Tangents and Normals

The gradient at the point

The derivative gives the gradient of a curve at every point. For $y=x^2$y = x² the derivative is $\frac{dy}{dx}=2x$dy/dx = 2x, so at the point where x = 2 the gradient is 2 × 2 = 4. The tangent there is the straight line through that point with the same gradient, 4.

Check it with a short chord. From x = 2 to x = 2.01 the curve rises from 4 to $2.01^2=4.0401$2.01² = 4.0401, so the chord gradient is (4.0401 − 4) ÷ 0.01 = 4.01, just above 4.

The curve $y=x^2$y = x² and its tangent at the point (2, 4). From that point, a step of 1 square across and 4 squares up lands on the tangent again at (3, 8): the gradient is 4, the value of 2x at x = 2.

The equation of the tangent

A straight line is fixed by its gradient and one point on it. The tangent has gradient m = 4, and it passes through the point on the curve, (2, 4). A line with gradient m through $\left(x_1,y_1\right)$(x₁, y₁) is $y-y_1=m\left(x-x_1\right)$y − y₁ = m(x − x₁), so the tangent is y − 4 = 4(x − 2). Multiply out: y = 4x − 8 + 4, which is y = 4x − 4.

Check it by substituting the point: at x = 2 the line gives 4 × 2 − 4 = 4, the height of the curve. The same working in the form y = mx + c gives 4 = 4 × 2 + c, so c = −4.

The y-value of the point always comes from the curve, never from the derivative. At x = 2 the curve has height $2^2=4$2² = 4 and gradient 2 × 2 = 4, two different quantities that happen to be equal here. At x = 3 they differ: the height is 9, the gradient is 6, and the tangent is y − 9 = 6(x − 3), which is y = 6x − 9.

A tangent can meet the curve again

On $y=x^3$y = x³ the derivative is $3x^2$3x². At (1, 1) the gradient is $3\times 1^2=3$3 × 1² = 3, so the tangent is y − 1 = 3(x − 1), which is y = 3x − 2. A chord from x = 1 to x = 1.01 has gradient (1.030301 − 1) ÷ 0.01 = 3.0301, close to 3.

This tangent touches the curve at (1, 1), and it also crosses the curve further left. Subtract the line from the curve: $x^3-\left(3x-2\right)=x^3-3x+2={\left(x-1\right)}^2\left(x+2\right)$x³ − (3x − 2) = x³ − 3x + 2 = (x − 1)²(x + 2). The squared factor ${\left(x-1\right)}^2$(x − 1)² is the touch at x = 1. The factor (x + 2) is a second meeting at x = −2, where the curve and the line both have height −8.

So a tangent is defined by touching the curve at one point with the same gradient. It does not have to avoid the curve everywhere else.

The curve $y=x^3$y = x³ and its tangent y = 3x − 2 at (1, 1). The line touches the curve at (1, 1) and crosses it at (−2, −8).

Perpendicular gradients multiply to −1

The normal to a curve at a point is the straight line through that point at right angles to the tangent. Its gradient comes from the tangent's gradient.

A line of gradient 4 goes 1 across and 4 up. Turn that step through a right angle, counterclockwise: the 1 across becomes 1 up, and the 4 up becomes 4 to the left. The turned line goes −4 across and 1 up, so its gradient is $1÷\left(-4\right)=-\frac{1}{4}$1 ÷ (−4) = −1/4.

The same turn works for any gradient m: 1 across and m up becomes m to the left and 1 up, a gradient of $-\frac{1}{m}$−1/m. This is the negative reciprocal of m: turn the gradient upside down and change its sign. The two gradients multiply to $m\times \left(-\frac{1}{m}\right)=-1$m × (−1/m) = −1, and here $4\times \left(-\frac{1}{4}\right)=-1$4 × (−1/4) = −1.

The steep line y = 4x goes 1 across and 4 up. Turned through a right angle, that step goes 1 up and 4 to the left, along the line $y=-\frac{x}{4}$y = −x/4, whose gradient is $-\frac{1}{4}$−1/4.

The equation of the normal

At (2, 4) on $y=x^2$y = x² the tangent has gradient 4, so the normal has gradient $-\frac{1}{4}$−1/4. It passes through the same point, so it is $y-4=-\frac{\left(x-2\right)}{4}$y − 4 = −(x − 2)/4. Multiply out: $y=-\frac{x}{4}+\frac{2}{4}+4$y = −x/4 + 2/4 + 4, which is $y=-\frac{x}{4}+\frac{9}{2}$y = −x/4 + 9/2.

Check it by substituting the point: $-\frac{2}{4}+\frac{9}{2}=-0.5+4.5=4$−2/4 + 9/2 = −0.5 + 4.5 = 4. Check the right angle: $4\times \left(-\frac{1}{4}\right)=-1$4 × (−1/4) = −1.

The normal also meets the curve again. Set $x^2=-\frac{x}{4}+\frac{9}{2}$x² = −x/4 + 9/2 and multiply by 4: $4x^2+x-18=0$4x² + x − 18 = 0, which factors as (x − 2)(4x + 9) = 0. The root x = 2 is the point itself; the other is $x=-\frac{9}{4}$x = −9/4, where the height is ${\left(-\frac{9}{4}\right)}^2=\frac{81}{16}$(−9/4)² = 81/16. On the normal, $-\left(-\frac{9}{4}\right)÷4+\frac{9}{2}=\frac{9}{16}+\frac{72}{16}=\frac{81}{16}$−(−9/4) ÷ 4 + 9/2 = 9/16 + 72/16 = 81/16 as well.

At the point (2, 4) on $y=x^2$y = x², the steep tangent y = 4x − 4 and the shallow normal $y=-\frac{x}{4}+\frac{9}{2}$y = −x/4 + 9/2 cross at a right angle. The normal meets the curve again at $\left(-\frac{9}{4},\frac{81}{16}\right)$(−9/4, 81/16).

A level tangent

When the gradient is 0 the tangent is horizontal, 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. At the bottom of $y=x^2$y = x², the origin, the tangent is the x-axis, y = 0, and the normal is the y-axis, x = 0.

The curve $y=\frac{x^3}{3}-x$y = x³/3 − x, whose derivative is $x^2-1$x² − 1. At x = 1.5 the tangent has gradient 1.25 and the normal −0.8, and 1.25 × (−0.8) = −1. Drag the point P: the product stays −1, and at x = 1 or x = −1 the tangent is level and the normal is vertical.

The usual mistakes

Using the tangent's gradient for the normal. A line of gradient 4 through (2, 4) is the tangent again. The normal's gradient is $-\frac{1}{4}$−1/4.

Changing only the sign. A gradient of −4 reflects the tangent in a vertical line; it is not at right angles to it. The gradient must also be turned upside down: $-\frac{1}{4}$−1/4.

Leaving x in the gradient. The tangent at x = 2 is not y = 2x: first put x = 2 into the derivative to get the number 4, then write the line through the point.

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

A skateboard ramp

In the application below, the side of a ramp is the curve $y=\frac{x^2}{4}$y = x²/4. A rider leaves its top edge along the tangent, and a support strut meets the ramp at right angles, so it lies along the normal.