10. Related Rates of Change

Video coming soon4 worked examples

Theory

In a related rates problem, two quantities are linked by an equation and both change with time. Differentiate the relationship with respect to time $t$ (using the chain rule) to connect their rates, then substitute the values at the instant asked.

The Golden Rule: write the equation relating the quantities, differentiate the whole equation with respect to $t$ , and only then substitute the given values.

⚠️ Common Examiner Traps

Differentiate with respect to time: the derivative is with respect to $t$ , so every variable picks up a rate (e.g. $\frac{dr}{dt}$ ).
Substituting too early: keep quantities as variables until after differentiating.
Chain of rates: link them correctly, e.g. $\frac{dA}{dt} = \frac{dA}{dr}\cdot\frac{dr}{dt}$ .
Constant quantities differentiate to zero: when a question says an area or volume “remains constant”, that is the whole point — the derivative of the right-hand side is $0$ , which is what lets you link the two rates.
Reading the sign: a negative rate means the quantity is decreasing. Say so in words; the sign alone is not the answer.

Worked examples

Example 1

The radius of a circle increases at $2$ cm/s. Find the rate at which the area is increasing when the radius is $5$ cm.

Step 1: The area is $A = \pi r^2$ . Differentiate with respect to $t$ :

\frac{dA}{dt} = 2\pi r \frac{dr}{dt}

Step 2: Substitute $r = 5$ and $\frac{dr}{dt} = 2$ :

\frac{dA}{dt} = 2\pi(5)(2) = 20\pi \ \text{cm}^2/\text{s}

Example 2

A spherical balloon is inflated so that its volume increases at $100$ cm³/s. Find the rate at which the radius is increasing when the radius is $5$ cm.

Step 1: The volume is $V = \frac{4}{3}\pi r^3$ . Differentiate with respect to $t$ :

\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}

Step 2: Substitute $\frac{dV}{dt} = 100$ and $r = 5$ :

100 = 4\pi(25)\frac{dr}{dt} = 100\pi\frac{dr}{dt}

Step 3: Solve for the rate:

\frac{dr}{dt} = \frac{100}{100\pi} = \frac{1}{\pi} \ \text{cm/s}

Example 3

A rectangle has sides $x$ cm and $y$ cm. Both are changing, but in such a way that the area stays constant at $48$ cm². If $x$ is increasing at $0.25$ cm/s, find the rate at which $y$ is changing when $x = 4$ .

Step 1: Write the relationship. The area is fixed, so the right-hand side is a constant:

xy = 48

Step 2: Differentiate with respect to $t$ , using the product rule on the left. A constant differentiates to zero:

y\frac{dx}{dt} + x\frac{dy}{dt} = 0

Step 3: Find the value of $y$ at the instant in question:

x = 4 \implies y = \frac{48}{4} = 12

Step 4: Substitute $x=4$ , $y=12$ and $\frac{dx}{dt} = 0.25$ :

12(0.25) + 4\frac{dy}{dt} = 0 \implies 3 + 4\frac{dy}{dt} = 0

Step 5: Solve for the rate:

\frac{dy}{dt} = -\frac{3}{4} = -0.75 \ \text{cm/s}

The negative sign means $y$ is decreasing, at $0.75$ cm/s — which makes sense, since the area must stay fixed as $x$ grows.

Example 4

An open-topped cylindrical tank of radius $r$ and height $h$ has a fixed total surface area of $27\pi$ cm². Find an expression for $\dfrac{dr}{dh}$ .

Step 1: With no lid, the surface area is one circular base plus the curved surface:

\pi r^2 + 2\pi rh = 27\pi

Step 2: Divide through by $\pi$ to simplify before differentiating:

r^2 + 2rh = 27

Step 3: Differentiate with respect to $h$ , treating $r$ as a function of $h$ . The $2rh$ term needs the product rule:

2r\frac{dr}{dh} + 2\left(r + h\frac{dr}{dh}\right) = 0

Step 4: Gather the $\frac{dr}{dh}$ terms and factorise:

\frac{dr}{dh}\bigl(2r + 2h\bigr) = -2r

Step 5: Divide, cancelling the factor of $2$ :

\frac{dr}{dh} = -\frac{2r}{2r + 2h} = -\frac{r}{r + h}