Rates of Change in Applied Contexts

Marginal analysis

Example 1

A company's cost of making

x

skateboards is

C(x) = 500 + 30x - 0.02x^2

dollars, and its revenue from selling them is

R(x) = 45x - 0.05x^2

dollars, for

0 \le x \le 400

. (a) Find

C'(200)

, interpret it, and compare it with the actual cost of the

201

st skateboard. (b) Find the marginal profit at

x = 200

and interpret it.

(a)

C'(x) = 30 - 0.04x

, so

C'(200) = 22

. When

200

skateboards have been made, the cost is increasing at

22

dollars per skateboard, so the

201

st costs about

22

dollars to make. The actual cost is

C(201) - C(200) = 21.98 \text{ dollars},

so the estimate is

2

cents more than the actual cost.

(b)

R'(x) = 45 - 0.1x

, so

R'(200) = 25

, and the marginal profit is

P'(200) = 25 - 22 = 3

dollars per skateboard. When

200

skateboards are made and sold, the

201

st adds about

3

dollars of profit.

A rate in and a rate out

Example 2

From 6 A.M. to noon, cars enter a parking garage at a rate of

E(t) = 30 + 12t - 2t^2

cars per hour. They leave at a rate of

D(t) = 8t

cars per hour, where

t

is hours after 6 A.M. Let

N(t)

be the number of cars in the garage. (a) Determine whether

N

is increasing or decreasing at

t = 4

, and at what rate. (b) Find the time when cars enter fastest, and the rate of change of

N

then. (c) Find every time at which the garage is gaining

30

cars per hour.

The rate of change of the number of cars is

\begin{aligned} N'(t) &= E(t) - D(t) \\ &= 30 + 4t - 2t^2. \end{aligned}

(a)

N'(4) = 30 + 16 - 32 = 14 > 0

, so at 10 A.M. the number of cars in the garage is increasing, at a rate of

14

cars per hour.

(b)

E(t) = 30 + 12t - 2t^2

is a parabola opening down, with its vertex at

t = \frac{12}{4} = 3

. So cars enter fastest at 9 A.M., when

N'(3) = 30 + 12 - 18 = 24

: the number of cars is increasing at

24

cars per hour.

(c) Solve

N'(t) = 30

4t - 2t^2 = 0

, so

t = 0

t = 2

. The garage is gaining

30

cars per hour at 6 A.M. and again at 8 A.M.

Cars in, E(t) (black), and cars out, D(t) (red). The garage gains cars while E > D and loses cars after they cross at t = 5.

Instantaneous and average rates

Example 3

The depth of water in a tide pool is

H(t) = t^3 - 6t^2 + 8t + 20

centimeters,

t

hours after noon, for

0 \le t \le 4

. (a) Find the average rate of change of the depth over

0 \le t \le 4

. (b) Find every time in

0 < t < 4

at which the instantaneous rate of change equals that average rate.

(a)

H(0) = 20

and

H(4) = 64 - 96 + 32 + 20 = 20

, so the average rate is

\frac{20 - 20}{4 - 0} = 0

centimeters per hour.

(b) Solve

H'(t) = 3t^2 - 12t + 8 = 0

\begin{aligned} t &= \frac{12 \pm \sqrt{144 - 96}}{6} = 2 \pm \frac{2}{\sqrt{3}} \\[4pt] &\approx 0.845 \ \text{ or } \ 3.155. \end{aligned}

Both lie in

0 < t < 4

: about

0.845

and

3.155

hours after noon. At those two times the tangent line is parallel to the secant line from

t = 0

t = 4

The secant from (0, 20) to (4, 20) (blue) is horizontal. The tangent lines at t ≈ 0.845 and t ≈ 3.155 (amber) are parallel to it.

Solving for a time from a rate

Example 4

A cup of tea cools in a room at

70

degrees Fahrenheit. Its temperature is

T(t) = 70 + 130e^{-t/10}

degrees Fahrenheit,

t

minutes after it's poured. (a) Find

T'(10)

and interpret it. (b) Find the temperature of the tea at the moment it's cooling at

5

degrees per minute.

(a)

T'(t) = -13e^{-t/10}

, so

T'(10) = -13e^{-1} \approx -4.782

. Ten minutes after it's poured, the tea's temperature is decreasing at a rate of about

4.782

degrees Fahrenheit per minute.

(b) Cooling at

5

degrees per minute means

T'(t) = -5

, so

e^{-t/10} = \frac{5}{13}

and

t = 10\ln\frac{13}{5} \approx 9.555

. The temperature then, in degrees Fahrenheit, is

T = 70 + 130 \cdot \frac{5}{13} = 120.

The tangent line at t ≈ 9.555 has slope −5. The dashed line T = 70 is the room temperature.