1.07m - Tangents and normals to curves

1.07m - Tangents and normals to curves

Use a derivative and a curve point to construct tangent and normal equations in 1.07m. You will handle horizontal and vertical exceptions, then work backwards from a specified direction to all possible points of contact.

A tangent needs a point and a gradient

How do you write the straight line which follows a curve's direction at one chosen point? For y=x3−2x+1y=x^3-2x+1 at x=2x=2, the curve point is (2,5)(2,5) and the gradient is 1010. Those two facts determine the tangent completely.

You already know

A line through (a,b)(a,b) with finite gradient mm has equation y−b=m(x−a)y-b=m(x-a). Differentiate a curve first, then substitute the input to obtain its numerical tangent gradient.

The input of the contact point is fixed. The coordinate xx in the line equation is free to move along the resulting straight line.

Tangent equation

For y=f(x)y=\mathrm f(x), at P=(a,f(a))P=(a,\mathrm f(a)) where a finite derivative exists,

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

Not in the formula sheet: learn it.

Find the point and slope separately before combining them. This also gives you two quick ways to check the result.

Worked example 1: a tangent to a cubic

Question. Find the tangent to y=x3−2x+1y=x^3-2x+1 at x=2x=2.

Plan. Evaluate the curve, evaluate its derivative, then use the point-gradient equation.

y(2)=8−4+1=5find the contact heightdydx=3x2−2differentiatem=3(2)2−2=10evaluate the slopey−5=10(x−2)use the fixed pointy=10x−15simplify\begin{aligned} y(2)&=8-4+1=5 && \text{find the contact height}\\ \frac{\mathrm dy}{\mathrm dx}&=3x^2-2 && \text{differentiate}\\ m&=3(2)^2-2=10 && \text{evaluate the slope}\\ y-5&=10(x-2) && \text{use the fixed point}\\ y&=10x-15 && \text{simplify} \end{aligned}

Why this step? The derivative gives the direction; the curve gives the point. Both are needed to select this particular line.

Answer. y=10x−15y=10x-15.

Check. At x=2x=2, the line gives 55, and its gradient is 1010.

Change only the contact input, keeping the same cubic. Its local direction need not be the same elsewhere.

Worked example 2: a different contact point

Question. Find the tangent to the same curve at x=−1x=-1.

Plan. Reuse the derivative, but recalculate both the point and slope.

y(−1)=−1+2+1=2new curve pointm=3(−1)2−2=1new tangent gradienty−2=x+1point-gradient formy=x+3simplify\begin{aligned} y(-1)&=-1+2+1=2 && \text{new curve point}\\ m&=3(-1)^2-2=1 && \text{new tangent gradient}\\ y-2&=x+1 && \text{point-gradient form}\\ y&=x+3 && \text{simplify} \end{aligned}

Why this step? A different point requires a new gradient evaluation, even though the derivative formula is unchanged.

Answer. y=x+3y=x+3.

Check. Substituting x=−1x=-1 gives y=2y=2, and the line's gradient is 11.

If coordinates are supplied, check they actually lie on the curve. A correct slope through an incorrect point produces a parallel line.

Watch out

Wrong: y−5=(3x2−2)(x−2)y-5=(3x^2-2)(x-2) for the tangent at x=2x=2. Right: evaluate the derivative first, giving the constant gradient 1010. A changing gradient does not describe a straight line.

Use a fresh point on the same cubic.

The normal is perpendicular to the tangent

A curve has another useful line at a point: its normal. The normal uses the same contact point but turns the direction through a right angle.

Normal

A normal to a curve at a point is the line through that point perpendicular to the tangent. It is not a line through the origin unless the geometry happens to require that.

The graph uses equal scales, so the perpendicular directions look perpendicular as well as satisfying the algebraic test.

Tangent and normal to the parabola at the same point

Normal with a finite gradient

If the tangent gradient m=f′(a)m=\mathrm f'(a) is finite and nonzero, the normal has gradient

n=−1m,y−f(a)=−1m(x−a).n=-\frac1m,\qquad y-\mathrm f(a)=-\frac1m(x-a).

The condition m≠0m\ne0 is essential.

Not in the formula sheet: learn it.

Use the negative reciprocal, not just a sign change. For example, tangent gradient −3/4-3/4 gives normal gradient 4/34/3; their product is −1-1.

Worked example 3: a normal to a parabola

Question. Find the normal to y=x2y=x^2 at (1,1)(1,1).

Plan. Find the tangent gradient, turn it into a normal gradient, then use the contact point.

dydx=2xdifferentiate the curvem=2,n=−12evaluate; take negative reciprocaly−1=−12(x−1)line through the contact pointy=−12x+32simplify\begin{aligned} \frac{\mathrm dy}{\mathrm dx}&=2x && \text{differentiate the curve}\\ m&=2,\qquad n=-\frac12 && \text{evaluate; take negative reciprocal}\\ y-1&=-\frac12(x-1) && \text{line through the contact point}\\ y&=-\frac12x+\frac32 && \text{simplify} \end{aligned}

Why this step? The derivative belongs to the tangent; the question asks for its perpendicular line.

Answer. y=−12x+32y=-\tfrac12x+\tfrac32.

Check. It contains (1,1)(1,1) and 2(−1/2)=−12(-1/2)=-1.

Change only the sign of the contact input. The normal gradient changes sign too.

Worked example 4: a normal on the other side

Question. Find the normal to y=x2y=x^2 at x=−1x=-1.

Plan. Recalculate the tangent gradient and use the unchanged height 11.

P=(−1,1)evaluate the curvem=2(−1)=−2,n=12negative reciprocaly−1=12(x+1)use Py=12x+32simplify\begin{aligned} P&=(-1,1) && \text{evaluate the curve}\\ m&=2(-1)=-2,\qquad n=\frac12 && \text{negative reciprocal}\\ y-1&=\frac12(x+1) && \text{use }P\\ y&=\frac12x+\frac32 && \text{simplify} \end{aligned}

Why this step? The normal's slope must reverse sign when the parabola is reflected in the vertical axis.

Answer. y=12x+32y=\tfrac12x+\tfrac32.

Check. At x=−1x=-1, the line gives 11, and (−2)(1/2)=−1(-2)(1/2)=-1.

A normal can meet the curve again. The first example's normal crosses y=x2y=x^2 at x=−3/2x=-3/2 as well as 11, since 2x2+x−3=(x−1)(2x+3)2x^2+x-3=(x-1)(2x+3). Perpendicular at one point does not mean “never intersects elsewhere”.

Exam tip

Label tangent and normal gradients separately before writing the line. Check their product is −1-1, then check the line contains the curve point, not a point copied from a different line.

Move the contact input once more.

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