Tangents, rates of change and small changes · Applications of differentiation

The derivative measures how fast one quantity changes as another changes. This lesson uses that three ways: the gradient of a curve at a point, rates of change over time, and the effect of a small change.

Tangents and normals

At a point on a curve, the derivative gives the tangent’s gradient $m$ , and the normal’s gradient is $-\frac1m$ (see differentiation). When the curve is an equation in $x$ and $y$ , differentiate implicitly first.

Tangent and normalGradients multiply to −1

Worked example · WAEC 2018

WAEC 2018 · Paper 2 · Q10 (a)

Find the equation of the normal to the curve $x^2 + xy + 2y^2 = 8$ at the point $(-3, 1)$ .

Differentiate implicitly
- ${2x + \left(y + x\frac{dy}{dx}\right) + 4y\frac{dy}{dx} = 0}$ .
- Collect: ${(x + 4y)\frac{dy}{dx} = -(2x + y)}$ .
- So ${\frac{dy}{dx} = -\frac{2x + y}{x + 4y}}$ .
Think first. xy needs the product rule; 2y² gives 4y·dy/dx.
The gradient at (−3, 1)
- ${\frac{dy}{dx} = -\frac{-6 + 1}{-3 + 4} = -\frac{-5}{1} = 5}$ .
The normal
- The normal’s gradient is ${-\frac15}$ .
- ${y - 1 = -\frac15(x + 3)}$ .
- Multiply by 5: ${5y - 5 = -x - 3}$ .
- Rearrange: ${x + 5y - 2 = 0}$ .
Think first. The normal's gradient is −1/m.

More: tangents and normals

Connected rates of change

When a quantity changes with time, so do the quantities that depend on it. The chain rule links their rates:

dA/dt = dA/dr × dr/dt

from the formula × the rate you are given

Connected ratesThe derivative from the formula, times the rate you are given

Write the formula connecting the two quantities, differentiate it, then multiply by the rate you know.

Worked example · WAEC 2017

WAEC 2017 · Paper 2 · Q3 (b)

The radius of a circle is $6\text{ cm}$ . If the area of the circle is increasing at the rate of $20\text{ cm}^2\text{ s}^{-1}$ , find, leaving the answer in terms of $\pi$ , the rate at which the radius is increasing.

The formula
- ${A = \pi r^2}$ , so ${\frac{dA}{dr} = 2\pi r}$ .
Think first. Area of a circle? Differentiate it with respect to r.
The chain rule
- ${\frac{dA}{dt} = \frac{dA}{dr} \times \frac{dr}{dt}}$ .
- Substitute: ${20 = 2\pi(6) \times \frac{dr}{dt}}$ .
- So ${\frac{dr}{dt} = \frac{20}{12\pi} = \frac{5}{3\pi}}$ cm s⁻¹.
Think first. dA/dt = dA/dr × dr/dt.

Small changes and percentages

For a small change $\delta x$ , the curve and its tangent are almost the same, so $\delta y \approx \frac{dy}{dx}\,\delta x$ . When $y = kx^n$ , this gives a quick rule for percentages:

δy ≈ (dy/dx) × δx

y = kxⁿ: % change in y ≈ n × % change in x

for small changes the tangent is a good copy of the curve

Small changesy = kxⁿ: the percentage change is about n times as big

Compare the estimate with the exact change:

Small changesPick a shape, change the percentage

6%estimate: 3 × 2%6.121%exact change

% increase in the side = 2

V = x³, so the percentage change is about 3 × 2% = 6%. Exactly, it is (1 + 0.02)³ − 1 = 6.121%. For a small change the two agree closely.

Worked example · WAEC 2013

WAEC 2013 · Paper 2 · Q3

A side of a rectangle is three times the other. If the perimeter increases by $2\%$ , find the percentage increase in the area of the rectangle.

Write P and A in one letter
- ${P = 2(x + 3x) = 8x}$ .
- ${A = x \times 3x = 3x^2}$ .
Think first. Sides x and 3x. What are the perimeter and the area?
The change in x
- $P$ is a fixed multiple of $x$ , so $x$ also grows by 2%.
Think first. P is 8 times x. If P grows by 2%, how much does x grow?
The change in A
- ${A = 3x^2}$ has ${n = 2}$ .
- So $A$ grows by about ${2 \times 2\% = 4\%}$ .
- (Exactly, ${1.02^2 - 1 = 4.04\%}$ .)
Think first. A = 3x². What is n?

More: small changes

WAEC 2020 · Paper 1 · Q33The variables $x$ and $y$ are such that $y = 2x^3 - 2x^2 - 5x + 5$ . Calculate the corresponding change in $y$ when $x$ changes …

Your turn

WAEC 2018 · Paper 2 · Q9 (a)

(a)
The radius of a sphere increased by $2\frac12\%$ . Find the percentage increase in the volume.

Worked solution (try it first)

(a)

$V = \frac43\pi r^3$ , so for a small change $\dfrac{\delta V}{V} \approx 3\dfrac{\delta r}{r}$ .
With $\dfrac{\delta r}{r} = 2\frac12\%$ : $\dfrac{\delta V}{V} \approx 3 \times 2\frac12\%$
$= 7\frac12\%$ .
(Exactly, $1.025^3 = 1.0769$ , an increase of about $7.69\%$ .)

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