Rates of change and connected rates of change

Every derivative is a rate of change. $\dfrac{dy}{dx}$ says how fast $y$ changes as $x$ changes; $\dfrac{dV}{dt}$ says how fast a volume changes as time passes. In a connected rates of change question you are told how fast one quantity is changing and asked how fast a related quantity is changing at a particular moment: a balloon is inflated at a known rate, so how fast is its radius growing? The chain rule links the two rates. Expect one of these on most Paper 1s, typically worth 3 to 5 marks.

A derivative is a rate

If $y$ is a function of $x$, then $\dfrac{dy}{dx}$ is the rate of change of $y$ with respect to $x$: the number of units $y$ increases by per unit increase in $x$, at that instant. When the variable on the bottom is time $t$, the derivative is a rate in the everyday sense.

Derivative	Meaning	Typical units
$\dfrac{dr}{dt}$	rate of change of a radius	$\text{cm s}^{-1}$
$\dfrac{dA}{dt}$	rate of change of an area	$\text{cm}^2\ \text{s}^{-1}$
$\dfrac{dV}{dt}$	rate of change of a volume	$\text{cm}^3\ \text{s}^{-1}$
$\dfrac{dh}{dt}$	rate of change of a depth	$\text{cm s}^{-1}$

The sign carries meaning. A positive rate means the quantity is increasing; a negative rate means it is decreasing. "Water leaks out at $4\ \text{cm}^3$ per second" translates to $\dfrac{dV}{dt} = -4$. "The radius is decreasing at $0.2\ \text{cm s}^{-1}$" means $\dfrac{dr}{dt} = -0.2$.

A "rate" in a P1 question always means a rate with respect to time unless it says otherwise, and the words "at a constant rate" mean the derivative is a fixed number.

Connecting two rates with the chain rule

Suppose a circular oil slick is spreading. You know how fast its radius grows, $\dfrac{dr}{dt}$, and want how fast its area grows, $\dfrac{dA}{dt}$. The area depends on the radius through $A = \pi r^2$, and the radius depends on time. So $A$ is a function of a function of $t$, and the chain rule gives

$$\frac{dA}{dt} = \frac{dA}{dr} \times \frac{dr}{dt}$$

The middle variable $r$ "cancels" on the right, which is a good way to check that you have written the chain the right way round. The factor $\dfrac{dA}{dr}$ comes from differentiating the formula that connects the two quantities, $A = \pi r^2$, giving $2\pi r$.

The method

Every connected rates question follows the same steps. Most marks are lost by substituting too early or by inverting a derivative incorrectly.

The value at the instant (such as "when the radius is $5$ cm") is substituted only after differentiating. If you put $r = 5$ into $V = \tfrac{4}{3}\pi r^3$ first, you have a constant, whose derivative is zero.

A point moving along a curve

A point $P(x, y)$ moving along a curve has both coordinates changing with time, linked by the equation of the curve. The derivative $\dfrac{dy}{dx}$ is the gradient of the curve, so the vertical speed is the gradient times the horizontal speed: $\dfrac{dy}{dt} = \dfrac{dy}{dx} \times \dfrac{dx}{dt}$.

On $y = 2x + \dfrac{5}{x}$ the gradient at $x = 1$ is $-3$ (the tangent shown). If $x$ increases at $0.02$ units per second there, $y$ changes at $-3 \times 0.02 = -0.06$ units per second: it is decreasing, because the curve slopes down at that point. This is Example 2.

Containers and similar triangles

Water filling a cone is the classic exam setting. The volume of water depends on both its depth $h$ and the radius $r$ of its surface, but these are linked by similar triangles, so $V$ can be written in terms of $h$ alone.

For a cone of radius $10$ cm and height $30$ cm, $\dfrac{r}{h} = \dfrac{10}{30}$, so $r = \dfrac{h}{3}$ and

$$V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \left(\frac{h}{3}\right)^2 h = \frac{\pi h^3}{27}$$

For a prism-shaped container (a trough), the volume is the cross-sectional area times the length, so again everything reduces to one variable. Often the question gives you the volume formula, as in Example 6, and the challenge is only the calculus.

Worked examples

Practice questions

Answers

1. $A = s^2$, $\dfrac{dA}{dt} = 2s \dfrac{ds}{dt} = 2(10)(2) = 40\ \text{cm}^2\ \text{s}^{-1}$.

2. $V = s^3$, $\dfrac{dV}{dt} = 3s^2 \dfrac{ds}{dt}$, so $-6 = 48\dfrac{ds}{dt}$ and $\dfrac{ds}{dt} = -0.125\ \text{cm s}^{-1}$ (decreasing).

3. $y = (2x + 1)^{\frac{1}{2}}$, $\dfrac{dy}{dx} = \tfrac{1}{2}(2x + 1)^{-\frac{1}{2}} \times 2 = \dfrac{1}{\sqrt{2x + 1}} = \dfrac{1}{3}$ at $x = 4$. So $\dfrac{dy}{dt} = \dfrac{1}{3} \times 3 = 1$.

4. $V = 10\pi r^2$, $\dfrac{dV}{dr} = 20\pi r = 60\pi$ at $r = 3$. $\dfrac{dV}{dt} = 60\pi \times 0.1 = 6\pi \approx 18.8\ \text{cm}^3\ \text{s}^{-1}$.

5. $\dfrac{dA}{dt} = 2\pi r \dfrac{dr}{dt}$, so $-6\pi = 8\pi \dfrac{dr}{dt}$, $\dfrac{dr}{dt} = -\tfrac{3}{4}\ \text{cm s}^{-1}$. $C = 2\pi r$, so $\dfrac{dC}{dt} = 2\pi \times \left(-\tfrac{3}{4}\right) = -\tfrac{3\pi}{2} \approx -4.71\ \text{cm s}^{-1}$.

6. $y = 3(2x - 1)^{-1}$, $\dfrac{dy}{dx} = -6(2x - 1)^{-2} = -\dfrac{6}{9} = -\dfrac{2}{3}$ at $x = 2$. With $\dfrac{dy}{dt} = -0.3$: $-0.3 = -\dfrac{2}{3}\dfrac{dx}{dt}$, so $\dfrac{dx}{dt} = 0.45$ units per second (increasing).

7. $\dfrac{dV}{dt} = 4\pi r^2 \dfrac{dr}{dt}$: $-10 = 100\pi\dfrac{dr}{dt}$, so $\dfrac{dr}{dt} = -\dfrac{1}{10\pi}$. Then $\dfrac{dS}{dt} = 8\pi r \dfrac{dr}{dt} = 40\pi \times \left(-\dfrac{1}{10\pi}\right) = -4$. The surface area is decreasing at $4\ \text{cm}^2$ per minute.

8. (a) Cross-section area $= \tfrac{1}{2} \times h \times h = \tfrac{1}{2}h^2$, so $V = 200 \times \tfrac{1}{2}h^2 = 100h^2$. (b) $\dfrac{dV}{dh} = 200h = 2000$ at $h = 10$. $500 = 2000 \dfrac{dh}{dt}$, so $\dfrac{dh}{dt} = 0.25\ \text{cm s}^{-1}$. (c) The surface is a rectangle $200$ by $h$, so $S = 200h$ and $\dfrac{dS}{dt} = 200 \times 0.25 = 50\ \text{cm}^2\ \text{s}^{-1}$.

9. $S = 4\pi r^2$, $\dfrac{dS}{dt} = 8\pi r \dfrac{dr}{dt}$: $20 = 48\pi \dfrac{dr}{dt}$, so $\dfrac{dr}{dt} = \dfrac{5}{12\pi}$. Then $\dfrac{dV}{dt} = 4\pi r^2 \dfrac{dr}{dt} = 144\pi \times \dfrac{5}{12\pi} = 60\ \text{cm}^3\ \text{s}^{-1}$.