MATH 100, Section 110 (CSP)
Week 3: Marked Homework Solutions
2010 Sep 30
1. (a) By Theorem 4, p. 171, the function must at least be continuous to be differentiable, so
from Week 2: Marked Homework Assignment, question 1 (see Solutions) we must have C = 0.
(0)
To check differentiability at x = 0 (using C = 0), we need to check: Does limh→0 f (0+h)−f
=
h
f (h)
limh→0 h exist (using f (0) = C = 0)? Since the formula for f (h) depends on whether h < 0
or h > 0, we need to consider left- and right-hand limits. (Notice that part of the solution of
this problem is realizing that you have to consider left- and right-hand limits, even though
one-sided limits were not explicitly mentioned in the question.) For h < 0,
lim−
h→0
f (0 + h) − f (0)
Ah2 + Bh
= lim−
= lim− (Ah + B) = B.
h→0
h→0
h
h
√
3/2
On the other hand, for h > 0 we have limh→0+ f (h)
= limh→0+ h cos(1/h)
= limh→0+ h cos(1/h),
h
h
and we have
√
√
√
−1 ≤ cos(1/h) ≤ 1,
− h ≤ h cos(1/h) ≤ h
for h > 0,
√
√
and the limits of both − h and h are 0, as h → 0+ . Therefore by the Squeeze Theorem
for limits,
√
f (0 + h) − f (0)
lim+
= lim+ h cos(1/h) = 0.
h→0
h→0
h
For the limit of the difference quotient to exist, the left- and right-hand limits must agree,
so
B=0
is required. The derivative f 0 (0) = 0 is the common value of the left- and right-hand limits
of the difference quotient when B = 0. In summary, for f 0 (0) to exist, we need
B = 0,
C = 0,
A can be any real number.
2. [10] Make a sketch of the landscape, i.e., the graph of the function given. The shortest
shadow corresponds to the maximum effective height of the sun in midsummer. Assuming
the sun is at (0, 15) on the y-axis, the edge of the mountain’s shadow is defined by the line
passing through the sun and tangent to the mountain. We want to find where this line
intersects the plain behind the mountain. So we should find the equation of the line, then
its x-intercept (see the figure below). The region in shadows all year round is between the
point of tangency and the x-intercept of the line that forms the edge of the shadow of the
mountain.
First we find the x-coordinate of the point of tangency, call it x = a. The unknown point
of tangency is
(a, f (a)) = (a, −a2 + 17a − 66).
The slope of the tangent line is f 0 (a) = −2a + 17, so the equation of the tangent line passing
through (a, f (a)) with slope f 0 (a) is y = f (a) + f 0 (a)(x − a), or
y = −a2 + 17a − 66 + (−2a + 17)(x − a).
1
Now we use the fact that the tangent line must pass through x = 0, y = 15, which means
these values must satisfy the equation of the line:
15 = −a2 + 17a − 66 + (−2a + 17)(0 − a).
Now simplify, to get a2 = 81, which has solutions a = ±9. Rejecting a = −9 as irrelevant
for this problem (what does it correspond to?), we take
a = 9,
and the corresponding y-coordinate on the landscape is
f (9) = −92 + 17 · 9 − 66 = 6.
The slope of the line is
f 0 (9) = −2 · 9 + 17 = −1,
so the equation of the tangent line is
y = 6 − 1 · (x − 9),
y = −x + 15.
Setting y = 0 in the equation of this line gives the x-intercept: 0 = −x + 15,
x = 15.
Thus the region corresponding to
9 ≤ x ≤ 15
is in the shadows, all year round.
3. [3] By the Product Rule (p. 184) and Power Rule (p. 175)
d 3/4
x · g(x)
dx
= 43 x−1/4 · g(x) + x3/4 · g 0 (x)
√
3
4
g(x)
+
x3 g 0 (x)
= √
4
4 x
f 0 (x) =
then setting x = 16 we get
√
3
4
f 0 (16) = √
g(16) + 163 g 0 (16) =
4
4 16
2
3
8
· 8 + 8(−1) = −5.
4. [3] By the Product Rule, differentiating gives
f 0 (x) = 1 · ex + x · ex = (1 + x)ex ,
and differentiating again (and using the Product Rule again) we get
f 00 (x) = 1 · ex + (1 + x) · ex = (1 + 2x)ex .
5. [3] Using the Quotient Rule, we get
d
cot x =
dx
=
=
=
=
d
dx
h cos x i
sin x
(− sin x)(sin x) − (cos x)(cos x)
[sin x]2
−1
(sin x)2
2
1
−
sin x
− csc2 x.
dy
6. [3] The tangent line to the curve y = x − sin x has slope dx
= 1 − cos x, and the tangent
dy
line is horizontal if and only if dx = 0 i.e. when cos x = 1. Therefore x = 0, ±2π, ±4π, · · · or
x = 2nπ where n is any integer. The corresponding y-values are y = 2nπ + sin(2nπ) = 2nπ,
so the points on the curve where the tangent line is horizontal are (2nπ, 2nπ), where n is
any integer.
4
4
4
4
7. [3 each part] (a) lims→0 sins4 2s = lims→0 sins2s = lims→0 22 sins2s = 24 lims→0 sin2s2s =
24 · 14 = 16
(b) Let x = θ − π/2, so that x → 0 iff θ → π/2. Then limθ→π/2
limx→0 33 sinx3x cos13x = limx→0 3 sin3x3x cos13x = 3 · 1 · 11 = 3
sin 8x
(c) limx→0 tan
= limx→0
4x
8
1
·
1
·
·
1
=
2.
4
1
8x 4x sin 8x
8x 4x tan 4x
= limx→0
8x sin 8x 4x
4x 8x tan 4x
tan 3(θ−π/2)
θ−π/2
= limx→0
= limx→0
8 sin 8x 4x
4 8x sin 4x
h(1+cos h)
h sin h
h sin h 1+cos h
(d) limh→0 1−cos
= limh→0 1−cos
= limh→0 h sin1−cos
= limh→0
2h
h
h 1+cos h
h(1+cos h)
h
1
limh→0 sin h = limh→0 sin h · limh→0 (1 + cos h) = 1 · 2 = 2
tan 3x
x
=
cos 4x =
h sin h(1+cos h)
sin2 h
=
(e) limt→∞ (1/t) = 0, so by the substitution law of limits (p. 70) and the continuity of the sine
function (let x = 1/t; then t → ∞ iff x → 0+) we have limt→∞ sin(1/t) = limx→0+ sin(x) =
sin(0) = 0
3