MATH 151: Calculus 1, SET8
SOLUTIONS
[Belmonte, 2017]
3 Differentiation Rules
2. [233/8] If a ball is thrown vertically upward with a velocity
of 80 ft/s, then its height after t seconds is s = 80t − 16t 2 .
(a) What is the maximum height reached by the ball?
(b) What is the velocity of the ball when it is 96 ft above the
ground on its way up? On its way down?
3.7 Rates of Change in the
Natural and Social Sciences
(a) The ball reaches maximum height when its velocity is 0.
Solve v (t) = s0 (t) = 80 − 32t = 0 to obtain t = 52 = 2.5 s,
for which its height is s 52 = 100 ft.
(b) Solve s (t) = 96 to get t = 2 or t = 3 s, at which times
the velocities are v (2) = 16 ft/s and v (3) = −16 ft/s.
1. [233/4] A particle’s position is given by s = f (t) = t 2 e−t ,
t ≥ 0, where s is in feet and t is in seconds.
(a) Find the velocity v (t) = f 0 (t).
(b) What is the velocity after 1 second?
(c) When is the particle at rest?
3. [234/13] Consider a circle of radius r.
(d) When is the particle moving in the positive direction?
(a) Find the average rate of change of the area of a circle
with respect to its radius r as r changes from
(i) 2 to 3
(ii) 2 to 2.5
(iii) 2 to 2.1
(e) Find the total distance traveled in the first 6 seconds.
(f) [SKIP, or rotate the graph in part (h).]
(b) Find the instantaneous rate of change when r = 2.
(g) Find the acceleration a (t) = v0 (t) and then a (1).
(c) Show the rate of change of area of a circle with respect
to its radius (at any r) is equal to the circumference of
the circle.
(h) Graph position, velocity, and acceleration for 0 ≤ t ≤ 6.
(i) When is particle speeding up? What is it slowing down?
(a) The velocity is v (t) = f 0 (t) = 2t − t 2 e−t .
(b) At 1 second, we havev (1) = e−1 ft/s, about 0.37 ft/s.
(c) Solve v (t) = 2t − t 2 e−t = 0 to get t = 0 s or t = 2 s.
(d) Solve v (t) > 0 to obtain 0 < t < 2 seconds.
(e) Total distance is | f (2) − f (0)| + | f (6) − f (2)|,
which is 8e−2 − 36e−6 ≈ 0.99 ft.
(f) [See graph in part (h) below and rotate.]
(g) The acceleration is a (t) = v0 (t) = t 2 − 4t + 2 e−t .
Therefore, a (1) = −e−1 ≈ −0.37 ft/s2 .
(i) The particle is speeding
up when
a (t) > 0; i.e, for
√
√
t ∈ [0, 2 − 2) ∪ 2 + 2, ∞ or approximately
t ∈ [0, 0.59) ∪ (3.41, ∞) seconds.
Theparticle is slowing
down when a (t) < 0; i.e., for
√
√ t ∈ 2 − 2, 2 + 2 ≈ (0.59, 3, 41) seconds.
(h) Herewith the promised plot.
(a) The area of a circle of radius r is A (r) = πr2 . The desired
A (r) − A (2)
. Hence
average rate of change is DQ (r) =
r−2
DQ (3.0) ≈ 15.71, DQ (2.5) ≈ 14.14, DQ (2.1) ≈ 12.88.
(b) The instantaneous rate of change is A0 (2) = 2πr r=2 or
4π ≈ 12.57.
(c) We have A0 (r) = 2πr, the circumference of the circle.
4. [234/17] The mass of the part of a metal rod that lies between
its left end and a point x meters to the right is 3x2 kg. Find the
linear density (see Example 2) when x is (a) 1 m, (b) 2 m, and
(c) 3 m. Where is the density the highest? The lowest?
If mass f (x) = 3x2 , linear density is ρ (x) = d f /dx = 6x.
Hence ρ ({1, 2, 3}) = {6, 12, 18} in kg/m. The density is
lowest at 1 m and highest at 3 m.
SET8, 233/4
2
position
velocity
acceleration
1.5
5. [234/18] If a tank holds 5000 gallons of water, which drains
from the bottom of the tank in 40 minutes, then Torricelli’s
Law gives the volume V of water remaining in the tank after t
1 2
minutes as V = 5000 1 − 40
t , 0 ≤ t ≤ 40. Find the rate at
which water is draining from the tank after (a) 5 min, (b)
10 min, (c) 20 min, and (d) 40 min. At what time is the water
flowing out fastest? slowest? Summarize your findings.
1
The rate at which water drains is V 0 (t) = −250 1 − 40
t . So
V 0 ({5, 10, 20, 40}) = {−250, −218.75, −187.5, −125} in
gal/min. Water flows out fastest at t = 0 (start) and slowest at
t = 40 (end). As time goes by, the speed (absolute value of
rate of flow) at which waters flows out decreases linearly.
1
0.5
0
-0.5
-1
0
1
2
3
4
Time t (seconds)
5
6
1
6. [234/19] The quantity of charge Q in coulombs (C) that has
passed through a point in a wire up to time t (measured in
seconds) is given by Q (t) = t 3 − 2t 2 + 6t + 2. Find the
current Q0 (t) when (a) t = 0.5 s and (b) t = 1 s. The unit of
current is an ampere (1 A = 1 C/s). At what time is the
current lowest?
Q0 (t) = 3t 2 − 4t + 6.
9. [236/31] The cost (in dollars) for a company to produce x
pairs of jeans is C (x) = 2000 + 3x + 0.01x2 + 0.0002x3 .
(a) Find the marginal cost C 0 (x).
(b) Find C 0 (100) and explain its meaning.
(c) Compare C 0 (100) with the cost of manufacturing the
101st item.
Q0 (0.5) = 4.75 A
The current is
So
and
Q0 (1) = 5 A. Determining when the current is lowest is an
optimization (max/min) problem dealt with in Chapter 4.
That said, the current is a quadratic function. So we can take
a page off the high school playbook and complete the square.
2
Suffice it to say, Q0 (t) = 3 t − 32 + 14
3 . So the current is
2
lowest when t = 3 s. [Looking ahead, set the derivative of
g (t) = Q0 (t) to zero and solve: g0 (t) = 6t − 4 = 0, whence
t = 23 . Note g0 < 0 for t < 23 and g0 > 0 for t > 32 . So we have
an absolute minimum at t = 32 .]
1
3
x2 + 50
x + 3.
(a) The marginal cost is C0 (x) = 5000
0
(b) So C (100) = $11 is the marginal cost when producing
100 pairs of jeans.
(c) The marginal cost predicts the approximate cost to
produce the 101st pair of jeans. The actual cost to
do so is C (101) −C (100) ≈ $11.07.
10. [236/37] The gas law for an ideal gas at absolute temperature
T (in kelvins), pressure P (in atmospheres), and volume V
(in liters) is PV = nRT , where n is the number of moles of
the gas and R = 0.0821 is the gas constant. At a certain time,
P = 8 atm and is increasing at a rate of 0.10 atm/min while
V = 10 L and is decreasing at a rate of 0.15 L/min. Find the
rate of change of T with respect to time at that instant given
that n = 10 mol.
PV
with respect
Given constants n and R , differentiate T =
nR
0
0
P V + PV
to time to obtain T 0 =
. Plugging in the data yields
nR
dT
= T 0 ≈ −0.2436 K/min at said instant.
dt
SET8, 234/19
12
11
Current (amps)
10
9
8
7
6
5
4
0
0.5
1
Time (seconds)
1.5
2
7. [234/22] Some of the highest tides in the world occur in the
Bay of Fundy on the Atlantic Coast of Canada. At Hopewell
Cape the water depth at low tide is about 2.0 m and at high
tide it is about 12.0 m. The natural period of oscillation is a
little more than 12 hours and on June, 20, 2009, high tide
occurred at 6:45 am. This helps explain the following model
for the water depth D (in meters) as a function of time t (in
hours after midnight) on that day.
D (t) = 7 + 5 cos (0.503 (t − 6.75))
How fast was the tide rising (or falling) at these times?
(a) 3:00 am
(b) 6:00 am
(c) 9:00 am
(d) Noon
503
503
13581
0
The derivative is D (t) = − 200 sin 1000 t − 4000 using
exact rationals. Accordingly, the rates at which the tide was
changing at these times is D0 ({3, 6, 9, 12}) or approximately
{−2.39, 0.93, −2.28, −1.21} in meters per hour.
8. [235/25] In Example 6 we considered a bacteria population
that doubles every hour. Suppose that another population of
bacteria triples every hour and starts with 400 bacteria. Find
an expression for the number n of bacteria after t hours and
use it to estimate the rate of growth of the bacteria population
after 2.5 hours.
In a manner similar to Example 6, n (t) = 400 (3t ). Thus
n0 (t) = 400 (3t ) ln 3. Hence n0 (2.5) ≈ 6850 bacteria per hour.
2