9F - Tangents and normals

Many straight lines pass through one point on a curve. You will use a derivative to choose the tangent, find its perpendicular normal, and locate points where a tangent or normal satisfies a given slope condition.

Equation of a tangent

Which line through $(0,1)$ follows the direction of the curve $y=e^x$ at that point?

The coordinates alone do not decide. Both $y=x+1$ and $y=2x+1$ pass through $(0,1)$ , but their gradients differ. The derivative of the curve supplies the missing gradient.

Tangent

At a point where a curve has a finite derivative, its tangent is the straight line through that point with gradient equal to the derivative there.

A tangent need not meet the curve only once; it can meet it again or cross it. Its defining feature here is its direction at the chosen point.

Tangent equation

For $y=f(x)$ , if $f(a)$ and the finite derivative $f'(a)$ exist, then the tangent at $(a,f(a))$ has equation

y-f(a)=f'(a)(x-a).

This uses the point-gradient form of a straight line.

Not on the formulae sheet: learn it.

Use the original function to find the point, and its derivative to find the gradient. Keep those two jobs distinct.

Worked example 1: a tangent to an exponential

Question. Find the equation of the tangent to $y=e^x$ at $x=0$ .

Plan. Find the point and derivative value, then use point-gradient form.

\begin{aligned} y(0)&=e^0=1 &&\text{point on the curve}\\ \frac{dy}{dx}&=e^x &&\text{differentiate}\\ m&=e^0=1 &&\text{evaluate the derivative}\\ y-1&=1(x-0) &&\text{line through }(0,1) \end{aligned}

Answer. $y=x+1$ .

Check. The line contains $(0,1)$ . At $x=0.001$ , the original curve gives approximately $1.001000500$ , while the line gives $1.001$ ; they agree closely near the point. The alternative line $y=2x+1$ gives $1.002$ , with the wrong local direction.

Change only the exponential's exponent from $x$ to $2x$ . The point at $x=0$ stays the same, but the chain factor changes the tangent's slope.

Worked example 2: the same point with a different slope

Question. Find the equation of the tangent to $y=e^{2x}$ at $x=0$ .

Plan. Include the inner derivative $2$ , then use the same point-gradient method.

\begin{aligned} y(0)&=e^0=1 &&\text{find the point}\\ \frac{dy}{dx}&=2e^{2x} &&\text{use the chain rule}\\ m&=2e^0=2 &&\text{evaluate at }x=0\\ y-1&=2(x-0) &&\text{form the tangent} \end{aligned}

Answer. $y=2x+1$ .

Check. The line contains $(0,1)$ . At $x=0.001$ , the curve is approximately $1.002002001$ and the line is $1.002$ , consistent with the calculated local slope.

The opening curve therefore selects $y=x+1$ . Sharing a point did not make the two candidate lines interchangeable.

Watch out

The gradient is $f'(a)$ , not $f(a)$ . The two numbers happen to agree in the first example, but are $2$ and $1$ respectively in the second.

Use a new chain factor in the next tangent calculation.

Equation of a normal

The normal passes through the same point as the tangent but meets it at a right angle.

Normal

A normal to a curve at a point is the line through that point perpendicular to its tangent, when the tangent is well defined.

For two perpendicular lines with finite gradients, the gradients multiply to $-1$ . Thus, if the tangent has finite nonzero gradient $m$ , the normal has gradient $-1/m$ .

[DIAGRAM: asset_name: Sloping and horizontal tangents with their normals; asset_slug: tangent-normal-cases; caption: A sloping tangent has a negative-reciprocal normal, while a horizontal tangent has a vertical normal; description: Two equal-scale coordinate panels show y equals x squared plus 1. The left marks P(1,2), tangent y equals 2x and normal y equals negative one half x plus five halves, with a right-angle marker. The right marks Q(0,1), horizontal tangent y equals 1 and vertical normal x equals 0, with a right-angle marker. The curve is grey, tangents pale blue-accent lines and normals sage-accent lines.]
A sloping tangent has a negative-reciprocal normal, while a horizontal tangent has a vertical normal

The negative-reciprocal calculation requires $m\ne0$ . If the tangent is horizontal, its perpendicular normal is vertical, so write an equation $x=a$ , rather than trying to divide by zero.

Worked example 3: a normal with a finite gradient

Question. Find the equation of the normal to $y=x^2+1$ at $x=1$ . Give the answer with integer coefficients in the form $ax+by+c=0$ .

Plan. Find the tangent gradient, take its negative reciprocal, then form the line through the curve point.

\begin{aligned} y(1)&=2 &&\text{point }(1,2)\\ \frac{dy}{dx}&=2x &m_{\mathrm{t}}&=2\\ m_{\mathrm{n}}&=-\tfrac12 &&\text{negative reciprocal}\\ y-2&=-\tfrac12(x-1) &&\text{normal through }(1,2)\\ x+2y-5&=0 &&\text{clear the fraction and rearrange} \end{aligned}

Answer. $x+2y-5=0$ .

Check. Substituting $(1,2)$ gives $1+4-5=0$ . The line's gradient is $-1/2$ , and $2(-1/2)=-1$ , so it is perpendicular to the tangent.

Keep the curve and change the point to $x=0$ . This gives the exceptional case that the reciprocal formula cannot handle.

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9F - Tangents and normals

Equation of a tangent

Equation of a normal

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