QCE Specialist Mathematics - Unit 4 - Rates of change and differential equations

Rates of Change | QCE Specialist Mathematics

Learn QCE Specialist implicit differentiation and related rates for changing geometry, motion and applied modelling.

Updated 2026-05-18 - 5 min read

Rates of change questions ask how fast one quantity changes when it is linked to another changing quantity. The main tools are implicit differentiation and the chain rule.

Implicit differentiation

An implicit relation connects $x$ and $y$ without necessarily making $y$ the subject. For example:

$ x^2+y^2=25. $

When differentiating with respect to $x$, treat $y$ as a function of $x$. That means:

$ \frac{d}{dx}(y^2)=2y\frac{dy}{dx}. $

So:

$ 2x+2y\frac{dy}{dx}=0 $

and

$ \frac{dy}{dx}=-\frac{x}{y}. $

Product terms

Terms like $xy$ need the product rule:

$ \frac{d}{dx}(xy)=x\frac{dy}{dx}+y. $

For example, if

$ x^2+4xy+2y^2=10, $

then:

$ 2x+4\left(x\frac{dy}{dx}+y\right)+4y\frac{dy}{dx}=0. $

Group the derivative terms:

$ (4x+4y)\frac{dy}{dx}=-(2x+4y) $

so:

$ \frac{dy}{dx}=-\frac{2x+4y}{4x+4y}. $

To find the gradient at a point, substitute the point after finding $\frac{dy}{dx}$. If the denominator becomes $0$, the tangent may be vertical or the derivative may be undefined at that point, so do not force a finite gradient.

Once the gradient is known, tangent and normal equations are ordinary line equations. If a point is $(x_1,y_1)$ and the tangent gradient is $m$, then:

$ y-y_1=m(x-x_1). $

The normal gradient is $-\frac1m$ when $m\ne0$. If the tangent is horizontal, the normal is vertical. If the tangent is vertical, the normal is horizontal. These edge cases are a common place to lose a mark by trying to force a reciprocal that does not exist.

Implicit differentiation is also useful when a parameter is involved. If $x=x(t)$ and $y=y(t)$, then:

$ \frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}. $

For example, if a moving point has coordinates $P(t)=(\sqrt{t},\sin t)$, then:

$ \frac{dx}{dt}=\frac{1}{2\sqrt t},\quad \frac{dy}{dt}=\cos t. $

The gradient of its path at time $t$ is:

$ \frac{dy}{dx}=\frac{dy/dt}{dx/dt}=2\sqrt t\cos t. $

Related rates

Related rates problems connect variables through a formula, then differentiate with respect to time.

For a sphere:

$ V=\frac43\pi r^3. $

Differentiate with respect to $t$:

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

Now if you know any two of $\frac{dV}{dt}$, $r$ and $\frac{dr}{dt}$, you can find the third.

Cones, pyramids and spheres appear often because their surface area and volume formulas connect dimensions. If two dimensions change together, use the geometry of similar shapes to reduce the number of variables before differentiating. For example, in a cone with a fixed shape, $h$ might be a constant multiple of $r$.

Related rates workflow

Most errors happen before the calculus. Use this order:

1. Draw or name the variables.
2. Write the equation connecting them.
3. Remove extra variables using fixed relationships.
4. Differentiate with respect to time.
5. Substitute the values at the instant in question.

Signs matter. If a radius is decreasing, $\frac{dr}{dt}$ is negative. If water height is rising, $\frac{dh}{dt}$ is positive. The sign should match the story, not just the algebra.

Angle-of-elevation rates

Trigonometric related-rates questions usually begin with a right triangle. Suppose a plane flies horizontally at height $h$ and its horizontal distance from an observer is $x(t)$. If the angle of elevation is $\theta(t)$, then:

$ \tan\theta=\frac{h}{x}. $

Differentiate with respect to time:

$ \sec^2\theta\frac{d\theta}{dt} =-\frac{h}{x^2}\frac{dx}{dt}. $

The answer for $\frac{d\theta}{dt}$ is in radians per unit time, because calculus derivatives of trigonometric functions assume radian measure.

You can substitute either a known angle or a known distance after differentiating. If the question says the angle is $45^\circ$, then $\theta=\frac{\pi}{4}$ and $x=h$. If it says the plane has been flying for $t$ seconds at constant horizontal speed, express $x$ in terms of $t$ before substituting.

Geometry and rates together

Some related-rates questions combine similar triangles with a volume or area formula. For a cone of water, the formula:

$ V=\frac13\pi r^2h $

has two changing variables, $r$ and $h$. If the cone keeps the same shape, similar triangles might give:

$ r=\frac{2}{5}h. $

Substitute before differentiating:

$ V=\frac13\pi\left(\frac{2h}{5}\right)^2h =\frac{4\pi}{75}h^3. $

Then:

$ \frac{dV}{dt}=\frac{4\pi}{25}h^2\frac{dh}{dt}. $

This avoids differentiating an equation with both $\frac{dr}{dt}$ and $\frac{dh}{dt}$ when only one rate is wanted.

For surface-area questions, be careful about what is actually changing. A balloon sphere might need:

$ S=4\pi r^2,\quad V=\frac43\pi r^3. $

If volume rate is given and surface-area rate is asked, first use:

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

to find $\frac{dr}{dt}$ at that instant, then substitute into:

$ \frac{dS}{dt}=8\pi r\frac{dr}{dt}. $

Quick check

Sources

• QCAA Specialist Mathematics subject page
• QCAA Specialist Mathematics 2025 syllabus
• OpenStax Calculus Volume 1: Implicit Differentiation
• OpenStax Calculus Volume 1: Related Rates