7.8 Connected rates of change and inverse functions — Practise with help

Work each question closed book first. Write the chain rule that links the two rates before you substitute any numbers.

Warm-up: remember

Differentiate $y = (2x + 1)^3$ .
Differentiate $y = e^{3x}$ .
Differentiate $y = x^3$ .

Answers

Chain rule: $3(2x + 1)^2 \times 2 = 6(2x + 1)^2$ .
$\dfrac{dy}{dx} = 3e^{3x}$ .
$\dfrac{dy}{dx} = 3x^2$ .

If question 1 was hard, revise the chain rule in Section 7.7.

Practise with help

1. A sphere grows so that its volume increases at $36\pi$ cm $^3$ per second. Find the rate at which the radius grows when $r = 3$ cm.

Hint

For a sphere, $V = \dfrac{4}{3}\pi r^3$ , so $\dfrac{dV}{dr} = 4\pi r^2$ . The chain rule is $\dfrac{dV}{dt} = \dfrac{dV}{dr} \times \dfrac{dr}{dt}$ .

Next step

Substitute $r = 3$ into $\dfrac{dV}{dr}$ to get $36\pi$ . Then solve $36\pi = 36\pi \times \dfrac{dr}{dt}$ .

Full working

\frac{dV}{dr} = 4\pi r^2 = 4\pi(9) = 36\pi, \qquad 36\pi = 36\pi \times \frac{dr}{dt}, \qquad \frac{dr}{dt} = 1 \text{ cm/s}.

The radius grows at $1$ cm/s when the radius is $3$ cm.

2. A cube expands so that each edge grows at $0.2$ cm/s. Find the rate at which the volume grows when the edge is $5$ cm.

Hint

The volume is $V = x^3$ , so $\dfrac{dV}{dx} = 3x^2$ . The rate you know is $\dfrac{dx}{dt}$ .

Next step

At $x = 5$ , $\dfrac{dV}{dx} = 3(25) = 75$ . Multiply by $\dfrac{dx}{dt} = 0.2$ .

Full working

\frac{dV}{dt} = \frac{dV}{dx} \times \frac{dx}{dt} = 75 \times 0.2 = 15 \text{ cm}^3\text{/s}.

The volume grows at $15$ cm $^3$ per second.

3. A circle's radius grows at $0.1$ cm/s. Find the rate at which its area grows when $r = 10$ cm.

Hint

The area is $A = \pi r^2$ , so $\dfrac{dA}{dr} = 2\pi r$ . Use $\dfrac{dA}{dt} = \dfrac{dA}{dr} \times \dfrac{dr}{dt}$ .

Next step

At $r = 10$ , $\dfrac{dA}{dr} = 20\pi$ . Multiply by $0.1$ .

Full working

\frac{dA}{dt} = 20\pi \times 0.1 = 2\pi \approx 6.28 \text{ cm}^2\text{/s}.

The area grows at about $6.28$ cm $^2$ per second.

4. The function $f(x) = x^3 + x$ has an inverse. Find $\left(f^{-1}\right)'(2)$ .

Hint

Find the value of $x$ for which $f(x) = 2$ . Then use the derivative of $f$ at that point.

Next step

$f(1) = 1 + 1 = 2$ , so the matching point is $x = 1$ . The derivative is $f'(x) = 3x^2 + 1$ , so $f'(1) = 4$ . Take the reciprocal.

Full working

f'(x) = 3x^2 + 1, \qquad f'(1) = 4, \qquad \left(f^{-1}\right)'(2) = \frac{1}{4}.

5. The function $f(x) = x^5$ has inverse $f^{-1}(y) = y^{1/5}$ . Find $\left(f^{-1}\right)'(32)$ .

Hint

Find $x$ such that $x^5 = 32$ , then differentiate $f$ at that point.

Next step

$2^5 = 32$ , so the matching point is $x = 2$ . Also $f'(x) = 5x^4$ , so $f'(2) = 80$ .

Full working

f'(2) = 5(2)^4 = 80, \qquad \left(f^{-1}\right)'(32) = \frac{1}{80}.

A direct check: the derivative of $y^{1/5}$ is $\tfrac{1}{5}y^{-4/5}$ , which at $y = 32$ gives $\tfrac{1}{5}(16) = \tfrac{1}{80}$ .

6. Two quantities are linked by $y = x^2$ , and $x$ grows at $3$ units per second. Find the rate at which $y$ grows when $x = 4$ .

Hint

Differentiate the linking formula to find $\dfrac{dy}{dx}$ . Then use $\dfrac{dy}{dt} = \dfrac{dy}{dx} \times \dfrac{dx}{dt}$ .

Next step

$\dfrac{dy}{dx} = 2x$ , which is $8$ at $x = 4$ . Multiply by $\dfrac{dx}{dt} = 3$ .

Full working

\frac{dy}{dt} = 2x \times \frac{dx}{dt} = 8 \times 3 = 24 \text{ units/s}.

The rate depends on $x$ , so it changes as $x$ grows even when $\dfrac{dx}{dt}$ is constant.