CA4 - Tangents and normals to curves

CA4 - Tangents and normals to curves

Differentiation supplies the direction of a curve at a point. CA4 combines that direction with familiar straight-line geometry to find tangents and normals, including points where the tangent is horizontal. You will also work backwards from a line's direction to find where it can be tangent to a curve.

Equations of tangents

A tangent matches a curve's direction at one point. Must it stay on one side of the curve everywhere? For y=x3y=x^3, the tangent at (1,1)(1, 1) turns out to meet the curve again. We will find that line first, then discover where the second meeting happens.

You already know

For y=f(x)y=\mathrm{f}(x), the tangent gradient at x=ax=a is f′(a)\mathrm{f}'(a). A line of gradient mm through (a,b)(a, b) has equation y−b=m(x−a)y-b=m(x-a).

Think of the calculation as three distinct jobs: find the point, find the gradient, then place the line. The derivative is usually an expression involving xx; the tangent gradient is one number obtained from it.

Tangent equation

y−f(a)=f′(a)(x−a)y-\mathrm{f}(a)=\mathrm{f}'(a)(x-a)

This is the tangent to the differentiable curve y=f(x)y=\mathrm{f}(x) at x=ax=a. Find the height from f\mathrm{f} and the gradient from f′\mathrm{f}'.
Not on the formulae sheet: learn it.

A gradient alone leaves the line free to slide up or down. Substituting the point fixes its position.

Worked example 1: a tangent at a positive coordinate

Question. Determine the tangent to y=x2+2x+3y=x^2+2x+3 at x=1x=1.

Plan. Calculate the point's height, differentiate, evaluate the gradient and use point-gradient form.

y(1)=1+2+3=6point on the curvedydx=2x+2gradient functionm=2(1)+2=4gradient at the pointy−6=4(x−1)line through (1,6)y=4x+2rearrange\begin{aligned} y(1)&=1+2+3=6 && \text{point on the curve}\\ \frac{\mathrm{d}y}{\mathrm{d}x}&=2x+2 && \text{gradient function}\\ m&=2(1)+2=4 && \text{gradient at the point}\\ y-6&=4(x-1) && \text{line through }(1,6)\\ y&=4x+2 && \text{rearrange} \end{aligned}

Why evaluate before forming the line? The line needs the fixed number 44 as its gradient, not the varying expression 2x+22x+2.

Answer. y=4x+2y=4x+2.

Check. At x=1x=1, the line gives y=6y=6. Subtracting it from the curve gives x2+2x+3−(4x+2)=(x−1)2x^2+2x+3-(4x+2)=(x-1)^2, with a repeated intersection at the contact point.

Change only the contact coordinate from 11 to −2-2. Both the point and gradient must be recalculated.

Worked example 2: a tangent at a negative coordinate

Question. Determine the tangent to y=x2+2x+3y=x^2+2x+3 at x=−2x=-2.

Plan. Repeat the point-and-gradient calculation at the new coordinate.

y(−2)=4−4+3=3new heightm=2(−2)+2=−2evaluate the derivativey−3=−2(x+2)through (−2,3)y=−2x−1rearrange\begin{aligned} y(-2)&=4-4+3=3 && \text{new height}\\ m&=2(-2)+2=-2 && \text{evaluate the derivative}\\ y-3&=-2(x+2) && \text{through }(-2,3)\\ y&=-2x-1 && \text{rearrange} \end{aligned}

Why x+2x+2? It is x−(−2)x-(-2), which vanishes at the contact point.

Answer. y=−2x−1y=-2x-1.

Check. The point (−2,3)(-2, 3) satisfies the line. Curve minus line is x2+4x+4=(x+2)2x^2+4x+4=(x+2)^2, again with the required repeated root.

Write the final line explicitly. OCR's tangent questions reward the derivative, its evaluation and a line through the point as different steps.

Watch out

Wrong: “The tangent is y=2x+2y=2x+2.” That copies the gradient function as a line equation. Right: evaluate the derivative at the contact coordinate, then use that numerical gradient and the point.

Now move the contact coordinate again.

Equations of normals

The normal passes through the contact point at right angles to the tangent. It shares the tangent's point, but its direction is different.

For example, a tangent gradient of 44 gives normal gradient −14-\frac14, since their product is −1-1. If the tangent is horizontal, the normal is vertical; dividing by the zero tangent gradient is not an available shortcut.

Normal equation

y−f(a)=−1f′(a)(x−a)y-\mathrm{f}(a)=-\frac{1}{\mathrm{f}'(a)}(x-a)

This applies when f′(a)≠0\mathrm{f}'(a)\ne0. When f′(a)=0\mathrm{f}'(a)=0, the tangent is y=f(a)y=\mathrm{f}(a) and the normal is x=ax=a.
Not on the formulae sheet: learn it.

Use the derivative to find the tangent direction first. Only then take the negative reciprocal for the normal.

Worked example 3: a normal to a quadratic

Question. Find the normal to y=x2+1y=x^2+1 at x=1x=1.

Plan. Find the point and tangent gradient, then use the perpendicular gradient through the same point.

y(1)=2point’s heightmT=2(1)=2differentiate and evaluatemN=−12negative reciprocaly−2=−12(x−1)normal through (1,2)x+2y=5clear the fraction\begin{aligned} y(1)&=2 && \text{point's height}\\ m_T&=2(1)=2 && \text{differentiate and evaluate}\\ m_N&=-\frac12 && \text{negative reciprocal}\\ y-2&=-\frac12(x-1) && \text{normal through }(1,2)\\ x+2y&=5 && \text{clear the fraction} \end{aligned}

Why keep the same point? Tangent and normal describe two directions at one contact point; the normal is not placed at an axis intercept instead.

Answer. x+2y=5x+2y=5.

Check. (1,2)(1, 2) satisfies the line, and 2×(−12)=−12\times(-\frac12)=-1.

Change only the contact coordinate to −1-1. The tangent's slope changes sign, so the normal's slope does too.

Worked example 4: a positive normal gradient

Question. Find the normal to y=x2+1y=x^2+1 at x=−1x=-1.

Plan. Evaluate the derivative at −1-1, take its negative reciprocal and use the point's height.

y(−1)=2height on the curvemT=2(−1)=−2tangent gradientmN=12negative reciprocaly−2=12(x+1)normal through (−1,2)2y=x+5rearrange\begin{aligned} y(-1)&=2 && \text{height on the curve}\\ m_T&=2(-1)=-2 && \text{tangent gradient}\\ m_N&=\frac12 && \text{negative reciprocal}\\ y-2&=\frac12(x+1) && \text{normal through }(-1,2)\\ 2y&=x+5 && \text{rearrange} \end{aligned}

Why is mNm_N positive? It must multiply the negative tangent gradient to give −1-1.

Answer. 2y=x+52y=x+5.

Check. At x=−1x=-1, the line gives y=2y=2, and (−2)(12)=−1(-2)(\frac12)=-1.

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