Tangents and Normals to Curves

Three steps

A tangent at a point needs two things: the point, and the gradient there. The point comes from the curve, by putting x = a into y = f(x). The gradient comes from the derivative, by putting x = a into f'(x). Then the line through $\left(x_1,y_1\right)$(x₁, y₁) with gradient m is $y-y_1=m\left(x-x_1\right)$y − y₁ = m(x − x₁).

On $y=x^2$y = x² at x = 2: the point is (2, 4), and the derivative 2x gives the gradient 4. So the tangent is y − 4 = 4(x − 2), which is y = 4x − 4. Check by substituting the point: 4 × 2 − 4 = 4.

The normal

The normal is the line through the same point at right angles to the tangent. Perpendicular gradients multiply to −1, so if the tangent has gradient m, the normal has gradient $-\frac{1}{m}$−1/m: turn m upside down and change its sign.

At (2, 4) on $y=x^2$y = x² the normal has gradient $-\frac{1}{4}$−1/4, so it is $y-4=-\frac{x-2}{4}$y − 4 = −(x − 2)/4, which is $y=-\frac{x}{4}+4.5$y = −x/4 + 4.5. Check by substituting the point: −0.5 + 4.5 = 4. And $4\times \left(-\frac{1}{4}\right)=-1$4 × (−1/4) = −1.

A cubic

Take $y=x^3-2x$y = x³ − 2x at x = 2. The point is (2, 8 − 4) = (2, 4). The derivative is $3x^2-2$3x² − 2, so the gradient is 3 × 4 − 2 = 10; the chord from x = 2 to x = 2.001 gives 10.006001.

The tangent is y − 4 = 10(x − 2), which is y = 10x − 16. Check: 10 × 2 − 16 = 4. The tangent meets the curve again: $x^3-2x=10x-16$x³ − 2x = 10x − 16 rearranges to $x^3-12x+16=0$x³ − 12x + 16 = 0, which is ${\left(x-2\right)}^2\left(x+4\right)=0$(x − 2)²(x + 4) = 0. The double root x = 2 is the point of contact, and at x = −4 the curve and the tangent both have height −56.

The normal has gradient $-\frac{1}{10}$−1/10, so it is $y-4=-\frac{x-2}{10}$y − 4 = −(x − 2)/10, which is $y=-\frac{x}{10}+4.2$y = −x/10 + 4.2. Check: −0.2 + 4.2 = 4, and $10\times \left(-\frac{1}{10}\right)=-1$10 × (−1/10) = −1. Setting $x^3-2x=-\frac{x}{10}+4.2$x³ − 2x = −x/10 + 4.2 gives $x^3-1.9x-4.2=0$x³ − 1.9x − 4.2 = 0, which is $\left(x-2\right)\left(x^2+2x+2.1\right)=0$(x − 2)(x² + 2x + 2.1) = 0. The quadratic factor has discriminant 4 − 8.4, which is negative, so the normal meets the curve only at the point itself.

The curve $y=x^3-2x$y = x³ − 2x with its tangent y = 10x − 16 and its normal $y=-\frac{x}{10}+4.2$y = −x/10 + 4.2 at (2, 4). The tangent is steep and the normal nearly level, at right angles to it.

A root curve

Take $y=\sqrt{x}$y = √x at x = 4. The point is (4, 2). The derivative is $\frac{1}{2\sqrt{x}}$1/(2√x), so the gradient is $\frac{1}{4}$1/4; the chord with h = 0.001 gives 0.249984. The tangent is $y-2=\frac{x-4}{4}$y − 2 = (x − 4)/4, which is $y=\frac{x}{4}+1$y = x/4 + 1. Check: 1 + 1 = 2.

The normal has gradient −4, so it is y − 2 = −4(x − 4), which is y = −4x + 18. Check: −16 + 18 = 2. The tangent crosses the x-axis where $\frac{x}{4}+1=0$x/4 + 1 = 0, at x = −4, and the normal where −4x + 18 = 0, at x = 4.5.

The curve $y=\sqrt{x}$y = √x with its tangent $y=\frac{x}{4}+1$y = x/4 + 1 and its normal y = −4x + 18 at (4, 2). The tangent crosses the x-axis at −4, and the normal at 4.5.

When there is no reciprocal

If f'(a) = 0, the tangent is level, the line y = f(a), and $-\frac{1}{m}$−1/m cannot be worked out because there is no dividing by 0. The normal is then vertical, the line x = a. On $y=x^3-3x$y = x³ − 3x at x = 1, the derivative $3x^2-3$3x² − 3 is 0, the point is (1, −2), the tangent is y = −2 and the normal is x = 1.

If f'(a) does not exist because the curve stands vertical there, the roles swap: the tangent is the vertical line x = a and the normal is level, y = f(a). On $y=\sqrt[3]{x}$y = ∛x at x = 0 the tangent is the y-axis, x = 0, and the normal is the x-axis, y = 0.

The curve $y=\frac{x^3}{3}-x$y = x³/3 − x, whose derivative is $x^2-1$x² − 1. At x = 0.5 the point is about (0.5, −0.458), the tangent has gradient −0.75 and the normal $\frac{4}{3}$4/3, about 1.33. Drag the point P to x = 1 or x = −1: the tangent is level and the normal has no gradient.

The usual mistakes

Changing only the sign. A tangent of gradient 4 has a normal of gradient $-\frac{1}{4}$−1/4, not −4; the check is that the two multiply to −1, and 4 × (−4) = −16.

Writing the tangent through the origin. y = 4x has the right gradient but misses (2, 4); the line must pass through the point, which gives y = 4x − 4.

Getting the intercept's sign wrong. y = 4x + 4 gives 12 at x = 2, not 4. Substituting the point into the finished line catches both slips.

Taking the y-value from the derivative. The height of the point comes from the curve: on $y=x^3-2x$y = x³ − 2x at x = 2 it is 4, while the gradient there is 10.

A solar mirror

In the application below, the cross-section of a mirror is the curve $y=\frac{x^2}{8}$y = x²/8. The tangent grazes the mirror where a strut is bolted, the strut lies along the normal, and putting x = 0 into the normal finds where the strut's line crosses the mirror's axis.