Rates of Change | CAIE 9709 Pure Mathematics 1 Notes

Pure Mathematics 1

Mathematics9709

Read a rate and its units

$dV/dt$ means the rate of change of volume with respect to time. “At this instant” asks for a derivative, not the average change over a whole time interval.

The numerator supplies the changing quantity’s unit; the denominator supplies the input unit. If volume is in cm³ and time in seconds, $dV/dt$ is in cm³/s. A negative rate means the quantity is decreasing.

Worked example

h = t²/5, where h is in metres and t in seconds. Find the rate at t = 2.

\begin{gathered}\frac{dh}{dt}=\frac{2t}{5}\\t=2\ \Rightarrow\ \frac{dh}{dt}=0.8\text{ m/s}\end{gathered}

Differentiate first, then substitute the time. Substituting first would leave only a number and remove the changing relationship.

Your turnPractice

$V=2t^2-3t+8$ cm³, with $t$ in seconds. Find the rate of change at $t=4$ , and state whether the volume is increasing.

Show worked answer

$dV/dt=4t-3=13\text{ cm}^3/\text{s}$ at $t=4$ . The rate is positive, so volume is increasing.

Connect rates through the intermediate variable

If time changes a circle’s radius, and the radius changes its area, the chain is $t\to r\to A$ . The chain rule connects the rates:

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

$dA/dr$ measures area gained per unit radius; $dr/dt$ measures radius gained per unit time. Their product gives area gained per unit time. It is the chain rule, not ordinary cancellation of separate $d$ s.

Worked example

A circular oil patch expands at 2 m per hour in radius. Find its area rate when r = 25 m.

The given rate is $dr/dt=2$ . Since $A=\pi r^2$ , $dA/dr=2\pi r$ .

\begin{aligned}\frac{dA}{dt}&=2\pi(25)(2)\\&=100\pi\text{ m}^2/\text{h}\end{aligned}

All factors describe the same instant. Check the units: $(\text{m}^2/\text{m})(\text{m}/\text{h})=\text{m}^2/\text{h}$ .

A constant radius rate does not make the area rate constant: $2\pi r$ increases as the radius grows.

Your turn9709/11 Oct/Nov 2023 Q3 [3 marks]

A cube has edge length $x$ cm and volume $V=x^3$ cm³. Its edges lengthen at $0.01\text{ cm/min}$ . Find the volume rate when $x=20$ .

Show worked answer

\begin{aligned}\frac{dV}{dt}&=\frac{dV}{dx}\frac{dx}{dt}\\&=3(20)^2(0.01)\end{aligned}

Answer

12\text{ cm}^3/\text{min}

Read a rate and its units

$dV/dt$ means the rate of change of volume with respect to time. “At this instant” asks for a derivative, not the average change over a whole time interval.

Worked example

h = t²/5, where h is in metres and t in seconds. Find the rate at t = 2.

\begin{gathered}\frac{dh}{dt}=\frac{2t}{5}\\t=2\ \Rightarrow\ \frac{dh}{dt}=0.8\text{ m/s}\end{gathered}

Differentiate first, then substitute the time. Substituting first would leave only a number and remove the changing relationship.

Your turnPractice

$V=2t^2-3t+8$ cm³, with $t$ in seconds. Find the rate of change at $t=4$ , and state whether the volume is increasing.

Show worked answer

$dV/dt=4t-3=13\text{ cm}^3/\text{s}$ at $t=4$ . The rate is positive, so volume is increasing.

Connect rates through the intermediate variable

If time changes a circle’s radius, and the radius changes its area, the chain is $t\to r\to A$ . The chain rule connects the rates:

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

Worked example

A circular oil patch expands at 2 m per hour in radius. Find its area rate when r = 25 m.

The given rate is $dr/dt=2$ . Since $A=\pi r^2$ , $dA/dr=2\pi r$ .

\begin{aligned}\frac{dA}{dt}&=2\pi(25)(2)\\&=100\pi\text{ m}^2/\text{h}\end{aligned}

All factors describe the same instant. Check the units: $(\text{m}^2/\text{m})(\text{m}/\text{h})=\text{m}^2/\text{h}$ .

A constant radius rate does not make the area rate constant: $2\pi r$ increases as the radius grows.

Your turn9709/11 Oct/Nov 2023 Q3 [3 marks]

A cube has edge length $x$ cm and volume $V=x^3$ cm³. Its edges lengthen at $0.01\text{ cm/min}$ . Find the volume rate when $x=20$ .

Show worked answer

\begin{aligned}\frac{dV}{dt}&=\frac{dV}{dx}\frac{dx}{dt}\\&=3(20)^2(0.01)\end{aligned}

Answer

12\text{ cm}^3/\text{min}