xS` 2
1
x 1 2 x2
(
5 lim
xS`
)
4
(
1
x 1 2 x2
lim (4)
)
xS`
5
((
1
lim x 1 2 x 2
xS`
))
lim (4)
5
xS`
(
1
lim (x) 3 lim 1 2 x 2
xS`
xS`
)
1
4
5 lim a b 3
x
120
5 0,
4x
and similarly lim x 2 2 1 5 0, the line y 5 4x is an
xS 2`
asymptote to the function y.
Local extrema can be found by examining the
derivative to determine critical points:
(x 2 2 1)(12x 2 ) 2 (4x 3 )(2x)
yr 5
(x 2 2 1)2
12x 4 2 12x 2 2 8x 4
5
(x 2 2 1)2
4x 4 2 12x 2
5
(x 2 2 1)2
Setting yr 5 0:
4x 4 2 12x 2 5 0
x 2 (x 2 2 3) 5 0
5-9
so x 5 0, x 5 6 !3 are the critical points of the
function
y9
x
x , 2 !3
x 5 2 !3
2 !3 , x , 0
x50
y9
1
0
2
0
Graph
Inc.
Local Max
Dec.
Horiz.
x
0 , x , !3
x 5 !3
x . !3
y9
2
0
2
Graph
Dec.
Local Min
Inc.
1
,x,1
2
2
2
x
(2!3, 26 !3) is a local maximum, ( !3, 6 !3) is
a local minimum, and (0, 0) is neither.
26. a. This function is continuous everywhere, so it
has no vertical asymptotes. To check for a horizontal
asymptote:
lim (4x 3 1 6x 2 2 24x 2 2)
xS`
5 lim x 3 a4 1
24
6
2
2 2 2 3)
x
x
x
6
2
24
5 lim (x 3 ) 3 lim a4 1 2 2 2 3 b
x
xS`
xS`
x
x
3
5 lim (x ) 3 (4 1 0 2 0 2 0)
xS`
x.1
0
1
Graph
Dec.
Local Min
Inc.
y0
1
1
1
Concavity
Up
Up
Up
y
30
20
10
x
–3 –2 –1 0
–10
1 2
y = 4x3 + 6x2 – 24x – 2
b. This function is discontinuous when
x2 2 4 5 0
(x 1 2)(x 2 2) 5 0
x 5 2 or x 5 22. The numerator is non-zero at
these points, so the function has vertical asymptotes
at both of them. The behaviour of the function near
these asymptotes is:
3x
x12
x22
y
2
,0
,0
,0
,0
1
x-values
xS`
x51
lim y
xS`
5`
Similarly,
lim (4x 3 1 6x 2 2 24x 2 2) 5 lim (x 3 ) 5 2`,
x S 22
,0
.0
,0
.0
1`
x S 22
.0
.0
,0
,0
2`
so this function has no horizontal asymptote.
The y-intercept can be found by letting x 5 0,
which gives y 5 4(0)3 1 6(0)2 2 24(0) 2 2
5 22
The derivative is of the function is
d
(4x 3 1 6x 2 2 24x 2 2)
yr 5
dx
5 12x 2 1 12x 2 24
5 12(x 1 2)(x 2 1), and the second derivative is
d
ys 5
(12x 2 1 12x 2 24)
dx
5 24x 1 12
Letting f r(x) 5 0 shows that x 5 22 and x 5 1 are
critical points of the function. Letting ys 5 0 shows
that x 5 2 12 is an inflection point of the function.
x S 21
.0
.0
.0
.0
1`
xS 2`
xS2`
x
x , 22
x 5 22
22 , x
y9
1
0
2
1
x52
2
2
Graph
Inc.
Local Max
Dec.
Dec.
y0
2
2
2
0
Concavity
Down
Down
Down
Infl.
5-10
x S 22
2`
To check for a horizontal asymptote:
3x
x(3)
lim 2
5 lim
xS` x 2 4
xS` 2
4
x 12 2
x
(
5 lim
xS`
)
3
(
4
x 1 2 x2
lim (3)
)
xS`
5
((
4
))
lim x 1 2 x 2
xS`
lim (3)
xS`
5
(
4
lim (x) 3 lim 1 2 x 2
xS`
xS`
1
3
5 lim 3
xS` x
120
50
)
3x
Similarly, lim x 2 2 4 5 0, so y 5 0 is a horizontal
xS`
asymptote of the function.
Cumulative Review of Calculus
This function has y 5 0 when x 5 0, so the origin is
both the x- and y-intercept.
The derivative is
(x 2 2 4)(3) 2 (3x)(2x)
yr 5
(x 2 2 4)2
2
23x 2 12
5
, and the second derivative is
(x 2 2 4)2
(x 2 2 4)2 (26x)
ys 5
(x 2 2 4)4
(23x 2 2 12)(2(x 2 2 4)(2x))
2
(x 2 2 4)4
26x 3 1 24x 1 12x 3 1 48x
5
(x 2 2 4)3
6x 3 1 72x
5
(x 2 2 4)3
The critical points of the function can be found by
letting yr 5 0, so
23x 2 2 12 5 0
x 2 1 4 5 0. This has no real solutions, so the
function y has no critical points.
The inflection points can be found by letting
ys 5 0, so
6x 3 1 72x 5 0
6x(x 2 1 12) 5 0
The only real solution to this equation is x 5 0, so
that is the only possible inflection point.
x
x , 22 22 , x , 0
x50
0,x,2
x.2
y9
2
2
2
2
2
Graph
Dec.
Dec.
Dec.
Dec.
Dec.
y0
2
1
0
2
1
Concavity
Down
Up
Infl.
Down
Up
6
4
2
y
–6 –4 –2 0
–2
–4
–6
3x
y = ——–—
x2 – 4
x
2 4 6
d
((24)e 5x11 )
dx
d
5 (24)e 5x11 3
(5x 1 1)
dx
5 (220)e 5x11
27. a. f r(x) 5
Calculus and Vectors Solutions Manual
d
(xe 3x )
dx
d
5 xe 3x 3
(3x) 1 (1)e 3x
dx
5 e 3x (3x 1 1)
d
c. yr 5 (63x28 )
dx
d
5 (ln 6)63x28 3
(3x 2 8)
dx
5 (3 ln 6)63x28
d
d. yr 5 (e sin x )
dx
d
5 e sin x 3
(sin x)
dx
5 (cos x)e sin x
28. The slope of the tangent line at x 5 1 can be
b. f r(x) 5
dy
found by evaluating the derivative dx for x 5 1:
d 2x21
dy
5
(e
)
dx
dx
d
5 e 2x21 3
(2x 2 1)
dx
5 2e 2x21
Substituting x 5 1 shows that the slope is 2e. The
value of the original function at x 5 1 is e, so the
equation of the tangent line at x 5 1 is
y 5 2e(x 2 1) 1 e.
29. a. The maximum of the function modelling the
number of bacteria infected can be found by
examining its derivative.
d
t
N r(t) 5 ((15t)e 2 5 )
dt
d
t
t
t
5 15te 25 3 a2 b 1 (15)e 25
dt
5
t
5 e25 (15 2 3t)
Setting Nr(t) 5 0 shows that t 5 5 is the only
critical point of the function (since the exponential
function is never zero). The maximum number of
infected bacteria therefore occurs after 5 days.
5
b. N(5) 5 (15(5))e 25
5 27 bacteria
dy
d
30. a.
5
(2 sin x 2 3 cos 5x)
dx
dx
d
5 2 cos x 2 3(2sin 5x) 3
(5x)
dx
5 2 cos x 1 15 sin 5x
5-11
b.
dy
d
5
(sin 2x 1 1)4
dx
dx
d
5 4(sin 2x 1 1)3 3
(sin 2x 1 1)
dx
d
5 4(sin 2x 1 1)3 3 (cos 2x) 3
(2x)
dx
5 8 cos 2x(sin 2x 1 1)3
1
c. y can be rewritten as y 5 (x 2 1 sin 3x)2 . Then,
dy
d
1
5
(x 2 1 sin 3x)2
dx
dx
1 2
d
1
5 (x 1 sin 3x)22 3
(x 2 1 sin 3x)
2
dx
1
1
5 (x 2 1 sin 3x)22
2
d
3 a2x 1 cos 3x 3
(3x)b
dx
2x 1 3 cos 3x
5
2 !x 2 1 sin 3x
dy
d
sin x
d.
5
a
b
dx
dx cos x 1 2
(cos x 1 2)(cos x) 2 (sin x)(2sin x)
5
(cos x 1 2)2
cos2 x 1 sin2 x 1 2 cos x
5
(cos x 1 2)2
1 1 2 cos x
5
(cos x 1 2)2
dy
d
5
(tan x 2 2 tan2 x)
e.
dx
dx
d
d
5
sec2 x 2 3
(x 2 )
dx
dx
d
2 2 tan x 3
(tan x)
dx
5 2x sec2 x 2 2 2 tan x sec2 x
dy
d
5
(sin (cos x 2 ))
f.
dx
dx
d
5 cos (cos x 2 ) 3
(cos x 2 )
dx
d
5 cos (cos x 2 ) 3 (2sin x 2 ) 3
(x 2 )
dx
5 22x sin x 2 cos(cos x 2 )
31.
l2
u
100
l1
250
u
As shown in the diagram, let u be the angle between
the ladder and the ground, and let the total length
of the ladder be l 5 l1 1 l2, where l1 is the length
from the ground to the top corner of the shed and
l2is the length from the corner of the shed to the
wall.
100
250
cos u 5
sin u 5
l1
l2
l2 5 100 sec u
l1 5 250 csc u
l 5 250 csc u 1 100 sec u
dl
5 2250 csc u cot u 1 100 sec u tan u
du
250 cos u
100 sin u
52
1
sin2 u
cos2 u
dl
To determine the minimum, solve du 5 0.
250 cos u
100 sin u
5
2
sin u
cos2 u
3
250 cos u 5 100 sin3 u
2.5 5 tan3 u
3
tan u 5 "
2.5
u 8 0.94
At u 5 0.94, l 5 250 csc 0.94 1 100 sec 0.94
8 479 cm
The shortest ladder is about 4.8 m long.
32. The longest rod that can fit around the corner is
determined by the minimum value of x 1 y. So,
determine the minimum value of l 5 x 1 y.
y
3
u
x
u
5-12
3
Cumulative Review of Calculus
3
3
From the diagram, sin u 5 y and cos u 5 x. So,
3
3
p
l 5 cos u 1 sin u, for 0 # u # 2 .
dl
3 sin u
3 cos u
5
2
2
du
cos u
sin2 u
3
3 sin u 2 3 cos3 u
5
cos2 u sin2 u
dl
Solving du 5 0 yields:
3
3
p 1
p
cos 4
sin 4
5 3 !2 1 3 !2
5 6 !2
So l 5
p
When u 5 0 or u 5 2 , the longest possible rod
would have a length of 3 m. Therefore the longest
rod that can be carried horizontally around the
corner is one of length 6 !2, or about 8.5 m.
3 sin3 u 2 3 cos3 u 5 0
tan3 u 5 1
tan u 5 1
p
u5
4
Calculus and Vectors Solutions Manual
5-13
Cumulative Review of Vectors
pp. 557–560
1. a. The angle, u, between the two
vectors is found
>
>
a?b
cos (u) 5 > > .
@a @ @b@
from the equation
> >
a ? b 5 (2, 21, 22) ? (3, 24, 12)
5 2(3) 2 1(24) 2 2(12)
5 214
>
0 a 0 5 "22 1 (21)2 1 (22)2
53
0 b 0 5 "32 1 (24)2 1 122
5 13
So u 5 cos21 ( 3 214
3 13 )
8 111.0°
>
>
>
b. The scalar projection of a on b is equal to
>
0 a 0 cos (u), where u is the angle between the two
vectors. So from the above work, cos (u) 5 3 214
3 13
>
>
>
and 0 a 0 5 3, so the scalar projection of a on b is
>
>
214
14
3 3 13 3 3 5 2 13 . The vector projection of a on b
is equal to the scalar projection multiplied by the unit
>
vector in the direction of b. So the vector projection
1
52 56
is 2 14
, , 2 168
13 3 13 (3, 24, 12) 5 (2
169 ).
> 169 169
>
c. > The scalar projection of b on a is equal to
0 b 0cos (u), where u is the angle between the two
vectors. So from the above work, cos (u) 5 3 214
3 > 13
>
>
and 0 b 0 5 13, so the scalar projection of a on b is
>
>
214
14
3 3 13 3 13 5 2 3 . The vector projection of b on a
is equal to the scalar projection multiplied by the
>
unit vector in the direction of a . So the vector
projection is 2 143 3 13 (2, 21, 22) 5 (2 289, 149, 289 ).
2. a. Since the normal of the first plane is (4, 2, 6)
and the normal of the second is (1, 21, 1), which
are not scalar multiples of each other, there is a line
of intersection between the planes.
The next step is to use the first and second equations
to find an equation with a zero for the coefficient of x.
The first equation minus four times the second
equation yields 0x 1 6y 1 2z 1 6 5 0. We may
divide by two to simplify, so 3y 1 z 1 3 5 0. If we
let y 5 t, then 3t 1 z 1 3 5 0, or z 5 23 2 3t.
Substituting these into the second equation yields
x 2 (t) 1 (23 2 3t) 2 5 5 0 or x 5 8 1 4t.
Calculus and Vectors Solutions Manual
So the equation of the line in parametric form is
x 5 8 1 4t, y 5 t, z 5 23 2 3t, tPR.
To check that this is correct, we substitute in the
solution to both initial equations
4x 1 2y 1 6z 2 14 5 4(8 1 4t) 1 2(t)
1 6(23 2 3t) 2 14
50
and x 2 y 1 z 2 5
5 (8 1 4t) 2 (t) 1 (23 2 3t) 2 5
5 0.
Hence the line given by the parametric equation
above is the line of intersection for the planes.
b. The angle between two planes is the same as the
angle between their corresponding normal vectors.
0 (4, 2, 6)0 5 "42 1 22 1 62
5 !56
0 (1, 21, 1) 0 5 "12 1 12 1 12
5 !3
(4, 2, 6) ? (1, 21, 1) 5 8, so the angle between the
planes is cos21 Q !38!56 R 8 51.9°.
> >
>
x?y
3. a. We have that cos (60°) 5 0 x> 0 0 y> 0 . Also since x
>
>
>
and y are unit vectors, 0 x 0 5 0 y 0 5 1, and moreover
> >
> >
1
x?y
1
cos (60°) 5 . So x ? y 5
5 .
2
131
2
b. Scalar multiples can be brought out to the front
>
>
> >
of dot products. Hence 2x ? 3y 5 (2)(3)(x ? y ),
>
>
and so by part a., 2x ? 3y 5 2 3 3 3 12 5 3.
c. The dot product is distributive,
>
>
>
>
so (2x 2 y ) ? (x 1 3y )
>
>
>
>
>
>
5 2x ? (x 1 3y ) 2 y ? (x 1 3y )
> >
>
>
> >
>
>
5 2x ? x 1 2x ? 3y 2 y ? x 2 y ? 3y
> >
>
>
> >
> >
5 2x ? x 1 2x ? 3y 2 x ? y 2 3y ? y
>
>
> >
> >
Since x and y are unit vectors, x ? x 5 y ? y 5 1,
and so by using the values found in part a. and b.,
>
>
>
>
(2x 2 y ) ? (x 1 3y ) 5 2(1) 1 (3) 2 A 12 B 2 3(1)
3
5
2 >
>
>
>
>
>
>
>
4. a. 2(i 2 2j 1 3k ) 2 4(2i 1 4j 1 5k ) 2 (i 2 j )
>
>
>
>
>
>
>
>
5 2i 2 4j 1 6k 2 8i 2 16j 2 20k 2 i 1 j
>
>
>
5 27i 2 19j 2 14k
9-1
>
>
>
>
>
>
>
>
b. 22(3i> 2 4j> 2 5k>) ? (2i> 1 3k> ) 1 2i
> ? (3j 2 2k )
5 22(3i 2 4j 2 5k ) ? (2i 1 0j 1 3k )
>
>
>
>
>
>
1 2(i 1 0j 1 0k ) ? (0i 1 3j 2 2k )
5 22(3(2) 2 4(0) 2 5(3)) 1 2(1(0)
1 0(3) 1 0(22))
5 22(29) 1 2(0)
5 18
5. The direction vectors for the positive x-axis,
y-axis, and z-axis are (1, 0, 0), (0, 1, 0), and (0, 0, 1),
respectively.
0 (4, 22, 23) 0 5 "42 1 (22)2 1 (23)2
5 !29,
and 0 (1, 0, 0) 0 5 0 (0, 1, 0) 0
5 0 (0, 0, 1) 0
5 !1
5 1.
(4, 22, 23) ? (1, 0, 0) 5 4, so the angle the vector
4
makes with the x-axis is cos21 Q 1 !29
R 8 42.0°.
(4, 22, 23) ? (0, 1, 0) 5 22, so the angle the vector
makes with the y-axis is cos21 Q 1 22
!29 R 5 111.8°.
(4,22,23) ? (0, 0, 1) 5 23, hence the angle the
vector makes with the z-axis is cos21 Q 1 23
!29 R 8 123.9°.
>
>
6. a. a 3 b 5 (1, 22, 3) 3 (21, 1, 2)
5 (22(2) 2 3(1), 3(21) 2 1(2),
1(1) 2 (22)(21))
5 (27, 25, 21)
b. By the> scalar law for> vector multiplication,
>
>
2a 3 3b 5 2(3)(a 3
> b)
>
5 6(a 3 b )
5 6(27, 25, 21) 5 (242, 230, 26)
>
c.> The area of a parallelogram determined by a and
b is equal
> to the magnitude of the cross product of
>
a and b.
A 5 area of> parallelogram
>
5 0a 3 b0
5 0 (27, 25, 21) 0
5 "(27)2 1 (25)2 1 (21)2
8 >8.66 square units >
>
>
d. (b 3 a ) 5 2 (a 3 b )
5 2 (27, 25, 21)
5 (7, 5, 1)
>
>
>
So c ? (b 3 a ) 5 (3, 24, 21) ? (7, 5, 1)
5 3(7) 2 4(5) 2 1(1)
50
>
>
7. A unit vector perpendicular to both a and b can
be determined from any vector perpendicular to
9-2
> >
>
>
both a and b>. a 3 b is a vector perpendicular to
>
both a and
b.
>
>
a 3 b 5 (1, 21, 1) 3 (2, 22, 3)
5 (21(3) 2 1(22), 1(2) 2 1(3),
1(22) 2 (21)(2))
5
(21, 21, 0)
>
>
0 a 3 b 0 5 0 (21, 21, 0) 0
5 "(21)2 1 (21)2 1 02
5 !2
1
1
1
So !2
(21, 21, 0) 5 Q 2 !2
, 2 !2
, 0 R is an unit vector
>
>
1
1
perpendicular to both a and b. Q !2
, !2
, 0 R is another.
8. a. Answers may vary. For example:
>
A direction
vector for the line is AB.
>
AB 5 (1, 2, 3) 2 (2, 23, 1)
5 (21, 5, 2)
Since A(2, 23, 1) is a point on the line,
>
r 5 (2, 23, 1) 1 t(21, 5, 2), tPR, is a vector
equation for a line and the corresponding parametric
equation is x 5 2 2 t, y 5 23 1 5t, z 5 1 1 2t,
tPR.
b. If the x-coordinate of a point on the line is 4, then
2 2 t 5 4, or t 5 22. At t 5 22, the point on the
line is (2, 23, 1) 2 2(21, 5, 2) 5 (4, 213, 23).
Hence C(4, 213, 23) is a point on the line.
9. The direction vector of the first line is (21, 5, 2),
while the direction vector for the second line is
(1, 25, 22) 5 2 (21, 5, 2). So the direction vectors
for the line are collinear. Hence the lines are parallel.
The lines coincide if and only if for any point on
the first line and any point on the second line, the
vector connecting the two points is a multiple of the
direction vector for the lines.
(2, 0, 9) is a point on the first line and (3, 25, 10) is
a point on the second line.
(2, 0, 9) 2 (3, 25, 10) 5 (21, 5, 21) 2 k(21, 5, 2)
for any kPR. Hence the lines are parallel and distinct.
10. The direction vector for the parallel line is
(0, 1, 1). Since parallel lines have collinear direction
vectors, (0, 1, 1) can be used as a direction vector
for the line. Since (0, 0, 4) is a point on the line,
>
r 5 (0, 0, 4) 1 t(0, 1, 1), tPR, is a vector equation
for a line and the corresponding parametric equation
is x 5 0, y 5 t, z 5 4 1 t, tPR.
11. The line is parallel to the plane if and only if the
direction vector for the line is perpendicular to the
normal vector for the plane. The normal vector for
the plane is (2, 3, c). The direction vector for the
line is (2, 3, 1). The vectors are perpendicular if and
only if the dot product between the two is zero.
Chapters 6–9: Cumulative Review
(2, 3, c) ? (2, 3, 1) 5 2(2) 1 3(3) 1 c(1)
5 13 1 c
So if c 5 213, then the dot product of normal
vector and the direction vector is zero. Hence for
c 5 213, the line and plane are parallel.
12. First put the line in its corresponding parametric
form. (3, 1, 5) is a direction vector and (2, 25, 3) is
the origin point, so a parametric equation for the
line is x 5 2 1 3s, y 5 25 1 s, z 5 3 1 5s, sPR.
If we substitute these coordinates into the equation
of the plane, we may find the s value where the line
intersects the plane.
5x 1 y 2 2z 1 2
5 5(2 1 3s) 1 (25 1 s) 2 2(3 1 5s) 1 2
5 10 1 15s 1 2 5 1 s 2 6 2 10s 1 2
5 1 1 6s
So if 5x 1 y 2 2z 1 2 5 0, then 1 1 6s 5 0 or
s 5 2 16. At s 5 2 16, the point on the line is ( 32, 2 316, 136) .
To check that this point is also on the plane, we
substitute the x, y, z values into the plane equation
and check that it equals zero.
3
31
13
5x 1 y 2 2z 1 2 5 5a b 1 a2 b 2 2a b 1 2
2
6
6
50
Hence ( 32, 2 316, 136) is the point of intersection between
b.
z
(–3, –2, 2)
(0, 0, 0)
(3, 2, 1)
y
x
Two direction vectors are:
(23, 22, 2) 2 (0, 0, 0) 5 (23, 22, 2)
and
(3, 2, 1) 2 (0, 0, 0) 5 (3, 2, 1).
c.
z
(0, 3, 6)
the line and the plane.
13. a.
(0, 0, 0)
z
x
(1, 1, –1)
y
(0, 0, 3)
x
(0, 3, 0)
y
(6, 0, 0)
Two direction vectors are:
(0, 3, 0) 2 (0, 0, 3) 5 (0, 3, 23)
and
(6, 0, 0) 2 (0, 0, 3) 5 (6, 0, 23).
Calculus and Vectors Solutions Manual
Two direction vectors are:
(0, 3, 6) 2 (0, 0, 0) 5 (0, 3, 6)
and
(1, 1, 21) 2 (0, 0, 0) 5 (1, 1, 21).
14. The plane is the right bisector joining
P(1, 22, 4) and its image. The line connecting the
two points has a direction vector equal to that of the
normal vector for the plane. The normal vector for
the plane is (2, 23, 24). So the line connecting the
two points is (1, 22, 4) 1 t(2, 23, 24), tPR, or in
9-3
corresponding parametric form is x 5 1 1 2t,
y 5 22 2 3t, z 5 4 2 4t, tPR.
The intersection of this line and the plane is the
bisector between P and its image. To find this point
we substitute the parametric equation into the plane
equation and solve for t.
2x 2 3y 2 4z 1 66
5 2(1 1 2t) 2 3(22 2 3t) 2 4(4 2 4t) 1 66
5 2 1 4t 1 6 1 9t 2 16 1 16t 1 66
5 58 1 29t
So if 2x 2 3y 2 4z 1 66 5 0, then 58 1 29t 5 0,
or t 5 22.
So the point of intersection is occurs at t 5 22, since
the origin point is P and the intersection occurs at the
midpoint of the line connecting P and its image, the
image point occurs at t 5 2 3 (22) 5 24.
So the image point is at x 5 1 1 2(24) 5 27,
y 5 22 2 3(24) 5 10, z 5 4 2 4(24) 5 20.
So the image point is (27, 10, 20).
15. Let (a, b, c) be the direction vector for this line.
>
So a line equation is r 5 (1, 0, 2) 1 t(a, b, c), tPR.
Since (1, 0, 2) is not on the other line, we may
choose a, b, and c such that the intersection occurs
at t 5 1. Since the line is supposed to intersect the
given line at a right angle, the direction vectors
should be perpendicular. The direction vectors are
perpendicular if and only if their dot product is zero.
The direction vector for the given line is (1, 1, 2).
(a, b, c) ? (1, 1, 2) 5 a 1 b 1 2c 5 0, so
b 5 2a 2 2c.
Also (1, 0, 2) 1 (a, b, c) 5 (1 1 a, b, 2 1 c) is the
point of intersection.
By substituting for b,
(1 1 a, b, 2 1 c) 5 (1 1 a, 2a 2 2c, 2 1 c).
So for some s value,
x 5 22 1 s 5 1 1 a
y 5 3 1 s 5 2a 2 2c
z 5 4 1 2s 5 2 1 c
Subtracting the first equation from the second yields
the equation, 5 1 0s 5 22a 2 2c 2 1.
Simplifying this gives 6 5 22a 2 2c or just
a 1 c 5 23.
Subtracting twice the first equation from the third
yields the equation, 8 5 22a 1 c.
So a 1 c 5 23 and 22a 1 c 5 8, which is two
equations with two unknowns. Twice the first plus
the second equations gives 0a 1 3c 5 2 or c 5 23.
Solving back for a gives 2 113 and since b 5 2a 2 2c,
b 5 73. Since a 1 b 1 2c 5 0, the direction vectors,
9-4
(1, 1, 2) and (a, b, c) are perpendicular. A direction
vector for the line is (211, 7, 2).
We need to check that
7 8
(1, 0, 2) 1 (a, b, c) 5 ( 28
3 , 3 , 3 ) is a point on the
given line.
x 5 22 1 s 5 2 83, at s 5 2 23. The point on the given
7 8
line at s 5 2 23 is Q 28
3 , 3 , 3 R . Hence
>
q 5 (1, 0, 2) 1 t(211, 7, 2), tPR, is a line that
intersects the given line at a right angle.
16. a. The Cartesian equation is found by taking
> the
cross
> product of the two direction vectors, AB and
AC.
>
AB 5 (22, 0, 0) 2 (1, 2, 3)
> 5 (23, 22, 23)
AC 5 (1, 4, 0) 2 (1, 2, 3) 5 (0, 2, 23)
>
>
AB 3 AC 5 (22(23) 2 (23)(2),
23(0) 2 (23)(23),
23(2) 2 (22)(0))
5 (12, 29, 26)
So 5 (12, 29, 26) is a normal vector for the
plane, so the plane has the form
12x 2 9y 2 6z 1 D 5 0, for some constant D. To
find D, we know that A(1, 2, 3) is a point on the
plane, so 12(1) 2 9(2) 2 6(3) 1 D 5 0. So
224 1 D 5 0, or D 5 24. So the Cartesian
equation for the plane is 12x 2 9y 2 6z 1 24 5 0.
b. Substitute into the formula to determine distance
between a point and a plane. So the distance, d, of
(0, 0, 0) to the plane 12x 2 9y 2 6z 1 24 5 0 is
equal to
@ 12 (0) 2 9 (0) 2 6 (0) 1 24 @
"122 1 (29)2 1 (26)2
24
!261 8 1.49.
.
So d 5
17. a. (3, 25, 4) is a normal vector for the plane, so
the plane has the form 3x 2 5y 1 4z 1 D 5 0, for
some constant D. To find D, we know that
A(21, 2, 5) is a point on the plane, so
3(21) 2 5(2) 1 4(5) 1 D 5 0. So 7 1 D 5 0,
or D 5 27. So the Cartesian equation for the plane
is 3x 2 5y 1 4z 2 7 5 0.
b. Since the plane is perpendicular to the line
connecting (2, 1, 8) and (1, 2, 24), a direction
vector for the line acts as a normal vector for the
plane. So (2, 1, 8) 2 (1, 2, 24) 5 (1, 21, 12) is a
normal vector for the plane. So the plane has the
form x 2 y 1 12z 1 D 5 0, for some constant D.
To find D, we know that K(4, 1, 2) is a point on the
plane, so (4) 2 (1) 1 12(2) 1 D 5 0. So
27 1 D 5 0, or D 5 227. So the Cartesian
equation for the plane is x 2 y 1 12z 2 27 5 0.
Chapters 6–9: Cumulative Review
c. Since the plane is perpendicular to the z-axis, a
direction vector for the z-axis acts as a normal vector
for the plane. Hence (0, 0, 1) is a normal vector for
the plane. So the plane has the form z 1 D 5 0, for
some constant D. To find D, we know that (3, 21, 3)
is a point on the plane, so
0(3) 1 0(21) 1 (3) 1 D 5 0. So 3 1 D 5 0, or
D 5 23. So the Cartesian equation for the plane is
z 2 3 5 0.
d. The Cartesian equation can be found by taking
the cross product of the two direction vectors for
the plane. Since (3, 1, 22) and (1, 3, 21) are two
points on the plane
(3, 1, 22) 2 (1, 3, 21) 5 (2, 22, 21) is a
direction vector for the plane. Since the plane is
parallel to the y-axis, (0, 1, 0) is also a direction
vector for the plane.
(2, 22, 21) 3 (0, 1, 0) 5 (22(0) 2
(21)(1), (21)(0)2 (2)(0), 2(1) 2 (22)(0))
5 (1, 0, 2)
So (1, 0, 2) is a normal vector for the plane, so the
plane has the form x 1 0y 1 2z 1 D 5 0, for some
constant D. To find D, we know that (3, 1, 22) is a
point on the plane, so
(3) 1 0(1) 1 2(22) 1 D 5 0. So 21 1 D 5 0,
or D 5 1. So the Cartesian equation for the plane is
x 1 2z 1 1 5 0.
18.
E
100 km/h
45°
F
sin 45°
3 100.
336.80
sin /EDF 8 0.2100.
Thus /EDF 8 12.1°, so the resultant velocity is
336.80 km> h, N 12.1° W.
19. a. The simplest way is to find the parametric
equation, then find the corresponding vector equation.
If we substitute x 5 s and y 5 t and solve for z, we
obtain 3s 2 2t 1 z 2 6 5 0 or z 5 6 2 3s 1 2t.
This yields the parametric equations x 5 s, y 5 t,
and z 5 6 2 3s 1 2t. So the corresponding vector
>
equation is r 5 (0, 0, 6) 1 s(1, 0, 23) 1 t(0, 1, 2),
s, tPR. To check that this is correct, find the
Cartesian equation corresponding to the above
vector equation and see if it is equivalent to the
Cartesian equation given in the problem. A normal
vector to this plane is the cross product of the two
directional
vectors.
>
n 5 (1, 0, 23) 3 (0, 1, 2) 5 (0(2) 2 (23)(1),
23(0) 2 1(2), 1(1) 2 0(0))
5 (3, 22, 1)
So (3, 22, 1) is a normal vector for the plane, so the
plane has the form 3x 2 2y 1 z 1 D 5 0, for some
constant D. To find D, we know that (0, 0, 6) is a point
on the plane, so 3(0) 2 2(0) 1 (6) 1 D 5 0.
So 6 1 D 5 0, or D 5 26. So the Cartesian equation
for the plane is 3x 2 2y 1 z 2 6 5 0. Since this is
the same as the initial Cartesian equation, the vector
equation for the plane is correct.
b.
z
sin /EDF 8
(0, 0, 6)
400 km/h
R
400 km/h
(0, –3, 0)
45°
D
100 km/h
Position Diagram
Vector Diagram
From
the triangle DEF and the cosine law, we have
>
0 R 0 2 5 4002 1 1002 2 2(400)(100) cos (45°)
8 336.80 km> h.
To find the direction of the vector, the sine law is
applied.
sin /DEF
sin /EDF
>
5
100
0R0
sin /EDF
sin 45°
8
.
336.80
100
Calculus and Vectors Solutions Manual
y
x
(2, 0, 0)
20. a. The angle, u, between the plane and the line
is the complementary angle of the angle between
the direction vector of the line and the normal
9-5
vector for the plane. The direction vector of the line
is (2, 21, 2) and the normal vector for the plane
is (1, 2, 1).
0 (2, 21, 2) 0 5 "22 1 (21)2 1 22
5 !9
5 3.
0 (1, 2, 1) 0 5 "12 1 22 1 12
5 !6
(2, 21, 2) ? (1, 2, 1) 5 2(1) 2 1(2) 1 2(1) 5 2
So the angle between the normal vector and the
2
direction vector is cos21 Q 3 !6
R 8 74.21°. So
From
the triangle DEF and the cosine law, we have
>
0 R 0 2 5 402 1 252 2 2(40)(25) cos (120°)
8 56.79 N.
To find the direction of the vector, the sine law is
applied.
sin /DEF
sin /EDF
>
5
100
0R0
sin 120°
sin /EDF
8
.
56.79
40
sin 120°
sin /EDF 8
3 40.
56.79
sin /EDF 8 0.610.
Thus /EDF 8 37.6°, so the resultant force is
approximately 56.79 N, 37.6° from the 25 N force
towards the 40 N force. The equilibrant force has
the same magnitude as the resultant, but it is in
the opposite direction. So the equilibrant is
approximately 56.79 N, 180° 2 37.6° 5 142.4°
from the 25 N force away from the 40 N force.
22.
u 8 90° 2 74.21° 5 15.79°.
To the nearest degree, u 5 16°.
b. The two planes are perpendicular if and only if
their normal vectors are also perpendicular.
A normal vector for the first plane is (2, 23, 1) and
a normal vector for the second plane is
(4, 23, 217). The two vectors are perpendicular if
and only if their dot product is zero.
a
b
–b
(2, 23, 1) ? (4, 23, 217) 5 2(4) 2 3(23)
1 1(217)
a.
–b
5 0.
a
Hence the normal vectors are perpendicular. Thus
a –b
the planes are perpendicular.
c. The two planes are parallel if and only if their
1
b.
b
2
normal vectors are also parallel. A normal vector for
1
the first plane is (2, 23, 2) and a normal vector for
b
2a
2
2a + 21 b
the second plane is (2, 23, 2). Since both normal
2a
vectors are the same, the planes are parallel. Since
2(0) 2 3(21) 1 2(0) 2 3 5 0, the point
>
23. a. The unit vector in the same direction of a is
(0, 21, 0) is on the second plane. Yet since
>
>
simply a divided by the magnitude of a .
2(0) 2 3(21) 1 2(0) 2 1 5 2 2 0, (0, 21, 0) is
>
0 a 0 5 "62 1 22 1 (23)2
not on the first plane. Thus the two planes are
parallel but not coincident.
5 "49
21.
57
>
So the unit vector in the same direction of a is
25 N
1 >
> a 5 17 (6, 2, 23) 5 ( 67, 27, 2 37 ).
0a0
>
60°
b. The unit vector in the opposite direction of a is
40 N
Position diagram
40 N
E
120°
25 N
60°
D
9-6
R
120°
F
simply the negative of the unit vector found in part
a. So the vector is 2 A 67, 27, 2 37 B 5 A2 67, 2 27, 37 B.
24. a. Since OBCD is a parallelogram, the point
C
>
occurs at (21, 7) 1 (9, 2) 5 (8, 9). So> OC is one
vector
equivalent to a diagonal and BD is the other.
>
OC> 5 (8, 9) 2 (0, 0) 5 (8, 9)
BD 5 (9, 2) 2 (1, 7) 5 (10, 25)
40 N
Vector diagram
Chapters 6–9: Cumulative Review
0 (8, 9) 0 5 "82 1 92
b.
5 "145
0 (10, 25) 0 5 "102 1 (25)2
5 "125
(8, 9) ? (10, 25) 5 8(10) 1 9(25)
5 235
So the angle between these diagonals is
235
cos21 A !145
!125 B 8 74.9°.
>
>
c. OB 5 (21, 7) and OD 5 (9, 2)
0 (21, 7) 0 5 "(21)2 1 72.
5 "50
0 (9, 2) 0 5 "92 1 22
5 "85
(21, 7) ? (9, 2) 5 2 (9) 1 7(2)
55
So the angle between these diagonals is
cos21 A !505!85 B 8 85.6°.
25. a. First step is to use the first equation to
remove x from the second and third.
1 x2y1z52
2 2x 1 y 1 2z 5 1
3 x 2 y 1 4z 5 5
So we have
4 0x 1 0y 1 3z 5 3, 1 1 2
5 0x 1 0y 1 3z 5 3, 21 3 1 1 3
Hence 3z 5 3, or z 5 1. Since both equations are
the same, this implies that there are infinitely many
solutions. Let x 5 t, then by substituting into the
equation 2, we obtain
2t 1 y 1 2(1) 5 1, or y 5 21 1 t.
Hence the solution to these equations is x 5 t,
y 5 21 1 t, z 5 1, tPR.
b. First step is to use the first equation to remove x
from the second and third.
1 22x 2 3y 1 z 5 211
2 x 1 2y 1 z 5 2
3 2x 2 y 1 3z 5 212
So we have
4 0x 1 1y 1 3z 5 27, 1 1 2 3 2
5 0x 2 1y 2 5z 5 13, 1 2 2 3 3
Now the fourth and fifth equations are used to
create a sixth equation where the coefficient of
y is zero.
6 0x 1 0y 2 2z 5 6, 4 1 5
So 22z 5 6 or z 5 23.
Calculus and Vectors Solutions Manual
Substituting this into equation 4 yields,
y 1 3(23) 5 27 or y 5 2. Finally substitute z and
y values into equation 2 to obtain the x value.
x 1 2(2) 1 (23) 5 2 or x 5 1.
Hence the solution to these three equations is
(1, 2, 23).
c. First step is to notice that the second equation is
simply twice the first equation.
1 2x 2 y 1 z 5 21
2 4x 2 2y 1 2z 5 22
3 2x 1 y 2 z 5 5
So the solution to these equations is the same as the
solution to just the first and third equations.
Moreover since this is two equations with three
unknowns, there will be infinitely many solutions.
4 4x 1 0y 1 0z 5 4, 1 1 3
Hence 4x 5 4 or x 5 1. Let y 5 t and solve for z
using the first equation.
2(1) 2 t 1 z 5 21, so z 5 23 1 t
Hence the solution to these equations is x 5 1,
y 5 t, z 5 23 1 t, tPR.
d. First step is to notice that the second equations
is simply twice the first and the third equation is
simply 24 times the first equation.
1 x 2 y 2 3z 5 1
2 2x 2 2y 2 6z 5 2
3 24x 1 4y 1 12z 5 24
So the solution to these equations is the same as the
solution to just the first equation. So the solution to
these equations is a plane. To solve this in parametric
equation form, simply let y 5 t and z 5 s and find
the x value.
x 2 t 2 3s 5 1, or x 5 1 1 t 1 3s
So the solution to these equations is x 5 1 1 3s 1 t,
y 5 t, z 5 s, s, tPR.
26. a. Since the normal of the first equation
is (1, 21, 1) and the normal of the second is
(1, 2, 22), which are not scalar multiples of each
other, there is a line of intersection between the
planes. The next step is to use the first and second
equations to find an equation with a zero for the
coefficient of x. The second equation minus the first
equation yields 0x 1 3y 2 3z 1 3 5 0. We may
divide by three to simplify, so y 2 z 1 1 5 0. If
we let z 5 t, then y 2 t 1 1 5 0, or y 5 21 1 t.
Substituting these into the first equation yields
x 2 (21 1 t) 1 t 2 1 5 0 or x 5 0. So the
equation of the line in parametric form is x 5 0,
y 5 21 1 t, z 5 t, tPR.
9-7
To check that this is correct, we substitute in the
solution to both initial equations
x 2 y 1 z 2 1 5 (0) 2 (21 1 t) 1 (t) 2 1
50
and
x 1 2y 2 2z 1 2 5 (0) 1 2(21 1 t) 2 2(t) 1 2
5 0.
Hence the line given by the parametric equation
above is the line of intersection for the planes.
b. The normal vector for the first plane is
(1, 24, 7), while the normal vector for the second
plane is (2, 28, 14) 5 2(1, 24, 7). Hence the
planes have collinear normal vectors, and so are
parallel.
The second equation is equivalent to
x 2 4y 1 7z 5 30, since we may divide the equation
by two. Since the constant on the right in the first
equation is 28, while the constant on the right in the
second equivalent equation is 30, these planes are
parallel and not coincident. So there is no intersection.
c. The normal vector for the first equation is
(1, 21, 1), while the normal vector for the second
equation is (2, 1, 1). Since the normal vectors are
not scalar multiples of each other, there is a line of
intersection between the planes.
The next step is to use the first and second equations
to find an equation with a zero for the coefficient of x.
The second equation minus twice the first equation
yields 0x 1 3y 2 z 1 0 5 0.
Solving for z yields, z 5 3y. If we let y 5 t, then
z 5 3(t) 5 3t.
Substituting these into the first equation yields
x 2 (t) 1 (3t) 2 2 5 0 or x 5 2 2 2t. So the
equation of the line in parametric form is x 5 2 2 2t,
y 5 t, z 5 3t, tPR.
To check that this is correct, we substitute in the
solution to both initial equations
x 2 y 1 z 2 2 5 (2 2 2t) 2 (t) 1 (3t) 2 2
50
and
2x 1 y 1 z 2 4 5 2(2 2 2t) 1 (t) 1 (3t) 2 4
5 0.
Hence the line given by the parametric equation
above is the line of intersection for the planes.
27. The angle, u, between the plane and the line is
the complementary angle of the angle between the
direction vector of the line and the normal vector
for the plane. The direction vector of the line is
9-8
(1, 21, 0) and the normal vector for the plane is
(2, 0, 22).
0 (1, 21, 0) 0 5 "12 1 (21)2 1 02
5 "2
0 (2, 0, 22) 0 5 "22 1 02 1 (22)2 5 "8
(1, 21, 0) ? (2, 0, 22) 5 1(2) 2 1(0) 1 0(22) 5 2
So the angle between the normal vector and the
direction vector is cos21 A !22!8 B 5 60°. So
u 5 90 260° 5 30°.
> >
a?b
28. a. We have that cos (60°) 5 > > . Also
0a0 0b0 >
>
>
>
since a and b are unit vectors, 0 a 0 5 0 b 0 5 1 and
>
>
> >
a ? a 5 b ? b> 5 1, and moreover cos (60°) 5 12. So
>
> >
a?b
a?b5
5 12.
131
The dot >product is distributive,
so >
>
>
>
>
>
(6a 1 b ) ? (a 2 2b ) 5 6a ? (a 2 2b )
>
>
>
1 b ? (a 2 2b )
>
> >
>
5 6a ? a 1 6a ? (22b )
>
>
>
>
1 b ? a 1 b ? (22b )
> >
> >
> >
5 6a ? a> 2> 12a ? b 1 a ? b
2 2b ? b
1
1
5 6(1) 2 12a b 1 a b
2
2
2 2(1)
3
52
2 > >
x?y
b. We have that cos (60°) 5 0 x> 0 0 y> 0 . Also since
>
>
0 x 0 5 3, 0 y 0 5 4, and cos (60°) 5 12,
> >
> >
>
x ? y 5 12 (4)(3) 5 6. Also x ? x 5 0 x 0 2 5 9
> >
>2
and y ? y 5 0 y 0 5 16.
The dot product is distributive, so
>
>
>
>
>
>
>
(4x 2 y ) ? (2x 1 3y ) 5 4x ? (2x 1 3y )
>
>
>
2 y ? (2x 1 3y )
> >
> >
> >
5 8x ? x 1 12x ? y 2 2y ? x
> >
2 3y ? y
5 8(9) 1 12(6) 2 2(6)
2 3(16)
5 84
29. The origin, (0, 0, 0), and (21, 3, 1) are two
points on this line. So (21, 3, 1) is a direction vector
for this line and since the origin is on the line, a
>
possible vector equation is r 5 t(21, 3, 1), tPR.
(21, 3, 1) is a normal vector for the plane. So the
equation of the plane is 2x 1 3y 1 z 1 D 5 0.
Chapters 6–9: Cumulative Review
(21, 3, 1) is a point on the plane. Substitute the
coordinates to determine the value of D.
119111D50
D 5 211
The equation of the plane is 2x 1 3y 1 z 2 11 5 0.
30. The plane is the right bisector joining
P(21, 0, 1) and its image. The line connecting the
two points has a direction vector equal to that of the
normal vector for the plane. The normal vector for
the plane is (0, 1, 21). So the line connecting the
two points is (21, 0, 1) 1 t(0, 1, 21), tPR, or in
corresponding
parametric form is x 5 21, y 5 t, z 5 1 2 t, tPR.
The intersection of this line and the plane is the
bisector between P and its image. To find this point
we plug the parametric equation into the plane
equation and solve for t.
0x 1 y 2 z 5 0(21) 1 (t) 2 (1 2 t)
5 21 1 2t
So if y 2 z 5 0, then 21 1 2t 5 0, or t 5 12.
So the point of intersection is occurs at t 5 12, since
the origin point is P and the intersection occurs at
the midpoint of the line connecting P and its image,
the image point occurs at t 5 2 3 12 5 1. So the
image point is at x 5 21, y 5 1, z 5 1 2 (1) 5 0.
So the image point is (21, 1, 0).
31. a. Thinking of the motorboat’s velocity vector
(without the influence of the current) as starting
at the origin and pointing northward toward the
opposite side of the river, the motorboat has velocity
vector (0, 10) and the river current has velocity
vector (4, 0). So the resultant velocity vector of the
motorboat is
(0, 10) 1 (4, 0) 5 (4, 10)
To reach the other side of the river, the motorboat
needs to cover a vertical distance of 2 km. So the
hypotenuse of the right triangle formed by the
marina, the motorboat’s initial position, and the
motorboat’s arrival point on the opposite side of
the river is represented by the vector
1
4
(4, 10) 5 a , 2b
5
5
(We multiplied by 15 to create a vertical component
of 2 in the motorboat’s resultant velocity vector,
the distance needed to cross the river.) Since this
new vector has horizontal component equal to 45,
this means that the motorboat arrives 45 5 0.8 km
downstream from the marina.
Calculus and Vectors Solutions Manual
b. The motorboat is travelling at 10 km> h, and in
part a. we found that it will travel along the vector
( 45, 2) . The length of this vector is
4
4 2
` a , 2b ` 5 a b 1 22
5
Å 5
5 "4.64
So the motorboat travels a total of !4.64 km to
cross the river which, at 10 km> h, takes
"4.64 4 10 8 0.2 hours
5 12 minutes.
32. a. Answers may vary. For example:
A direction
vector for this line is
>
AB 5 (6, 3, 4) 2 (2, 21, 3)
5 (4, 4, 1)
So, since the point B(6, 3, 4) is on this line, the
vector equation of this line is
>
r 5 (6, 3, 4) 1 t(4, 4, 1), tPR.
The equivalent parametric form is
x 5 6 1 4t
y 5 3 1 4t
z 5 4 1 t, tPR.
b. The line found in part a. will lie in the plane
x 2 2y 1 4z 2 16 5 0 if and only if both points
A(2, 21, 3) and B(6, 3, 4) lie in this plane.
We verify this by substituting these points into the
equation of the plane, and checking for consistency.
For A:
2 2 2(21) 1 4(3) 2 16 5 0
For B:
6 2 2(3) 1 4(4) 2 16 5 0
Since both points lie on the plane, so does the line
found in part a.
33. The wind velocity vector is represented by (16, 0),
and the water current velocity vector is represented
by (0, 12). So the resultant of these two vectors is
(16, 0) 1 (0, 12) 5 (16, 12).
Thinking of this vector with tail at the origin and
head at point (16, 12), this vector forms a right
triangle with vertices at points (0, 0), (0, 12), and
(16, 12). Notice that
0 (16, 12) 0 5 "162 1 122
5 "400
5 20
This means that the sailboat is moving at a speed
of 20 km> h once we account for wind and water
velocities. Also the angle, u, this resultant vector
makes with the positive y-axis satisfies
9-9
cos u 5
12
20
u 5 cos 21 a
12
b
20
8 53.1°
So the sailboat is travelling in the direction
N 53.1° E, or equivalently E 36.9° N.
34. Think of the weight vector for the crane with tail
at the origin at head at (0, 2400) (we use one unit
for every kilogram of mass). We need to express this
weight vector as the sum of two vectors: one that is
parallel to the inclined plane and pointing down this
>
incline (call this vector x 5 (a, b)), and one that is
perpendicular to the inclined plane and pointing
>
toward the plane (call this vector y 5 (c, d)). The
>
angle between x and (0, 2400) is 60° and the angle
>
>
>
between y and (0, 2400) is 30°. Of course, x and y
are perpendicular. Using the formula for dot product,
we get
>
>
y ? (0, 2400) 5 0 y 0 0 (0, 2400) 0cos 30°
2400d 5 400a
"3
b"c 2 1 d 2
2
22d 5 "3 ? "c 2 1 d 2
4d 2 5 3(c 2 1 d 2 )
d 2 5 3c 2
So, since c is positive and d is negative (thinking of
the inclined plane as moving upward from left to
>
right as we look at it means that y points down and
d
to the right), this last equation means that c 5 2"3
>
So a vector in the same direction as y is (1, 2"3).
>
We can find the length of y by computing the scalar
projection of (0, 2400) on (1, 2 !3), which equals
(0, 2400) ? (1, 2"3)
400"3
5
2
0 (1, 2"3) 0
5 200"3
>
That is, 0 y 0 5 200"3. Now we can find the length
>
of x as well by using the fact that
>
>
0 x 0 2 1 0 y 0 2 5 0 (0, 2400) 0 2
>
0 x 0 2 1 (200"3)2 5 4002
>
0 x 0 5 "160 000 2 120 000
5 "40 000
5 200
9-10
So we get that
>
>
0 x 0 5 200 and 0 y 0 5 200"3. This means that the
component of the weight of the mass parallel to the
inclined plane is
>
9.8 3 0 x 0 5 9.8 3 200
5 1960 N,
and the component of the weight of the mass
perpendicular to the inclined plane is
>
9.8 3 0 y 0 5 9.8 3 200"3
8 3394.82 N.
35. a. True; all non-parallel pairs of lines intersect
in exactly one point in R2. However, this is not
the case for lines in R3 (skew lines provide a
counterexample).
b. True; all non-parallel pairs of planes intersect in a
line in R3.
c. True; the line x 5 y 5 z has direction vector
(1, 1, 1), which is not perpendicular to the normal
vector (1, 22, 2) to the plane x 2 2y 1 2z 5 k,
k any constant. Since these vectors are not
perpendicular, the line is not parallel to the plane,
and so they will intersect in exactly one point.
d. False; a direction vector for the line
z11
x
5 y 2 1 5 2 is (2, 1, 2). A direction vector
2
x21
y21
z11
for the line 24 5 22 5 22 is (24, 22, 22),
or (2, 1, 1) (which is parallel to (24, 22, 22)).
Since (2, 1, 2) and (2, 1, 1) are obviously not
parallel, these two lines are not parallel.
36. a. A direction vector for
y22
L1: x 5 2,
5z
3
is (0, 3, 1), and a direction vector for
z 1 14
L2: x 5 y 1 k 5
k
is (1, 1, k). But (0, 3, 1) is not a nonzero scalar
multiple of (1, 1, k) for any k since the first
component of (0, 3, 1) is 0. This means that the
direction vectors for L1 and L2 are never parallel,
which means that these lines are never parallel for
any k.
b. If L1 and L2 intersect, in particular their
x-coordinates will be equal at this intersection point.
But x 5 2 always in L1 so we get the equation
25y1k
y522k
Chapters 6–9: Cumulative Review
y22
Also, from L1 we know that z 5 3 , so substituting
this in for z in L2 we get
2k 5 z 1 14
y22
2k 5
1 14
3
3(2k 2 14) 5 y 2 2
y 5 6k 2 40
So since we already know that y 5 2 2 k, we
now get
2 2 k 5 6k 2 40
7k 5 42
k56
Calculus and Vectors Solutions Manual
So these two lines intersect when k 5 6. We have
already found that x 5 2 at this intersection point,
but now we know that
y 5 6k 2 40
5 6(6) 2 40
5 24
y22
z5
3
24 2 2
5
3
5 22
So the point of intersection of these two lines is
(2, 24, 22), and this occurs when k 5 6.
9-11
Vector Appendix
Gaussian Elimination, pp. 588–590
1. a. First write the system in matrix form (omitting
the variables and using only coefficients of each
equation). The coefficients of the unknowns are
entered in columns on the left side of a matrix with
a vertical line separating the coefficients from the
numbers on the right side.
1
x 1 2y 2 z 5 21
2 2x 1 3y 2 2z 5 21
3
3y 2 2z 5 23
3. Answers may vary. For example:
2
£0
3
1 6 0
22 1 † 0 §
1 0 1
1
£0
3
0.5
22
1
1 2 21 21
£ 21 3 22 † 21 §
0 3 22 23
1
£0
0
0.5
22
20.5
2x 2 z 5 1
2y 2 z 5 16
23x 1 y 5 10
1
£0
0
0.5
22
0
3 0
1 † 0§
1
29.25 1 2 (row 2) 1 row 3
4
2
£0
0
1
22
0
6 0 2 (row 1)
1 † 0§
237 4 4 (row 3)
b.
1
2
3
2 0 21 1
£ 0 2 21 † 16 §
23 1
0 10
c.
1
2
3
2x 2 y 2 z 5 22
x 2 y 1 4z 5 21
2x 2 y 5 13
2 21 21 22
£ 1 21
4 † 21 §
21 21
0 13
1
(row 1)
3 0
2
1 † 0§
0 1
3 0
1 † 0§
29 1 23 (row 1) 1 row 3
4. a. Answers may vary. For example:
21
0
1 2
0 21
2 0
≥
∞ ¥
1
3
1
2
22
2
4
3
3 0
` d
21 1
1
£0
6
0
21
29
21 22 21 (row 1)
2 † 0§
224
4 12 (row 3)
1
c
3
1
1.5 0
(row 1)
` d 2
21 1
1
£0
0
0
21
29
21 22
2 † 0§
218 16 26 (row 1) 1 row 3
c
1
0
1.5 0
` d
25.5 1 23 (row 1) 1 row 2
c
2
0
3 0 2 (row 1)
` d
25.5 1
1
£0
0
0
21
0
21 22
2 † 0§
236 16 29 (row 2) 1 row 3
2. Answers may vary. For example:
c
2
3
The last two matrices are both in row-echelon form
and are equivalent.
Calculus and Vectors Solutions Manual
1
0
21 22
b. £ 0 21
2 † 0§
0
0 236 16
1
236z 5 16 or z 5 2
4
9
2y 1 2z 5 0
8
4
2y 1 2a2 b 5 0 or y 5 2
9
9
x 2 z 5 22
22
4
x 2 a2 b 5 22 or x 5 2
9
9
22
8
4
x52 ,y52 ,z52
9
9
9
5. a. Each row of an augmented matrix corresponds
to an equation in a system. The numbers on the left
of the vertical line correspond to the coefficients
of the unknowns, while the number to the right of
the vertical line correspond to numerical answer to
the equation.
22 21
23 † 1 §
21
0
1
£2
2
22
0 21
0
b. £ 1 22
0 † 4§
0
1
2 23
22x 2 z 5 0
x 2 2y 5 4
y 1 2z 5 23
22
0
0 21
0
0
0 † 22 §
1
1
0
1 6
` d
25 15
25y 5 15 or y 5 23
22x 1 y 5 6
22x 2 3 5 6 or x 5 2
9
2
9
x 5 2 , y 5 23
2
2
21
2
0
1
11
3 †
0§
6 236
6z 5 236 or z 5 26
2y 1 3z 5 0
2y 1 3(26) 5 0 or y 5 9
2x 2 y 1 z 5 11
2x 2 (9) 1 (26) 5 11 or x 5 13
x 5 13, y 5 9, z 5 26
c. A solution does not exist to this system, because
the last row has no variables, but is still equal to a
non-zero number, which is not possible.
21
£ 0
0
3
1
0
1
2
22 †
1§
0 213
4 21 21 0
£ 0 21
0 † 4§
0
0 21 5
2z 5 5 or z 5 25
2y 5 4 or y 5 24
4x 2 y 2 z 5 0
4x 2 (24) 2 (25) 5 0 or x 5 2
2z 5 0
x 5 22
y1z50
6. a. This system can be solved using back
substitution.
c
2
£0
0
d. This system can be solved using back
substitution.
x 2 2y 5 21
2x 2 3y 5 1
2x 2 y 5 0
0
c. £ 1
0
b. This system can be solved using back
substitution.
9
4
9
x 5 2 , y 5 24, z 5 25
4
1 21 3 2
e. £ 0
0 0 † 0§
0
0 0 0
x 2 y 1 3z 5 2
Set z 5 t and y 5 s,
x 2 (s) 1 3(t) 5 2 or x 5 2 2 3t 1 s
x 5 2 2 3t 1 s, y 5 s, z 5 t, s, tPR
f. This system can be solved using back
substitution.
21
£ 0
0
0
1
0
0 24
2 † 4§
21
2
2z 5 2 or z 5 22
y 1 2z 5 4
y 1 2(22) 5 4 or y 5 8
2x 5 24 or x 5 4
x 5 4, y 5 8, z 5 22
Vector Appendix
7. a. This matrix is in row-echelon form, because it
satisfies both properties of a matrix in row-echelon
form.
1. All rows that consist entirely of zeros must
be written at the bottom of the matrix.
2. In any two successive rows not consisting
entirely of zeros, the first nonzero number in
the lower row must occur further to the right
that the first nonzero number in the row
directly above.
b. A solution does not exist to this system, because
the second row has no variables, but is still equal to
a nonzero number, which is not possible.
c. Answers may vary. For example:
21 1 1
3
£ 0 0 0 † 23 §
0 0 0
0
21 1 1 3
£ 22 2 2 † 3 § row 2 1 2(row 1)
0 0 0 0
21 1 1 3
£ 22 2 2 † 3 §
21 1 1 3 row 1 1 row 3
8. a. This matrix is not in row-echelon form.
Answers may vary.
21 0 1 3
£ 0 1 0 † 0§
0 1 2 1
21
£ 0
0
0
1
0
1 3
0 † 0§
2 1 row 3 2 row 2
b. This matrix is not in row-echelon form. Answers
may vary.
1 0
2 23
£ 3 1 24 † 2 §
0 0
3
6
1
£0
0
0
1
0
2 23
210 † 11 § row 2 2 3(row 1)
3
6
c. This matrix is not in row-echelon form. Answers
may vary. Switch row 2 and row 3 to obtain
row-echelon form.
21 2 1
0
£ 0 0 0 † 0 § Interchange row 2 & row 3
0 0 1 26
Calculus and Vectors Solutions Manual
£
21
0
0
2
0
0
1
0
1 † 26 §
0
0
d. This matrix is in row-echelon form, because it
satisfies both conditions of a matrix in row-echelon
form.
1. All rows that consist entirely of zeros must be
written at the bottom of the matrix.
2. If any two successive rows not consisting entirely
of zeros, the first nonzero number in the lower
row must occur further to the right that the first
nonzero number in the row directly above.
21 0 1 3
9. a. i. £ 0 1 0 † 0 §
0 0 2 1
1
2z 5 1 or z 5
2
y50
Using back substitution,
2x 1 z 5 3
5
1
2x 1 5 3 or x 5 2
2
2
5
1
x 5 2 , y 5 0, z 5
2
2
1 0
2 23
ii. £ 0 1 210 † 11 §
0 0
3
6
3z 5 6 or z 5 2
Using back substitution,
y 2 10z 5 11
y 2 10(2) 5 11 or y 5 31
x 1 2z 5 23
x 1 2(2) 5 23 or x 5 27
x 5 27, y 5 31, z 5 2
21
iii. £ 0
0
2
0
0
1
0
1 † 26 §
0
0
z 5 26
2x 1 2y 1 z 5 0
Set y 5 t and z 5 26,
2x 1 2(t) 1 (26) 5 0
x 5 2t 2 6
x 5 2t 2 6, y 5 t, z 5 26, tPR
3
1 24 1
0
1 2 † 23 §
iv. £ 0
0
0 0
0
y 1 2z 5 23
Set z 5 t,
y 1 2(t) 5 23 or y 5 23 2 2t
x 2 4y 1 z 5 0
x 2 4(23 2 2t) 1 t 5 0
x 5 212 2 9t
x 5 212 2 9t, y 5 23 2 2t, z 5 t, tPR
b. i. The solution is the point at which the three
planes meet.
ii. The solution is the point at which the three
planes meet.
iii. The solution is the line at which the three planes
meet. There is one parameter t.
iv. The solution is the line at which the three planes
meet. There is one parameter t.
10. a. 1 2x 1 y 1 z 5 9
2 x 2 2y 1 z 5 15
3 2x 2 y 2 z 5 212
21
£ 1
2
1
22
21
21
£ 0
0
1 1 9
21 2 † 24 § row 1 1 row 2
1 1 6 row 3 1 2(row 1)
£
21
0
0
1
9
1 † 15 §
21 212
1 1 9
21 2 † 24 §
0 3 30 row 2 1 row 3
3z 5 30 or z 5 10,
2y 1 2z 5 24
2y 1 2(10) 5 24 or y 5 24
2x 1 y 1 z 5 9
2x 1 (24) 1 (10) 5 9 or x 5 23
x 5 23, y 5 24, z 5 10
These three planes meet at the point (23, 24, 10).
b. 1
x1y1z50
2
2x 1 3y 1 z 5 0
3 23x 2 2y 2 4z 5 0
1
1
£ 2
3
23 22
4
1 0
1 † 0§
24 0
1
£0
0
1
1
1
1 0
21 † 0 § row 2 2 2(row 1)
21 0 row 3 1 3(row 1)
1 1
1 0
£ 0 1 21 † 0 §
0 0
0 0 row 2 1 row 3
y2z50
Set z 5 t,
y5t
x1y1z50
x1t1t50
x 5 22t
x 5 22t, y 5 t, z 5 t, tPR
These three planes meet at this line.
c. 1 x 2 y 1 3z 5 21
2 5x 1 y 2 3z 5 25
3 2x 1 y 2 3z 5 22
1 21
3 21
£5
1 23 † 25 §
2
1 23 22
1 21
3 21
£0
6 218 † 0 § row 2 2 5(row 1)
0
3
29
0 row 3 2 2(row 1)
1 21
3 21
£0
6 218 † 0 §
1
0
0
0
0 row 3 2 (row 2)
2
6y 2 18z 5 0
Set z 5 t,
6y 2 18t 5 0 or y 5 3t
x 2 y 1 3z 5 21
Set z 5 t and y 5 3t,
x 2 (3t) 1 3t 5 21 or x 5 21
x 5 21, y 5 3t, z 5 t, tPR
These three planes meet at this line.
d. 1 x 1 3y 1 4z 5 4
2 2x 1 3y 1 8z 5 24
3
x 2 3y 2 4z 5 24
1
£ 21
1
1
£0
0
3
3
23
3
6
26
4
4
8 † 24 §
24 24
4
4
12 † 0 § row 2 1 row 1
28 28 row 3 2 row 1
Vector Appendix
1
£0
0
3
6
0
4
4
12 † 0 §
4 28 row 3 1 row 2
4z 5 28 or z 5 22
6y 1 12z 5 0
6y 1 12(22) 5 0
y54
x 1 3y 1 4z 5 4
x 1 3(4) 1 4(22) 5 4
x50
x 5 0, y 5 4, z 5 22
The three planes meet at the point (0, 4, 22).
e. 1
2x 1 y 1 z 5 1
2 4x 1 2y 1 2z 5 2
3 22x 1 y 1 z 5 3
2
£ 4
22
2
£0
0
2
£0
0
1
2
1
1
0
2
1
2
0
1 1
2 † 2§
1 3
1
0
2
1
2
0
1
† 0 § row 2 2 2(row 1)
4 row 3 1 row 1
1
† 4 § Interchange row 2 & row 3
0
2y 1 2z 5 4
Set z 5 t,
2y 1 2t 5 4 or y 5 2 2 t
2x 1 y 1 z 5 1
2x 1 (2 2 t) 1 t 5 1
1
x52
2
1
x 5 2 , y 5 2 2 t, z 5 t, tPR
2
The three planes meet at this line.
f. 1 x 2 y 5 2500
2 2y 2 z 5 3500
3
x 2 z 5 2000
1 21
0 2500
£0
2 21 † 3500 §
1
0 21 2000
1 21
0 2500
£0
2 21 † 3500 §
0
1 21 2500 row 3 2 row 1
Calculus and Vectors Solutions Manual
1 21
£0
2
0
0
0 2500
21 † 3500 §
1
20.5
750 row 3 2 (row 2)
2
20.5z 5 750 or z 5 21500
2y 2 z 5 3500
2y 2 (21500) 5 3500 or y 5 1000
x 2 y 5 2500
x 2 (1000) 5 2500 or x 5 500
x 5 500, y 5 1000, z 5 21500
The three planes meet at the point
(500, 1000, 21500).
11. a 5 x 1 2y 2 z
b 5 x 2 y 1 2z
c 5 3x 1 3y 1 z
1
£1
3
2 21 a
21
2 † b§
3
1 c
1
£0
0
2 21
a
23
3 † 2a 1 b § row 2 2 row 1
23
4 23a 1 c row 3 2 3(row 1)
1
£0
0
2 21
a
23
3 †
2a 1 b §
0
1 22a 2 b 1 c row 3 2 row 2
z 5 22a 2 b 1 c
23y 1 3z 5 2a 1 b
23y 1 3(22a 2 b 1 c) 5 2a 1 b
25a 2 4b 1 3c
y5
3
x 1 2y 2 z 5 a
25a 2 4b 1 3c
x 1 2a
b 2 (22a 2 b 1 c) 5 a
3
7a 1 5b 2 3c
x5
3
25a 2 4b 1 3c
7a 1 5b 2 3c
x5
,y5
,
3
3
z 5 22a 2 b 1 c
12. First write an equation corresponding to each
point using the given equation y 5 ax 2 1 bx 1 c.
Then create a matrix corresponding to these three
equations, and solve for a, b, and c.
A(21, 27):27 5 a(21)2 1 b(21) 1 c
27 5 a 2 b 1 c
B(2, 20) : 20 5 a(2)2 1 b(2) 1 c
20 5 4a 1 2b 1 c
C(23, 25) : 25 5 a(23)2 1 b(23) 1 c
25 5 9a 2 3b 1 c
5
1 21 1 27
£4
2 1 † 20 §
9 23 1 25
1 21
1 27
£0
6 23 † 48 § row 2 2 4(row 1)
0
6 28 58 row 3 2 9(row 1)
1 21
1 27
£0
6 23 † 48 §
0
0 25 10 row 3 2 row 2
25c 5 10 or c 5 22
6b 2 3c 5 48
6b 2 3(22) 5 48 or b 5 7
a 2 b 1 c 5 27 or a 5 2
a 5 2, b 5 7, c 5 22
y 5 2x 2 1 7x 2 2
13. First change the variables in the equations to
easier variables to work with, and then use matrices
to solve for the unknowns. Once you have figured
out the unknowns you created, use back substitution
to solve for p, q, and r.
1 pq 2 2"q 1 3rq 5 8
2 2pq 2 "q 1 2rq 5 7
3 2pq 1 "q 1 2rq 5 4
Let x 5 pq, y 5 "q, and z 5 rq.
1 x 2 2y 1 3z 5 8
2 2x 2 y 1 2z 5 7
3 2x 1 y 1 2z 5 4
1 22 3 8
£ 2 21 2 † 7 §
21
1 2 4
1 22
3
8
£0
3 24 † 29 § row 2 2 2(row 1)
0 21
5 12 row 3 1 row 1
8
1 22
3
≥0
3 24 ∞ 29 ¥
1
11
0
0
9 row 3 1 (row 2)
3
3
11
27
z 5 9 or z 5
3
11
3y 2 4z 5 29
27
3y 2 4a b 5 29
11
3
y5
11
6
x 2 2y 1 3z 5 8
3
27
x 2 2a b 1 3a b 5 8
11
11
13
x5
11
13
3
27
x5 ,y5 ,z5
11
11
11
9
3
y5
5 "q or q 5
11
121
13
9
x5
5 pq 5 pa
b
11
121
143
p5
9
27
9
z5
5 rq 5 r a
b
11
121
r 5 33
9
143
p5
,q5
, r 5 33
9
121
a 1 1 a
14. £ 1 a 1 † a §
1 1 a a
1
£1
a
1
a
1
a a Interchange row 1 & row 3
1 † a§
1 a
1
1
£0 a 2 1
0 12a
a
a
1 2 a † 0 § row 2 2 row 1
1 2 a 2 a 2 a 2 row 3 2 a (row 1)
1
1
£0 a 2 1
0
0
a
a
12a
† 0 §
(2a 2 2)(a 2 1) a 2 a 2
row 2 1 row 3
Analyzing this matrix, you can tell that when a 5 1,
the matrix has a row of all zeros, which means that
this matrix has an infinite number of solutions. If
you substitute in a 5 22, the matrix has a row of
variables with the coefficient zero, but a non zero
number, which means this system has no solution.
Any other number that is substituted for a gives a
unique solution.
a. a 5 22
b. a 5 1
c. a 2 22 or a 2 1
Vector Appendix
Gauss-Jordan Method for Solving
Systems of Equations, pp. 594–595
1. a. c
3 1
` d
21 2
21
0
c
21
0 7 row 1 1 3 (row 2)
` d
0 21 2
c
1
0
0 27 21 (row 1)
`
d
1 22 21 (row 2)
1
b. £ 0
0
0
1
0
2 3
2 † 2§
21 0
1
£0
0
0
1
0
0 3 row 1 1 2 (row 3)
0 † 2 § row 2 1 2 (row 3)
21 0
2. The solution to each reduced row-echelon matrix
is the numbers corresponding to each leading 1.
a. (27, 22)
b. (3, 2, 0)
c. (1, 1, 0)
d. (8, 21, 24)
3. a. 1
x2y1z50
2
x 1 2y 2 z 5 8
3 2x 2 2y 1 z 5 211
1 21
1
0
£1
2 21 †
8§
2 22
1 211
1 21
1
0
£0
3 22 †
8 § row 2 2 row 1
0
0 21 211 row 3 2 2 (row 1)
1
0
21
0
6
1 row 1 1 row 2
2 † 21 §
1
0
1
£0
0
0
21
0
0
1 row 1 2 6 (row 3)
0 † 21 § row 2 2 2 (row 3)
1
0
1
£0
0
0
1
0
0 1
0 † 1 § 21 (row 2)
1 0
0
d. £ 1
0
0
0
1
21
4
2 † 0§
0 21
1
£0
0
0
1
0
2
0
0 † 21 § Rearrange rows
21
4
1
£0
0
0
1
0
0
8 row 1 1 2 (row 3)
0 † 21 §
1 24 21 (row 3)
Calculus and Vectors Solutions Manual
1
0
2
8 1
(row 2)
1 2
¥
3
3 3
0 21 211
0
1
1 4
2
c. £ 0 21 2 † 21 §
0
0 1
0
1
£0
0
21
≥0
1 0 0 3
£0 1 0 † 2§
0 0 1 0 21 (row 3)
1
0
0
1
0
0
1
8
row 1 1 row 2
3
3
2 ∞
8¥
2
3
3
21 211
1
0
0
≥0
1
0
0
≥
1
row 3
3
0 ∞ 10 ¥ row 2 2 2 row 3
3
1 11 21 (row 3)
21
row 1 1
The solution to this system of equations is x 5 21,
y 5 10, and z 5 11.
b. 1 3x 2 2y 1 z 5 6
2 x 2 3y 2 2z 5 226
3 2x 1 y 1 z 5 9
3 22
1
6
£ 1 23 22 † 226 §
21
1
1
9
£
1 23 22 226 Interchange rows 1 & 2
3 22
1 †
6§
21
1
1
9
1 23 22 226 row 2 2 3(row 1)
£0
7
7 † 84 § row 3 1 row 1
0 22 21 217
7
1 23 22 226
1
£0
1
1 † 12 § (row 2)
7
0 22 21 217
d.
2
3
1 21 23
3
£0
4
7 † 27 § row 2 2 2(row 1)
0 22 22
2 row 3 1 row 1
1 10 row 1 1 3 (row 2)
1 † 12 §
1 7
0
1
0
x 2 y 2 3z 5 3
2x 1 2y 1 z 5 21
2x 2 y 1 z 5 21
1 21 23
3
£ 2
2
1 † 21 §
21 21
1 21
1 23 22 226
£0
1
1 † 12 §
0
0
1
7 row 3 1 2(row 2)
1
£0
0
1
1 21 23
3
0
4
7 27
≥
∞
¥
1
3
3
row 3 1 (row 2)
0
0
2
2
2
2
1 0 0 3 row 1 2 row 3
£ 0 1 0 † 5 § row 2 2 row 3
0 0 1 7
The solution to this system of equations is x 5 3,
y 5 5, and z 5 7.
c. 1 2x 1 2y 1 5z 5 214
2
2x 1 z 5 25
3
y2z56
2
£ 21
0
2
0
1
5 214
1 † 25 §
21
6
1 21 23
3
£0
4
7 † 27 §
2
(row 3)
0
0
1 21
3
1 21 23
3
£0
4
0 † 0 § row 2 2 7(row 3)
0
0
1 21
21
£ 2
0
0
2
1
1 25 Interchange rows 1 & 2
5 † 214 §
21
6
1 21 23
3
1
£0
1
0 † 0 § (row 2)
4
0
0
1 21
21
£ 0
0
0
2
1
1 25
7 † 224 § row 2 1 2(row 1)
21
6
21
0
0
2
0
0
1 25
7 224
¥
∞
1
9
18
row 3 2 (row 2)
2
2
2
21
0
0
0
2
0
≥
£
1 25
7 † 224 §
2
1 24 2 (row 3)
9
0 0 21 row 1 2 row 3
2 0 † 4 § row 2 2 7(row 3)
0 1 24
21
£ 0
0
1
0
0
1
≥0
1
0∞
2¥
0
0
1 24
21 (row 1)
1
(row 2)
2
The solution to this system of equations is x 5 1,
y 5 2, and z 5 24.
8
1 0 23
3 row 1 1 row 2
£0 1
0 † 0§
0 0
1 21
1 0 0
0 row 1 1 3(row 3)
£0 1 0 † 0§
0 0 1 21
The solution to this system of equations is x 5 0,
y 5 0, and z 5 21.
e.
1
2
3
1
2
≥
1
2
1
x 1 9y 2 z 5 1
2
x 2 6y 1 z 5 26
2x 1 3y 2 z 5 27
9
26
3
21
∞
1
1 26
21 27
¥
1 26
1 26 Interchange rows 1 & 2
1
≥
9 21 ∞ 1 ¥
2
2
3 21 27
Vector Appendix
1 26
1 26
£ 1 18 22 † 2 § 2(row 2)
2
3 21 27
1 26
1 26
£ 0 24 23 † 8 § row 2 2 row 1
0 15 23
5 row 3 2 2(row 1)
1
1
24 row 1 1 (row 2)
1
0
4
4
≥ 0 24 23 ∞ 8 ¥
9
15
0 row 3 2
0
0 2
(row 2)
8
24
1
24
4
≥
∞
¥
0 224 23
8
8
0
0
1
0 2 (row 3)
9
1
0
1
0
≥0
2
0
0
1
0
≥0
2
0
0
1
0
≥0
1
0
0
1
0 24 row 1 2 (row 3)
4
2
3 ∞
¥ row 2 1 3(row 3)
3
0
1
0 24
2
0 ∞
¥ row 2 2 3(row 3)
3
0
1
0 24
1 1
0 ∞
¥ (row 2)
3 2
0
1
The solution to this system of equations is x 5 24,
y 5 13, and z 5 0.
f. 1 2x 1 3y 1 6z 5 3
2
x2y2z50
3 4x 1 3y 2 6z 5 2
2
3
6 3
£ 1 21 21 † 0 §
4
3 26 2
1 21 21 0 Interchange row 1 & row 2
£2
3
6 † 3§
4
3 26 2
1 21 21 0
£0
5
8 † 3 § row 2 2 2(row 1)
0
7 22 2 row 3 2 4(row 1)
Calculus and Vectors Solutions Manual
≥
1 21
0
5
0
0
21
3
8
¥
∞
11
7
66
2
row 3 2 (row 2)
2
5
5
5
0
5
3
5
8
0
0
1
1
0
0
≥0
5
0
0
0
1
1
≥0
0
∞
3
1
row 1 1 (row 2)
5
5
3¥
5
1
2 (row 3)
66
6
∞
1
3
row 1 2 (row 3)
2
5
5
¥ row 2 2 8(row 3)
3
1
6
1
2
1 1
≥ 0 1 0 ∞ 3 ¥ 5 (row 2)
1
0 0 1
6
The solution to this system of equations is x 5 12,
y 5 13, and z 5 16.
4. a. 1 2x 1 y 2 z 5 26
2 x 2 y 1 2z 5 9
3 2x 1 y 1 z 5 9
1
0
0
2
1 21 26
£ 1 21
2 † 9§
21
1
1
9
1 21
2
9 Interchange row 1 & row 2
£ 2
1 21 † 26 §
21
1
1
9
1 21
2
9
£0
3 25 † 224 § row 2 2 2(row 1)
0
0
3
18 row 3 1 row 1
1
1
1 row 1 1 (row 2)
3
3
≥
∞
¥
0 3 25 224
1
0 0
1
6
(row 3)
3
1
1 0 0 21 row 1 2 (row 3)
3
£ 0 3 0 † 6 § row 2 1 5(row 3)
0 0 1
6
1
0
9
1
£0
0
0 21
1
0 † 2 § (row 2)
3
1
6
0
1
0
The solution to this system of equations is x 5 21,
y 5 2, and z 5 6.
b. 1
x 2 y 1 z 5 230
2 22x 1 y 1 6z 5 90
3
2x 2 y 2 z 5 220
1 21
1 230
£ 22
1
6 † 90 §
2 21 21 220
1 21 1 230
£ 22
1 6 † 90 §
0
0 5
70 row 2 1 row 3
1
21
230
1
≥0
21
8
0
0
1
∞
30 ¥
row 2 1 2(row 1)
14
1
(row 3)
5
1
0 27 260 row 1 2 1(row 2)
£ 0 21
8 † 30 §
0
0
1
14
1
0
£ 0 21
0
0
1
£0
0
0
1
0
0
38
0 † 282 §
1
14
row 1 1 7(row 3)
row 2 2 8(row 3)
0 38
0 † 82 § 21(row 2)
1 14
The solution to this system of equations is x 5 38,
y 5 82, and z 5 14.
5. 1
x 1 y 1 z 5 21
2
x2y1z52
3 3x 2 y 1 3z 5 k
1
1
£ 1 21
3 21
1 21
1 † 2§
3
k
1
1
£ 0 22
0 24
1
21
0 †
3 § row 2 2 row 1
0 k 1 3 row 3 2 3(row 1)
1
1
£ 0 22
0
0
1
21
0 †
3§
0 k 2 3 row 3 2 2(row 2)
10
a. There is an infinite number of solutions for this
system when k 5 3, because this creates a zero
row.
b. The system will have no solution when k 2 3,
kPR because there will be a row of zeros equal to a
nonzero number, which is not possible.
c. The system cannot have a unique solution,
because the matrix cannot be put in reduced
row-echelon form.
6. a. Every homogeneous system has at least one
solution, because (0, 0, 0) satisfies each equation.
b. 1 2x 2 y 1 z 5 0
2
x1y1z50
3 5x 2 y 1 3z 5 0
2 21 1 0
£1
1 1 † 0§
5 21 3 0
1
£2
5
1 1 0 Interchange row 1 and row 2
21 1 † 0 §
21 3 0
1
1
1 0 row 2 2 2(row 1)
£ 0 23 21 † 0 § row 3 2 5(row 1)
0 26 22 0
1
1
1 0
£ 0 23 21 † 0 §
0
0
0 0 row 3 2 2(row 2)
2
1
0
row 1 1 (row 2)
3
3
1 ∞ ¥
1
≥
0 1
0
2 (row 2)
3
3
0 0 0 0
The reduced row-echelon form shows that the
intersection of these planes is a line that goes
through the point (0, 0, 0). x 5 2 23 t, y 5 2 13 t,
z 5 t, tPR
1
2
6
5
7. 1
2 1 5
x
y
z
6
2
3
12
2
2 1
52
x
y
z
3
6
2
23
3
1 1 5
x
y
z
6
1
0
Vector Appendix
1 1
1
Let the variables be x, y, and z
1
≥2
3
1
≥0
0
1
≥0
0
22
6
23
12
6
2
∞
5
6
2¥
23
6
5
22
6
6
1
1
0 ∞ ¥
3
4
12 216
3
3
0
6
2
1
1
0 ∞
¥
3
8
0 216 2
3
row 2 2 2(row 1)
row 3 2 3(row 1)
1
0
0
≥0
1
0
0
0
1
∞
6
1
row 1 1 (row 3)
16
2
1
¥
3
1
1
2 (row 3)
16
6
1
1
5 ,x52
x
2
1
1
5 ,y53
y
3
1
1
5 ,z56
z
6
(2, 3, 6)
row 1 1 2(row 2)
row 3 2 12(row 2)
Calculus and Vectors Solutions Manual
11