MATH 2250 - 23228 MIDTERM EXAM 2
SPRING 2012 - MOON
Write your answer neatly and show intermediate steps.
Calculator, computer and any electronic devices are not allowed.
(1) (1pt) Write any suggestion for improving this class. (For instance, give written
homework twice in a week, give more examples in class, the tempo of this class
is too fast, ...)
dy
of
dx
y = (x2 + 1)(3x − 2).
(2) (4pts) Compute the derivative
Solution:
dy
d
d
= (x2 + 1) (3x − 2) + (3x − 2) (x2 + 1)
dx
dx
dx
2
= (x + 1) · 3 + (3x − 2) · 2x
= 9x2 − 4x + 3
d
d
• Writing step (x2 + 1) (3x − 2) + (3x − 2) (x2 + 1): 2pts.
dx
dx
• Computing the answer (x2 + 1) · 3 + (3x − 2) · 2x: 3pts.
• Simplifying and getting answer 9x2 − 4x + 3: 4pts.
Date: March 1, 2012.
1
MATH 2250 - 23228 MIDTERM EXAM 2
(3) (5pts) Compute the derivative
2
dy
of
dx
3x − 1
y=
.
2x + 5
Solution:
d
d
(2x + 5) (3x − 1) − (3x − 1) (2x + 5)
dy
dx
dx
=
dx
(2x + 5)2
(2x + 5) · 3 − (3x − 1) · 2
17
=
=
(2x + 5)2
(2x + 5)2
d
d
(2x + 5) (3x − 1) − (3x − 1) (2x + 5)
dx
dx
• Writing step
: 3pts.
(2x + 5)2
(2x + 5) · 3 − (3x − 1) · 2
: 4pts.
• Computing
(2x + 5)2
17
• Simplifying and getting
: 5pts.
(2x + 5)2
(4) (5pts) Compute the derivative
Solution:
y=
dy
of
dx
√
3
y = tan x.
√
1
3
tan x = (tan x) 3 .
1
u = tan x ⇒ y = u 3 .
2
1
dy du
1 2
dy
=
·
= u− 3 sec2 x = (tan x)− 3 sec2 x.
dx
du dx
3
3
• Finding appropriate intermediate variable u = tan x and express y as a
1
function with respect u, y = u 3 : 2pts.
1 2
• Applying chain rule and obtaining u− 3 sec2 x: 4pts.
3
1
− 23
• Getting correct answer (tan x) sec2 x: 5pts.
3
MATH 2250 - 23228 MIDTERM EXAM 2
(5) (5pts) By using logarithmic differentiation, compute the derivative
y=
3
dy
of
dx
(3x − 1)5 (x + 4)
.
(x2 + 3)4
Solution:
(3x − 1)5 (x + 4)
= ln(3x − 1)5 + ln(x + 4) − ln(x2 + 3)4
ln y = ln
(x2 + 3)4
= 5 ln(3x − 1) + ln(x + 4) − 4 ln(x2 + 3).
d
d
ln y =
5 ln(3x − 1) + ln(x + 4) − 4 ln(x2 + 3) .
dx
dx
1 dy
3
1
2x
15
1
8x
=5·
+
−4· 2
=
+
− 2
.
y dx
3x − 1 x + 4
x +3
3x − 1 x + 4 x + 3
dy
15
1
8x
1
8x
(3x − 1)5 (x + 4)
15
=y
+
−
+
−
=
.
dx
3x − 1 x + 4 x2 + 3
(x2 + 3)4
3x − 1 x + 4 x2 + 3
• Computing logarithm of right side 5 ln(3x − 1) + ln(x + 4) − 4 ln(x2 + 3):
2pts.
1
8x
(3x − 1)5 (x + 4)
15
+
−
• Getting correct answer
by com(x2 + 3)4
3x − 1 x + 4 x2 + 3
15
1
8x
1 dy
=
+
− 2
: 5pts.
puting
y dx
3x − 1 x + 4 x + 3
dy
of
dx
y = ln(sin(x2 + 1)).
(6) (5pts) Compute the derivative
Solution:
u = sin(x2 + 1) ⇒ y = ln u.
dy
dy du
1 du
1
du
=
·
=
=
.
2
dx
du dx
u dx
sin(x + 1) dx
v = x2 + 1 ⇒ u = sin v.
du dv
du
=
·
= (cos v) · 2x = (cos(x2 + 1)) · 2x.
dx
dv dx
dy
1
du
1
cos(x2 + 1)
2
=
=
cos(x
+
1)(2x)
=
2x
.
dx
sin(x2 + 1) dx
sin(x2 + 1)
sin(x2 + 1)
dy
1
du
• Computing
=
: 3pts.
2
dx
sin(x + 1) dx
du
• Computing
= (cos(x2 + 1)) · 2x: + 2pts.
dx
MATH 2250 - 23228 MIDTERM EXAM 2
4
(7) (5pts) Find the tangent line of the curve
x2 + 2y 2 = 6y
at (2, 1).
Solution:
dy
dy
=6 .
dx
dx
dy
dy
−2x
(4y − 6)
= −2x ⇒
=
.
dx
dx
4y − 6
dy −2 · 2
= 2.
x=2 =
dx y=1 4 · 1 − 6
The equation of tangent line: y = 2(x − 2) + 1 = 2x − 3.
dy
−2x
=
: 2pts.
• Computing
dx 4y − 6
dy • Computing
= 2: 3pts.
dx x=2
y=1
• Finding the equation of tangent line: 5pts.
x2 + 2y 2 = 6y ⇒ 2x + 4y
(8) (5pts) For y = f (x) = ex +x3 , let f −1 (y) be the inverse function of f (x). Compute
0
f −1 (e + 1).
Solution:
f −1 (e + 1) = x ⇔ f (x) = e + 1 ⇒ ex + x3 = e + 1 ⇒ x = 1.
f 0 (x) = ex + 3x2 ⇒ f 0 (1) = e + 3.
By the inverse function theorem,
1
1
0
f −1 (e + 1) = 0
=
.
f (1)
e+3
• Writing precise statement of inverse function theorem: + 2pts.
• Finding x such that f (x) = e + 1: + 2pts.
1
0
• Finding f 0 (1) = e + 3 and computing f −1 (e + 1) =
: + 1pt.
e+3
MATH 2250 - 23228 MIDTERM EXAM 2
5
(9) The following figure is the graph of velocity v(t) of a car moving along a straight
road. Answer the following questions. You don’t need to write the reason.
(a) (2pts) When does the car move forward?
(0, 2) and (7, 9).
(b) (2pts) When does the car reverse direction?
At 2 and 7.
(c) (2pts) When is the car moving at a constant speed?
(3, 5).
(d) (2pts) When does the car move at its greatest speed?
At 6.
(e) (2pts) When is the car’s acceleration negative?
(1, 3), (5, 6) and (8, 9).
MATH 2250 - 23228 MIDTERM EXAM 2
(10) (5pts) By using linearization, compute an approximation value of
√
1.04.
Solution:
f (x) =
f (1) =
√
1
x = x2 ⇒
√
1 = 1,
√
1.04 = f (1.04).
1
1 1
f 0 (x) = x− 2 ⇒ f 0 (0) = .
2
2
So the linearization L(x) at 1 is
1
L(x) = f (1) + f 0 (1)(x − 1) = 1 + (x − 1).
2
Therefore the linear approximation of f (1.04) is
1
L(1.04) = 1 + (1.04 − 1) = 1.02.
2
•
•
•
•
Introducing appropriate function f (x): + 1pt.
Knowing the formula of linearization: + 1pt.
Computing linearization L(x) at 1: + 2pts.
Calculating approximation value 1.02: + 1pt.
6
MATH 2250 - 23228 MIDTERM EXAM 2
7
(11) (10pts) You are recording a race from a stand 100 ft from the track, following a
car that is moving at 200 ft/sec. How fast will your camera angle θ be changing
π
when θ = ?
4
1
1
π
π
π
Note that sin = √ , cos = √ , tan = 1.
4
4
4
2
2
Solution:
x: the length of the base of above right triangle.
θ: the camera angle.
dθ Goal:
=?
dt θ= π4
Known facts:
x
.
• tan θ =
100
dx
•
= 200.
dt
π
x
θ= ⇒1=
⇒ x = 100.
4
100
x
dθ
1 dx
dθ
1 dx
tan θ =
⇒ sec2 θ =
⇒
= cos2 θ ·
.
100
dt
100 dt
dt
100 dt
2
dθ 1
1
·
· 200 = 1(rad/sec).
π = √
dt θ= 4
100
2
dθ • Writing precise goal π : + 2pts.
dt θ= 4
x
dx
• Writing related equations tan θ =
and
= 200: + 2pts for each.
100
dt
dθ
1 dx
• Computing relation
= cos2 θ ·
: + 2pts.
dt
100 dt
• Getting correct answer 1: + 2pts.