7.8 Connected rates of change and inverse functions — Practise

Work through these on your own. Write the chain rule that links the rates before you substitute. For an inverse function, find the matching point on the original function first.

Both the questions and the answers below come from a set that is checked by computer every time the book is built.

A sphere grows so that $\dfrac{dV}{dt} = 100$ cm $^3$ /s. Find $\dfrac{dr}{dt}$ when $r = 5$ cm.
A sphere grows so that $\dfrac{dV}{dt} = 4\pi$ cm $^3$ /s. Find $\dfrac{dr}{dt}$ when $r = 1$ cm.
A cube expands so that each edge grows at $0.5$ cm/s. Find $\dfrac{dV}{dt}$ when the edge is $4$ cm.
A circle's area grows at $2$ cm $^2$ /s. Find $\dfrac{dr}{dt}$ when $r = 3$ cm.
A circle's radius grows at $0.05$ cm/s. Find $\dfrac{dA}{dt}$ when $r = 6$ cm.
The function $f(x) = x^3$ has an inverse. Find $\left(f^{-1}\right)'(27)$ .
The function $f(x) = x^3 + 3x$ has an inverse. Find $\left(f^{-1}\right)'(0)$ .
The function $f(x) = x^5 + x$ has an inverse. Find $\left(f^{-1}\right)'(2)$ .
Two quantities are linked by $y = x^2$ , and $x$ grows at $4$ units per second. Find $\dfrac{dy}{dt}$ when $x = 5$ .
A cylinder has fixed height $10$ cm and volume $V = \pi r^2 h$ . Its radius grows at $0.1$ cm/s. Find $\dfrac{dV}{dt}$ when $r = 2$ cm.
A spherical droplet grows so that $\dfrac{dV}{dt} = 5$ cm $^3$ /s. Find $\dfrac{dr}{dt}$ when $r = 2$ cm.
The function $f(x) = x^3 + 4x$ has an inverse. Find $\left(f^{-1}\right)'(5)$ .

Answers and concise solutions

1. The rate

Answer: $V = \dfrac{4}{3}\pi r^3$ , so $\dfrac{dV}{dr} = 4\pi r^2 = 100\pi$ at $r = 5$ . Then $\dfrac{dV}{dt} = \dfrac{dV}{dr} \times \dfrac{dr}{dt}$ gives $100 = 100\pi \dfrac{dr}{dt}$ , so $\dfrac{dr}{dt} = \dfrac{1}{\pi} \approx 0.318$ cm/s.

2. The rate

Answer: $\dfrac{dV}{dr} = 4\pi r^2 = 4\pi$ at $r = 1$ . Then $\dfrac{dr}{dt} = \dfrac{4\pi}{4\pi} = 1$ cm/s.

3. The rate

Answer: $V = x^3$ , so $\dfrac{dV}{dx} = 3x^2 = 48$ at $x = 4$ . Then $\dfrac{dV}{dt} = 48 \times 0.5 = 24$ cm $^3$ /s.

4. The rate

Answer: $A = \pi r^2$ , so $\dfrac{dA}{dr} = 2\pi r = 6\pi$ at $r = 3$ . Then $2 = 6\pi \dfrac{dr}{dt}$ , so $\dfrac{dr}{dt} = \dfrac{1}{3\pi} \approx 0.106$ cm/s.

5. The rate

Answer: $\dfrac{dA}{dr} = 2\pi r = 12\pi$ at $r = 6$ . Then $\dfrac{dA}{dt} = 12\pi \times 0.05 = \dfrac{3\pi}{5} \approx 1.88$ cm $^2$ /s.

6. The gradient

Answer: $f(3) = 27$ , so the matching point is $x = 3$ . Then $f'(x) = 3x^2$ and $f'(3) = 27$ , so the inverse gradient is $\dfrac{1}{27}$ .

7. The gradient

Answer: $f(0) = 0$ , so the matching point is $x = 0$ . Then $f'(x) = 3x^2 + 3$ and $f'(0) = 3$ , so the inverse gradient is $\dfrac{1}{3}$ .

8. The gradient

Answer: $f(1) = 1 + 1 = 2$ , so the matching point is $x = 1$ . Then $f'(x) = 5x^4 + 1$ and $f'(1) = 6$ , so the inverse gradient is $\dfrac{1}{6}$ .

9. The rate

Answer: $\dfrac{dy}{dx} = 2x = 10$ at $x = 5$ , so $\dfrac{dy}{dt} = 10 \times 4 = 40$ units/s.

10. The rate

Answer: With $h = 10$ , $V = 10\pi r^2$ and $\dfrac{dV}{dr} = 20\pi r = 40\pi$ at $r = 2$ . Then $\dfrac{dV}{dt} = 40\pi \times 0.1 = 4\pi \approx 12.6$ cm $^3$ /s.

11. The rate

Answer: $\dfrac{dV}{dr} = 4\pi r^2 = 16\pi$ at $r = 2$ . Then $5 = 16\pi \dfrac{dr}{dt}$ , so $\dfrac{dr}{dt} = \dfrac{5}{16\pi} \approx 0.0995$ cm/s.

12. The gradient

Answer: $f(1) = 1 + 4 = 5$ , so the matching point is $x = 1$ . Then $f'(x) = 3x^2 + 4$ and $f'(1) = 7$ , so the inverse gradient is $\dfrac{1}{7}$ .