4.4 - Tangents and normals to curves

4.4 - Tangents and normals to curves

Turn a curve's local gradient into the equation of a straight line. You will differentiate to obtain tangent gradients, use perpendicular gradients for normals and handle the special case of a horizontal tangent with a vertical normal.

Writing the tangent equation

Can a curve have a perfectly ordinary tangent but a normal whose gradient cannot be written as a number? We will reach that case after building the usual tangent and normal methods.

You already know

The derivative gives the tangent gradient at a point. A line through (a,b)(a,b) with finite gradient mm has equation y−b=m(x−a)y-b=m(x-a). This straight-line equation is on the formulae sheet.

Use the curve equation to work out the point's height, and the derivative to work out its tangent gradient. A point and a gradient then determine the line. The differentiation step is not on the formulae sheet: learn it.

Worked example 1: point and tangent gradient

Question. A curve has equation y=x2−2x+3y=x^2-2x+3. Work out the equation of the tangent at x=2x=2, in the form y=mx+cy=mx+c.

Plan. Work out the height and derivative at x=2x=2, then use the point-gradient line equation.

y(2)=4−4+3=3point is (2,3)dydx=2x−2differentiatem=2(2)−2=2gradient at x=2y−3=2(x−2)line through the pointy=2x−1requested form\begin{aligned} y(2)&=4-4+3=3 &&\text{point is }(2,3)\\ \frac{\mathrm{d}y}{\mathrm{d}x}&=2x-2 &&\text{differentiate}\\ m&=2(2)-2=2 &&\text{gradient at }x=2\\ y-3&=2(x-2) &&\text{line through the point}\\ y&=2x-1 &&\text{requested form} \end{aligned}

Why this step? The height 33 and gradient 22 come from different functions; both are needed to locate the tangent.

Answer. y=2x−1y=2x-1.

Check. At x=2x=2, the line gives y=3y=3, and its gradient 22 equals the curve's gradient there.

Change only the sign of the input, from 22 to −2-2. Both the point's height and its tangent gradient must be recalculated.

Worked example 2: a different point on the same curve

Question. For y=x2−2x+3y=x^2-2x+3, work out the tangent at x=−2x=-2, in the form y=mx+cy=mx+c.

Plan. Substitute −2-2 into the curve and its derivative, then build the line.

y(−2)=4+4+3=11point is (−2,11)dydx=2x−2same gradient functionm=2(−2)−2=−6new inputy−11=−6(x+2)subtract the negative coordinatey=−6x−1expand and collect\begin{aligned} y(-2)&=4+4+3=11 &&\text{point is }(-2,11)\\ \frac{\mathrm{d}y}{\mathrm{d}x}&=2x-2 &&\text{same gradient function}\\ m&=2(-2)-2=-6 &&\text{new input}\\ y-11&=-6(x+2) &&\text{subtract the negative coordinate}\\ y&=-6x-1 &&\text{expand and collect} \end{aligned}

Why this step? In x−ax-a, the coordinate a=−2a=-2 gives x−(−2)=x+2x-(-2)=x+2, not x−2x-2.

Answer. y=−6x−1y=-6x-1.

Check. At x=−2x=-2, the line gives 12−1=1112-1=11, with the required gradient −6-6.

Here is a useful exact fact about a parabola. The tangent to y=x2y=x^2 at (a,a2)(a,a^2) is y=2ax−a2y=2ax-a^2, so the vertical gap between the curve and that line is

x2−(2ax−a2)=(x−a)2x^2-(2ax-a^2)=(x-a)^2

At horizontal distance hh from contact, the curve is exactly h2h^2 above its tangent, wherever the contact point is. This fact is specific to this parabola; tangents to other curves can cross their curves.

Exam tip

Check which line the question asks for before writing its gradient. Using the negative reciprocal gives the normal, even if all your later straight-line work is correct.

The matched question uses a new quadratic and input; give the equation, not just the gradient.

Writing the normal equation

A normal is the line through the point of contact perpendicular to the tangent. It passes through the same point as the tangent, but usually has a different gradient.

[DIAGRAM: asset_name: A tangent and its perpendicular normal; asset_slug: l2fm-tangent-normal; caption: The tangent and normal both pass through P and their gradients multiply to minus 1; description: Equal-scale axes show the parabola y equals x squared and P at 1 comma 1. A sage tangent y equals 2x minus 1 and a clay normal y equals minus one half x plus three halves meet at P at a right angle. Separate labels show tangent gradient 2 and normal gradient minus one half. The curve and both line equations are identified]
The tangent and normal both pass through P and their gradients multiply to minus 1

The perpendicular-gradient relationship applies when both gradients are finite and non-zero. Remember the negative sign as well as the reciprocal.

Normal gradient

If the tangent gradient at a point is a finite non-zero number mm, the normal gradient is

mnormal=−1mm_{\mathrm{normal}}=-\frac1m

The normal passes through the same point on the curve.
Not on the formulae sheet: learn it.

Use this gradient in the same straight-line equation as before. The point's coordinates stay unchanged.

Worked example 3: a fractional normal gradient

Question. Work out the normal to y=x2y=x^2 at x=1x=1, in the form y=mx+cy=mx+c.

Plan. Work out the tangent gradient, take its negative reciprocal and use the point (1,1)(1,1).

y(1)=1height of the contact pointdydx=2x⟹mtangent=2differentiate and substitutemnormal=−12negative reciprocaly−1=−12(x−1)line through (1,1)y=−12x+32collect constants\begin{aligned} y(1)&=1 &&\text{height of the contact point}\\ \frac{\mathrm{d}y}{\mathrm{d}x}&=2x\Longrightarrow m_{\mathrm{tangent}}=2 &&\text{differentiate and substitute}\\ m_{\mathrm{normal}}&=-\frac12 &&\text{negative reciprocal}\\ y-1&=-\frac12(x-1) &&\text{line through }(1,1)\\ y&=-\frac12x+\frac32 &&\text{collect constants} \end{aligned}

Why this step? The derivative supplies the tangent gradient, so an extra perpendicular-gradient step is required for the normal.

Answer. y=−12x+32y=-\frac12x+\frac32.

Check. The gradients multiply to 2(−1/2)=−12(-1/2)=-1; the line gives y=1y=1 at x=1x=1.

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