Related rates

What you will be able to do

• set up a related rates problem from a picture;
• differentiate with respect to time with the chain rule;
• put in the numbers at the right moment, after differentiating;
• solve problems with a ladder, a sphere, a cone and a shadow.

Quantities that change together

A ladder slides down a wall. Its foot moves out, and its top moves down. The two motions are tied by Pythagoras: $x^2 + y^2 = 13^2$. If we know how fast the foot moves, we can find how fast the top moves.

This is a related rates problem. The quantities $x$ and $y$ both depend on the time $t$, and one equation links them.

The method

1. Draw a picture. Name every quantity that changes, with its units.
2. Write the link. Find an equation that is true at every moment, not only at the moment asked about.
3. Differentiate with respect to $t$. Use the chain rule: $\dfrac{d}{dt}(x^2) = 2x\,\dfrac{dx}{dt}$.
4. Put in the numbers. Only now use the values at the moment asked about.
5. Answer with units and a sign. A quantity that is shrinking has a negative rate.

Step 4 is where most mistakes happen. Putting in $x = 5$ before step 3 turns $x^2$ into the constant $25$, and its derivative is $0$.

The ladder

A $13\ \text{m}$ ladder leans on a wall. Its foot is $5\ \text{m}$ from the wall and slides away at $0.6\ \text{m/s}$. How fast does the top come down?

Differentiate $x^2 + y^2 = 169$ with respect to $t$:

When $x = 5$, then $y = \sqrt{169 - 25} = 12$. So $\dfrac{dy}{dt} = -\dfrac{5}{12} \times 0.6 = -0.25\ \text{m/s}$. The top comes down at $0.25\ \text{m/s}$. The minus sign means $y$ is decreasing.

A balloon

A pump fills a spherical balloon at $200\ \text{cm}^3/\text{s}$. How fast does the radius grow when $r = 5\ \text{cm}$?

The link is $V = \tfrac{4}{3}\pi r^3$. Differentiating,

Then $200 = 4\pi \times 25 \times \dfrac{dr}{dt}$, so $\dfrac{dr}{dt} = \dfrac{200}{100\pi} = \dfrac{2}{\pi} \approx 0.637\ \text{cm/s}$.

A cone and the second variable

Sand pours at $3\ \text{m}^3/\text{min}$ and makes a cone whose height always equals its radius. How fast does the height grow when $h = 2\ \text{m}$?

The volume is $V = \tfrac{1}{3}\pi r^2 h$. But $r = h$, so we write the volume with one variable: $V = \tfrac{1}{3}\pi h^3$. Then

So $3 = \pi \times 4 \times \dfrac{dh}{dt}$, and $\dfrac{dh}{dt} = \dfrac{3}{4\pi} \approx 0.239\ \text{m/min}$.

When a cone fills with water, the radius of the water surface changes as the depth changes. Use similar triangles to write $r$ in terms of $h$ before you differentiate.

A shadow

A street lamp is $5\ \text{m}$ high. A girl $1\ \text{m}$ tall walks away from it at $1.2\ \text{m/s}$. How fast does the tip of her shadow move?

Let $x$ be her distance from the lamp and $s$ the distance of the tip of the shadow from the lamp. Similar triangles give $\dfrac{s}{5} = \dfrac{s - x}{1}$, so $s = \tfrac{5}{4}x$. Then $\dfrac{ds}{dt} = \tfrac{5}{4} \times 1.2 = 1.5\ \text{m/s}$. The tip of the shadow moves faster than the girl.

A worked example

A square metal plate expands when heated. Its side grows at $0.02\ \text{cm/s}$. How fast does its area grow when the side is $12\ \text{cm}$?

1. The link. $A = x^2$.
2. Differentiate. $\dfrac{dA}{dt} = 2x\,\dfrac{dx}{dt}$.
3. Put in the numbers. $2 \times 12 \times 0.02 = 0.48$.
4. Answer. The area grows at $0.48\ \text{cm}^2/\text{s}$.

Practise with new numbers

Try it in Python

The program finds the speed of the top of the ladder twice: by the formula, and by moving the foot a tiny distance and measuring the change in height.

Common mistakes

• Putting in numbers before differentiating. A fixed number has derivative $0$.
• Treating a changing quantity as constant. In the cone, $r$ changes with $h$. The ladder length does not change.
• Forgetting the chain rule factor. $\dfrac{d}{dt}(r^3) = 3r^2\,\dfrac{dr}{dt}$, not $3r^2$.
• Losing the sign. A ladder top coming down has $\dfrac{dy}{dt} < 0$. Say which way it moves.
• Leaving out units. Write $\text{m/s}$ or $\text{cm}^3/\text{s}$.

Remember these

• Link the quantities with an equation true at every moment, then differentiate with respect to $t$.
• Use the chain rule: $\dfrac{d}{dt}f(r) = f'(r)\,\dfrac{dr}{dt}$.
• Put in the numbers for the moment asked about only after differentiating.
• Use similar triangles to remove an extra variable.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. The radius of a circle grows at $3\ \text{cm/s}$. How fast does the area grow when $r = 10\ \text{cm}$?

2. The edge of a cube grows at $2\ \text{cm/s}$. How fast does its volume grow when the edge is $5\ \text{cm}$?

3. A point moves along $y = x^2$ with $\dfrac{dx}{dt} = 4$ units per second. Find $\dfrac{dy}{dt}$ when $x = 3$.

4. The foot of a ladder is pulled towards the wall. Is $\dfrac{dy}{dt}$ positive or negative, with $y$ measured upwards?

5. Sand makes a cone with $V = \tfrac{1}{3}\pi h^3$ at $\dfrac{dV}{dt} = 32\pi\ \text{cm}^3/\text{s}$. Find $\dfrac{dh}{dt}$ when $h = 4\ \text{cm}$.

References

• NCERT, Mathematics Textbook for Class XII, Part I, Chapter 6 (Application of Derivatives), the section on the rate of change of quantities.
• OpenStax, Calculus Volume 1, Section 4.1 (Related Rates).