Forming differential equations and connected rates of changeEdexcel International A Level Maths: Revision notes

A differential equation contains a derivative. To form one from a description, name the variables and translate each phrase. 'The rate of change of $y$ with respect to $t$' is $\frac{dy}{dt}$. 'Is proportional to $y$' means $=ky$ for a constant $k$. So 'the rate of increase of $P$ is proportional to $P$' gives $\frac{dP}{dt}=kP$. A decrease needs a minus sign: if $V$ is decreasing at a rate proportional to $\sqrt V$ then $\frac{dV}{dt}=-k\sqrt V$ with $k>0$. 'Inversely proportional' gives $\frac{dy}{dt}=\frac ky$. Data given later, such as $\frac{dV}{dt}=-5$ when $V=200$, is used to find $k$.

When a quantity changes because of several processes, add the rates with signs: $\frac{dV}{dt}=\text{rate in}-\text{rate out}.$ If water is poured in at $20$ cm$^3$ s$^{-1}$ and leaks at $kh$ cm$^3$ s$^{-1}$, then $\frac{dV}{dt}=20-kh$. The quantity stops changing, an equilibrium, when $\frac{dV}{dt}=0$. If the result is positive the quantity is increasing; if negative it is decreasing.

When $y$ depends on $x$ and $x$ depends on $t$, the chain rule links the rates: $\frac{dy}{dt}=\frac{dy}{dx}\times\frac{dx}{dt}.$ Example: a sphere's radius grows at $0.5$ cm s$^{-1}$, with $V=\frac43\pi r^3$. Then $\frac{dV}{dr}=4\pi r^2$ and $\frac{dV}{dt}=4\pi r^2\times0.5$. When $r=6$, $\frac{dV}{dt}=72\pi$ cm$^3$ s$^{-1}$. Rearranging gives $\frac{dr}{dt}=\frac{dV}{dt}\div\frac{dV}{dr}$. You may need to find a missing rate, such as the surface area change $\frac{dS}{dt}=8\pi r\frac{dr}{dt}$.

Cube: the edge $x$ cm grows at $0.2$ cm s$^{-1}$. $V=x^3$, so $\frac{dV}{dt}=3x^2\times0.2$. When $x=4$: $9.6$ cm$^3$ s$^{-1}$. Sphere: the surface area grows at $10$ cm$^2$ s$^{-1}$. $\frac{dS}{dt}=8\pi r\frac{dr}{dt}$, so at $r=5$: $10=40\pi\frac{dr}{dt}$ and $\frac{dr}{dt}=\frac{1}{4\pi}\approx0.0796$ cm s$^{-1}$. Cone filling: liquid enters at $20$ cm$^3$ s$^{-1}$ and $V=\frac13\pi h^3$. $\frac{dV}{dh}=\pi h^2$, so $\frac{dh}{dt}=\frac{20}{\pi h^2}$. At $h=5$: $\frac{4}{5\pi}\approx0.255$ cm s$^{-1}$. As $h$ grows the depth rises more slowly.

Define your letters and the sign convention before you write the equation. In 'show that' questions the final form is given, so display each step: the rate statement, the relation between $V$ and $h$, and the chain rule. When a question asks 'at what rate', include units and say whether the quantity is increasing or decreasing. Give decimals to 3 significant figures unless an exact answer is requested.