18.01 (Fall 2014) Solutions to Problem Set 1
Part II
1. solution
a) For any x, we have E(x) + O(x) = 0. By the condition that E(x) is even and
O(x) is odd, we have 0 = E(−x) + O(−x) = E(x) − O(x). Using these 2
equations, we can solve for E(x) and O(x) and get E(x) = O(x) = 0 for any x.
b) From part a), since A sin(ax) is odd and B cos(bx) is even, we deduce that
A sin(ax) = 0 and B cos(bx) = 0. Putting x = 0 into the second equation, we
have B = 0. For the equation A sin(ax) = 0, either A = 0 or sin(ax) = 0 for any
x. Therefore, we can get either A = 0 or a = 0.
2. solution
a) We have V = π9 h3 . Thus, h(V ) =
have
q
3
9
πV
1
3
. From the definition of derivative, we
h(V + 4V ) − h(V )
h0 (V ) = lim
4V →0
4V
r
1
1
(V + 4V ) 3 − V 3
3 9
lim
=
π 4V →0
4V
r
1
2
1
1
2
1
((V + 4V ) 3 − V 3 )((V + 4V ) 3 + (V + 4V ) 3 V 3 + V 3 )
3 9
=
lim
2
1
1
2
π 4V →0
4V ((V + 4V ) 3 + (V + 4V ) 3 V 3 + V 3 )
r
(V + 4V ) − V
3 9
lim
=
π 4V →0 4V ((V + 4V ) 32 + (V + 4V ) 13 V 13 + V 23 )
r
1
3 9
=
lim
2
π 4V →0 ((V + 4V ) 3 + (V + 4V ) 13 V 13 + V 23 )
r
r
1
1 −2
3 9
3
=
V 3
2 =
π 3V 3
3π
b) We first note that V0 = π9 h30 = π9 . Using 4h = h0 (V0 )4V , we have
4h = h0 (V0 )4V =
3
4V
π
c) We substitute 4V = 0.1 into the previous equation. We find that 4h =
0.02. Therefore, the tool is not good enough.
0.3
π
>
d) We have V 0 (h) = π3 h2 . Therefore, using the corresponding approximate error
formula, we have 4V = V 0 (h0 )4h = π3 4h. This is exactly the same equation
as in b). Therefore, it will not change the answer in c).
1
3. solution
Section 3.1: 14 Both curves must pass through (3, 3). We can get the equations
3 = 9 + 3a + b
3 = 3c − 9
Thus, c = 4. Now, both curves must have the same slope when x = 3. Differentiating
both equations with respect to x yields
y 0 = 2x + a
y 0 = c − 2x
Plugging c = 4, x = 3 into the above equations and setting the equations equal to
each other, we find that a = −8. Solving for b, we find that b = 18.
Section 3.1: 18 The slope of the curve at (x0 , x30 ) is given by y 0 (x0 ) = 3x20 . Therefore, the tangent line at (x0 , x30 ) is given by y − x30 = 3x20 (x − x0 ). Plugging in
(x, y) = (0, 2) yields 2 − x30 = −3x30 . We have 2 = −2x30 , x0 = −1. The desired
tangent is given by y + 1 = 3(x + 1) or y = 3x + 2.
Section 3.1: 21
1 2
a) The equation of the parabola is y = 4p
x . Differentiation with respect to x
x
0
yields y = 2p . The tangent at (x0 , y0 ) is given by y − y0 = x2p0 (x − x0 ). Plugging
x2
in x = 0, we find that the y-intercept is y = − 2p0 + y0 = −2y0 + y0 = −y0 .
b) The distance from (0, p) to (0, −y0 ) is p + y0 . The distance from (0, p) to (x0 , y0 )
is
q
q
x20 + (y0 − p)2 = 4py0 + y02 − 2py0 + p2
q
= y02 + 2py0 + p2 = y0 + p
Therefore, the triangle with these vertices is isosceles.
c) We name following points: X = (x0 , y0 ), F = (0, p), Z = (0, −y0 ). Any isosceles
triangle has the same base angles. The angle of incidence ∠F XZ = ∠F ZX is
equal to the angle between the tangent and the vertical line. Since the angle of
incidence is equal to the angle of reflection, after reflection, the ray should be
vertical, that is, point upward.
4. solution
Section 2.5: 19f )
sin2 x
1
sin x
sin x
1
1
= ( lim
)( lim
)= ·1·1=
x→0
x→0
x→0 3x2
3
x
x
3
3
lim
2
Section 2.5: 19g)
lim
x→0
2
sin 2x
3x
2
2
sin 2x
= ( lim
)( lim
)= ·1·1=
x→0 sin 3x
sin 3x
3 x→0 2x
3
3
Section 2.5: 20a)
√
sin x
1
1
sin x
lim √ = ( lim
)( lim x) = · 1 · 0 = 0
x→0 3 x
x→0
3 x→0 x
3
Section 2.5: 20b)
sin2 x
sin x
= ( lim
)( lim sin x) = 1 · 0 = 0
x→0
x→0
x→0
x
x
lim
Section 2.5: 20g)
3x2
4x
3x2 + 4x
= lim
+ lim
x→0 sin 2x
x→0 sin 2x
x→0 sin 2x
3
2x
2x
= ( lim
)( lim x) + 2 lim
x→0 sin 2x
2 x→0 sin 2x x→0
3
= ·1·0+2·1=2
2
lim
Section 2.5: 22a) Let f (θ) =
θ
1
0.1
0.01
0.001
1−cos θ
θ2
f (θ)
0.45969769
0.49958347
0.49999583
0.49999996
We can make a conjecture that limθ→0
1−cos θ
θ2
= 12 .
Section 2.5: 22b) Using the trig identity cos θ = 1 − 2 sin2 ( 2θ ), we find that
2 sin2 ( 2θ )
1 − cos θ
=
lim
θ→0
θ→0
θ2
θ2
sin( θ )
sin( θ )
1
1
1
= (lim θ 2 )(lim θ 2 ) = · 1 · 1 = .
θ→0
2 θ→0 2
2
2
2
lim
5. solution
a)
d −1
d 1
x =
( )=
dx
dx x
d
dx 1
3
d
· x − dx
x·1
−1
= 2 = −x−2
2
x
x
d −n
d −1
x
= −x−2 . Now, suppose we know that dx
x
=
b) From a), we have dx
−(n+1)
−(n+1)
−n
−1
−nx
. For the case n + 1, we have x
= x x . Using the product
rule, we have
d −(n+1)
d −n −1
d −n −1
d −1
x
=
(x x ) =
x · x + x−n ·
x
dx
dx
dx
dx
= −nx−(n+1) · x−1 + x−n · (−x−2 ) = −nx−(n+2) − x−(n+2)
= −(n + 1)x−(n+2)
Therefore, by the method of induction, the formula
for all positive integers n.
d −n
dx x
= −nx−(n+1) holds
d
d n
c) When n = 1, we have dx
x = 1. Now, suppose we know that dx
x = nxn−1 . For
n+1
n
the case n + 1, we have x
= x · x. Using the product rule, we have
d n
d n
d
d n+1
x
=
(x · x) =
x · x + xn ·
x
dx
dx
dx
dx
= nxn−1 · x + xn · 1 = nxn + xn = (n + 1)xn .
Therefore, by the method of induction, the formula
positive integers n.
4
d n
dx x
= nxn−1 holds for all