Rates of Change

Video lesson · from 31:463 worked examples

One lesson video covers all of Differentiation, so it opens at 31:46 for this topic — not from the beginning.

Theory

The derivative of a function describes its rate of change. This can be evaluated for specific values by substituting these into the derivative.

Displacement, Velocity & Acceleration

The velocity ( $v$ ) of an object is defined as the rate of change of displacement ( $s$ ) with respect to time ( $t$ )

v(t) = s'(t)

The acceleration ( $a$ ) of an object is defined as the rate of change of velocity ( $v$ ) with respect to time ( $t$ )

a(t) = v'(t)

As we already know, the gradient of a straight line is constant. We can determine the gradient of a curve, at a particular point, by differentiating i.e. finding the rate of change.

⚠️ Common Examiner Traps

Differentiate with respect to the right variable: in a context question the letters change. If the formula is in terms of $t$ , you want $\frac{dV}{dt}$ , not $\frac{dy}{dx}$ .
Rate of change means the derivative, evaluated: substitute the given value in. An unsubstituted derivative does not answer the question.
Substituting negative numbers: brackets are essential when substituting negative values into a formula. Write $3(-2)^2$ , not $3-2^2$ .
Say what it means: if the rate is negative, the quantity is decreasing. Context questions usually want that interpretation, with units.

Worked examples

Example 1

Given $f(x) = 2x^5$ for $x \in \mathbb{R}$ , find the rate of change of $f$ when $x = 3$ .

First find the derivative:

f'(x) = 10x^4

Then substitute $x = 3$ :

f'(3) = 10(3)^4 = 10 \times 81 = 810

Example 2

A ball thrown so that its displacement $s$ after $t$ seconds is given by $s(t) = 12t - 5t^2$ . Find its velocity after 2 seconds.

Velocity is the derivative of displacement:

v(t) = s'(t) = 12 - 10t

Substitute $t = 2$ :

v(2) = 12 - 10(2) = 12 - 20 = -8

Velocity is -8.

Example 3

Find the gradient of the curve with equation $y = 3x^2$ at the point $(5, 2)$ .

To find the gradient, we need to find the rate of change (or derivative) and evaluate it at the x-coordinate of the point (which is $x = 5$ ).

\frac{dy}{dx} = 6x

When $x = 5$ :

m = 6(5) = 30