APPM 1340
1.
Final Exam Solutions
Fall 2014
(a) (10 points) Find dy/dx given y sin x = x + y 2 .
Solution:
y sin x = x + y 2
dy
dy
y cos x + sin x
= 1 + 2y
dx
dx
dy
dy
sin x
− 2y
= 1 − y cos x
dx
dx
dy
= 1 − y cos x
(sin x − 2y)
dx
dy
1 − y cos x
=
dx
sin x − 2y
(b) (16 points) Let f (x) = csc2 (2x) and g(x) = cot2 (2x).
i. Find an equation for the line tangent to y = f (x) at x = π4 .
ii. Find an equation for the line tangent to y = g(x) at x = π4 .
Solution:
i. f 0 (x) = 2 csc(2x)(−2 csc(2x) cot(2x)) = −4 csc2 (2x) cot(2x)
f 0 ( π4 ) = −4 csc2 ( π2 ) cot( π2 ) = −4(1)(0) = 0
f ( π4 ) = csc2 ( π2 ) = 1
Since the slope f 0 ( π4 ) = 0, the tangent line is y = f ( π4 ) or y = 1 .
ii. g 0 (x) = 2 cot(2x)(−2 csc2 (2x)) = −4 cot(2x) csc2 (2x) = f 0 (x)
g 0 ( π4 ) = f 0 ( π4 ) = 0
g( π4 ) = cot2 ( π2 ) = 0
Since the slope g 0 ( π4 ) = 0, the tangent line is y = g( π4 ) or y = 0 .
2. (22 points) Let f (x) = x2 − 4.
(a) Sketch a graph of y = f (x).
(b) What is the domain and range of f ?
(c) Find the derivative of f at x = 1.
(d) Sketch a graph of y = h(x) = −(x − 2)2 + k for any constant k.
(e) Suppose there is a function g(x) such that h(x) ≤ g(x) ≤ f (x) for all x, where f and h are the
functions defined above. The Squeeze Theorem can be used to find lim g(x) for what values of
x→a
a and k? Justify your answer using appropriate limits.
Solution:
(a)
y
f HxL
-2
x
2
hHxL
(b) The domain of f is R and the range is [0, ∞) .
(c) Since f (x) = −(x2 − 4) at x = 1, f 0 (x) = −2x and f 0 (1) = −2 .
(d) See the graph of y = h(x) above with k = 0.
(e) The functions f (x) and h(x) meet if a = 2 and k = 0 . Then by the Squeeze Theorem
lim x2 − 4 = lim −(x − 2)2 = 0 ⇒ lim g(x) = 0.
x→2
x→2
x→2
3. (18 points) Consider the function f defined below. Is f continuous at x = 0? State the definition of
continuity and use it to justify your answer.

1


x<0

2 − 3 − x ,
f (x) = 5/3,
x=0



cot(3x) sin(5x), x > 0
Solution: f is continuous at x = 0 if lim f (x) = lim f (x) = f (0). First evaluate the limit as
x→0−
x → 0 from the left.
lim f (x) = lim 2 −
x→0−
x→0−
1
3−x
x→0+
=2−
1
5
=
3−0
3
sin θ
= 1 for the limit as x → 0 from the right.
θ→0 θ
Now use the theorem lim
cos(3x) sin(5x)
·
sin(3x)
1
3x sin(5x) 5x
·
·
3x
1
5x
sin(5x) 5x
5
5
·
=1·1·1· =
5x
3x
3
3
lim f (x) = lim cot(3x) sin(5x) = lim
x→0+
x→0+
x→0+
1
·
sin(3x)
3x
= lim cos(3x) ·
·
+
sin(3x)
x→0
= lim cos(3x) ·
x→0+
Since lim f (x) = lim f (x) = f (0) = 5/3, f is continuous at x = 0.
x→0−
4.
x→0+
(a) (10 points)
q Let L(x) equal the linearization of
value of
3
√
3
x + 1 at x = 0. Use L(x) to approximate the
9
10 .
Solution:
1
f (x) = (x + 1)1/3 , f 0 (x) = (x + 1)−2/3
3
f (0) = 1, f 0 (0) =
1
3
1
L(x) = f (0) + f 0 (0)(x − 0) = 1 + x
3
r
9
1
1
1
29
1
3
≈L −
=1+
−
=
=f −
10
10
10
3
10
30
t2
1
+ 2t + .
10
6
i. Use the definition of derivative to find H 0 (t).
ii. Suppose the length of Rapunzel’s hair after t years is H(t) feet. By how much does her hair
grow during the week after her 10th birthday? Use differentials to approximate the answer.
(You may assume that there are 52 weeks in a year.)
(b) (16 points) Let H(t) =
Solution:
i.
H(t + h) − H(t)
h→0
h
1
1 2
(t + h)2 + 2(t + h) + 16 − ( 10
t + 2t + 16)
= lim 10
h→0
h
H 0 (t) = lim
2
2
1
1 2
t
t
2t + 2h − 10
−
2t − 16
10 + 5 th + 10 h + = lim
h
h→0
= lim
h→0
h
t
+
+2
5 10
=
t
+2
5
ii. Given t = 10 and dt = 1/52, find dH.
dH
t
= + 2 ⇒ dH =
dt
5
t
10
1
4
1
+ 2 dt =
+2 ·
=
=
ft
5
5
52
52
13
√
5. (16 points) Let g(x) = x − 4 x, x ≥ 0.
(a) Show that g satisfies the hypotheses of Rolle’s Theorem on the interval [0, k] for some value of
k. What must k equal? What conclusion can be made about g?
(b) Use the value of k found in part (a) to find the absolute maximum and minimum values of g on
[0, k].
Solution:
√
(a) Since g 0 (x) = 1 − 2/ x, g is continuous for x ≥ 0 and differentiable for x > 0, satisfying two
of the hypotheses for Rolle’s Theorem. The third hypothesis can be met if g(k) = g(0) = 0.
Solve g(x) = 0 for k.
√
√ √
g(x) = 0 ⇒ x − 4 x = 0 ⇒ x x − 4 = 0 ⇒ x = 0, 16
Since f (0) = f (16), the hypotheses of Rolle’s Theorem are met if k = 16 . We can conclude
that there is a c in (0, 16) such that g 0 (c) = 0.
√
(b) First find the critical numbers of g. The derivative g 0 = 1 − 2/ x is undefined at the endpoint
x = 0 and equals 0 if
√
2
1 − √ = 0 ⇒ x = 2 ⇒ x = 4.
x
Check the critical points and endpoints for absolute extrema. There is an absolute minimum at
f (4) = −4 and absolute maxima at f (0) = 0 and f (16) = 0 .
6. (12 points) An object hanging from a spring is stretched 6 units beyond its rest position and released
at time t = 0 to bob up and down. Its position at time t is s(t) = 6 cos t, 0 ≤ t ≤ 3π.
(a)
(b)
(c)
(d)
Find the velocity at time t.
When is the object moving downward (in a positive direction)?
When is the acceleration negative?
When does the object attain maximum speed?
Solution:
(a) v(t) = s0 (t) = −6 sin t
(b) The object is moving in a positive direction when v(t) > 0 on (π, 2π) .
(c) a(t) = v 0 (t) = −6 cos t. The acceleration is negative on the intervals (0, π/2) and (3π/2, 5π/2) .
(d) The object reaches maximum speed at t = π/2, 3π/2, 5π/2 .
7. (14 points) A baseball diamond is√a square with side length 90 ft. A batter hits the ball and runs
toward first base with a speed of 10 5 ft/s. When he is halfway to first base, his distance from second
base is changing at a rate of −k ft/s. Find integer k.
Solution:
Let x represent the distance between the batter and first base.
Let s represent the distance between
√ the batter and second base.
We are given that dx/dt = −10 5 ft/s. Find ds/dt = −k when x = 45 ft.
90
s
x
Use the Pythagorean Theorem. At this moment
p
p
p
√
s = x2 + 902 = 452 + 902 = 452 (12 + 22 ) = 45 5.
x2 + 902 = s2
dx
ds
2x
+ 0 = 2s
dt
dt
√ √
−10
5 =
45
5(−k)
45
k = 10
8. (16 points)
(a) Sketch below the graph of a function f (x) that satisfies all of the following conditions.
No explanation is necessary.
lim f (x) = ∞
f is odd
x→−2
x = 3 is a root of f
lim f (x) = −3
x→∞
(b) Sketch below the graph of a function g(x) that satisfies all of the following conditions.
No explanation is necessary.
g(0) and g 0 (2) do not exist
g 0 (−1) =
− 21
lim g(x) = −1
x→0
lim g(x) = g(2)
x→2
Solution: Here are two possible solutions.
y ‡ f HxL
y ‡ gHxL
y
y
3
-3-2
1
2 3
-3
x
-2
-1
1
-1
-2
2
3
x