Tangents and Normals | CAIE 9709 Pure Mathematics 1 Notes

Pure Mathematics 1

Mathematics9709

Use one point and one numerical gradient

A tangent follows the curve's direction at the contact point. The passes through the same point at a right angle to the tangent.

At $P=(a,f(a))$ , the tangent gradient is the number $m=f'(a)$ . A line through $(x_1,y_1)$ has equation $y-y_1=m(x-x_1)$ : the vertical change is the gradient multiplied by the horizontal change.

For perpendicular lines with finite gradients, $m\,m_{\text{normal}}=-1$ . Thus the normal gradient is $-1/m$ when $m\ne0$ . Both lines below move through the same point.

Point position a1Show the normal

P = (1, 1)

Curve y = x²Tangent at PNormal at P

Tangent gradient: 2
Normal gradient: -0.5

The axes use equal scales, so the right angle is preserved. At a = 0 the tangent is horizontal and the normal is vertical.

Calculate the point before the line

Worked example

Find the tangent and normal to $y=x^2$ at $x=1$ .

The point is $(1,1)$ . Since $dy/dx=2x$ , the tangent gradient is $2$ and the normal gradient is $-1/2$ .

\begin{aligned}y-1&=2(x-1),\\y&=2x-1.\\[6pt]y-1&=-\tfrac12(x-1),\\y&=-\tfrac12x+\tfrac32.\end{aligned}

Both equations give $y=1$ at $x=1$ . Their gradients multiply to $-1$ . Do not use the whole function $2x$ as the gradient in a straight-line equation.

A zero gradient needs a different normal

If $f'(a)=0$ , the tangent is horizontal: $y=f(a)$ . The normal is vertical: $x=a$ . A vertical line has no finite gradient, so $-1/0$ is not a valid calculation. Set $a=0$ in the graph to see this case.

If a normal has a given non-zero gradient $n$ , reverse the relation: the tangent gradient must be $-1/n$ . Use that value in $f'(x)$ .

Keep exact working

Your turn9709/12 Feb/Mar 2024 Q5 [6 marks]

For $y=3/(2x^2-5)$ , find the normal at $(2,1)$ in the form $ax+by+c=0$ , with integer coefficients.

Show worked answer

Rewrite as $3(2x^2-5)^{-1}$ and apply the chain rule:

\frac{dy}{dx}=-12x(2x^2-5)^{-2}.

At $x=2$ , $m=-24/9=-8/3$ . Hence the normal gradient is $3/8$ .

\begin{aligned}y-1&=\tfrac38(x-2),\\8y-8&=3x-6,\\3x-8y+2&=0.\end{aligned}

The last line has the required form and still passes through $(2,1)$ .