Rates of Change in Context

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$P'(t) = kP(t)$ when $P(t) = P_0 e^{kt}$

What to Remember

The model is already a function of one variable

Differentiate that function. Then substitute the given input. The derivative is the rate of change with respect to that input.

Keep the units in the sentence

If $C$ is dollars and $x$ is the number of items, $C'(x)$ is dollars per item.

A positive rate can still be getting smaller

If $V'(t) > 0$, the quantity is still increasing. Compare $V'$ at two times to see whether that increase is faster or slower.

An exponential model

If $P(t) = P_0 e^{kt}$, then $P'(t) = k P_0 e^{kt} = k\, P(t)$.

Level 1

Easy

Differentiate the given model and read the rate.

Example

The cost of producing $x$ items is $C(x) = 5x + 20$ dollars. Find $C'(x)$ and state its meaning.

$C'(x) = 5$

The cost increases by $5 for each additional item, at every production level.

Answer: $C'(x) = 5$ dollars per item

The area of a circle of radius $r$ is $A = \pi r^2$. How fast does the area change with respect to the radius when $r = 3$?

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$\dfrac{dA}{dr} = 2\pi r$

Substitute $r = 3$.

$6\pi$

When the radius is 3, area increases at $6\pi$ square units per unit of radius.

Answer: $6\pi$

The depth of water in a tank, in centimeters, is $d(t) = 12 - 0.3t$, where $t$ is minutes. Find $d'(t)$.

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$d'(t) = -0.3$

The depth is decreasing at 0.3 centimeters per minute.

Answer: $-0.3$ cm per minute

A population is $P(t) = 200 + 12t - t^2$ for $0 \le t \le 6$, with $t$ in hours. Find $P'(2)$.

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$P'(t) = 12 - 2t$

Substitute $t = 2$.

$12 - 4$

$8$

Answer: 8 people per hour

The volume of water poured into a cistern, in liters after $t$ minutes, is $V(t) = 4t - 0.1t^2$. Find $V'(10)$.

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$V'(t) = 4 - 0.2t$

$V'(10) = 4 - 2$

$2$

At 10 minutes, water is still entering at 2 liters per minute.

Answer: $2$

Which sentence matches $C'(x) = 5$ when $C$ is dollars and $x$ is the number of items?

A. Each additional item adds $5 to the cost.

B. The total cost of 5 items is $1.

C. The cost is $5.

D. The cost is decreasing by 5 items.

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$C'(x) = 5$ dollars per item.

Answer: A. Each additional item adds $5 to the cost.

Level 2

Medium

A quadratic model or a quotient. Compute the rate, then say what it means at that input.

Example

Cost, in dollars, is $C(x) = 0.01x^2 + 4x + 50$. Find the marginal cost at $x = 40$.

$C'(x) = 0.02x + 4$

Substitute $x = 40$.

$0.8 + 4$

$4.8$

Answer: $4.80 per item

Revenue, in dollars, for selling $x$ items is $R(x) = 30x - 0.5x^2$. Find $R'(12)$.

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$R'(x) = 30 - x$

$R'(12) = 18$

At 12 items, revenue is increasing at $18 per additional item.

Answer: $18 per item

The number of animals in a reserve is $N(t) = \dfrac{500}{1 + t}$. Find $N'(4)$.

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$N(t) = 500(1 + t)^{-1}$

$N'(t) = -500(1 + t)^{-2}$

Substitute $t = 4$.

$-\dfrac{500}{25}$

$-20$

Answer: $-20$

A circular spill has radius $r(t) = 3 + 0.4t$ meters, with $t$ in minutes. How fast is the area changing when $t = 5$?

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$A = \pi r^2$

$\dfrac{dA}{dt} = 2\pi r\,\dfrac{dr}{dt}$

$r(5) = 5$ and $r'(t) = 0.4$

$2\pi(5)(0.4)$

$4\pi$

Answer: $4\pi$ square meters per minute

Profit, in dollars, for $x$ items is $G(x) = -0.002x^2 + 6x - 400$. Find $G'(500)$.

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$G'(x) = -0.004x + 6$

$G'(500) = -2 + 6$

$4$

Answer: $4 per item

For the same profit function, find $G'(2000)$ and state whether profit is increasing or decreasing at 2000 items.

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$G'(x) = -0.004x + 6$

$G'(2000) = -8 + 6$

$-2$

At 2000 items, profit is decreasing at $2 per additional item.

Answer: decreasing at $2 per item

Level 3

Hard

An exponential model, a comparison of two rates, and an estimate from the rate you found.

Example

A population is $P(t) = 800e^{0.02t}$. Find $P'(25)$ and interpret it.

$P'(t) = 800\cdot 0.02\, e^{0.02t}$

$P'(t) = 16e^{0.02t}$

Substitute $t = 25$.

$16e^{0.5}$

At $t = 25$, the population is increasing at a rate of $16e^{0.5}$ individuals per unit of time.

Answer: $16e^{0.5}$

Light intensity is $I(d) = \dfrac{50}{d^2}$ at a distance of $d$ meters from a lamp. Find $I'(5)$.

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$I(d) = 50d^{-2}$

$I'(d) = -100d^{-3}$

$I'(5) = -\dfrac{100}{125}$

$-\dfrac{4}{5}$

Answer: $-\dfrac{4}{5}$

For $G(x) = -0.002x^2 + 6x - 400$, use $G'(500)$ to estimate the change in profit from 500 items to 501 items.

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$G'(500) = 4$

$\Delta G \approx 4\cdot 1$

Profit increases by about $4.

Answer: about $4

The volume of water in a cistern, in liters after $t$ minutes, is $V(t) = 4t - 0.1t^2$. Compare $V'(0)$ and $V'(10)$.

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$V'(t) = 4 - 0.2t$

$V'(0) = 4$

$V'(10) = 2$

Both rates are positive, so the volume is still increasing at $t = 10$.

The rate at $t = 10$ is smaller.

Answer: both positive; slower at $t = 10$

Which value is $N'(4)$ for $N(t) = \dfrac{500}{1 + t}$?

A. $-20$

B. $20$

C. $-25$

D. $-\dfrac{500}{4}$

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$N'(t) = -\dfrac{500}{(1 + t)^2}$

$N'(4) = -\dfrac{500}{25} = -20$

Answer: A. $-20$

At $t = 5$, a spill has radius 5 meters and the radius is increasing at 0.4 meters per minute. The area is $A = \pi r^2$. Find $dA/dt$ at that moment.

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$\dfrac{dA}{dt} = 2\pi r\,\dfrac{dr}{dt}$

$2\pi(5)(0.4)$

$4\pi$

Answer: $4\pi$ m²/min

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