Tangents and normals

What you will be able to do

• find the gradient of a curve at a point from its derivative;
• write the equations of the tangent and the normal at a point;
• handle a horizontal tangent and a vertical tangent;
• find the gradient for an implicit curve and for a parametric curve.

The gradient of a curve

A straight line has one gradient everywhere. A curve has a different gradient at every point. We measure it with a tangent, the line that touches the curve at that point and has the same direction as the curve there.

The gradient of the tangent is the derivative $f'(x_0)$. The word gradient is used in A-Level and IB. In NCERT and AP the same number is called the slope. We use both words.

These topics are examined in JEE Main, A-Level Pure Maths, IB Analysis and Approaches and AP Calculus. Check the current syllabus of your own board for tangents, normals and approximations.

The tangent and the normal

Let $P(x_0, y_0)$ be a point on the curve $y = f(x)$, and let $m = f'(x_0)$.

The tangent at $P$ is the line through $P$ with gradient $m$:

The normal at $P$ is the line through $P$ at a right angle to the tangent. Two perpendicular lines have gradients whose product is $-1$, so the normal has gradient $-\dfrac{1}{m}$:

Take $y = x^3 - 2x + 1$ at $x = 1$. Then $y_0 = 1 - 2 + 1 = 0$ and $f'(x) = 3x^2 - 2$, so $m = 1$. The tangent is $y = x - 1$. The normal has gradient $-1$ and is $y = -x + 1$.

Two special cases

• A horizontal tangent. If $m = 0$, the tangent is $y = y_0$. The normal is the vertical line $x = x_0$. The formula $-\dfrac{1}{m}$ does not work, because we cannot divide by $0$.
• A vertical tangent. If $\dfrac{dy}{dx}$ is infinite, the tangent is the vertical line $x = x_0$, and the normal is the horizontal line $y = y_0$. The curve $y = \sqrt{x}$ has a vertical tangent at the origin.

Implicit and parametric curves

For a circle $x^2 + y^2 = 25$, we cannot solve for $y$ easily. Differentiate both sides with respect to $x$:

At the point $(3, 4)$ the gradient is $-\dfrac{3}{4}$. The tangent is $3x + 4y = 25$. The normal has gradient $\dfrac{4}{3}$ and passes through the origin, which is the centre. The normal of a circle always passes through the centre.

For a parametric curve $x = x(t)$, $y = y(t)$, the chain rule gives

For $x = t^2$ and $y = 2t$, we get $\dfrac{dy}{dx} = \dfrac{2}{2t} = \dfrac{1}{t}$. At $t = 1$ the point is $(1, 2)$ and the gradient is $1$. The tangent is $y = x + 1$.

A worked example

Find the tangent and the normal to $y = \sqrt{x}$ at $x = 4$.

1. The point. $y_0 = \sqrt{4} = 2$, so $P = (4, 2)$.
2. The gradient. $\dfrac{dy}{dx} = \dfrac{1}{2\sqrt{x}}$, so $m = \dfrac{1}{4}$.
3. The tangent. $y - 2 = \tfrac{1}{4}(x - 4)$, that is, $y = \tfrac{1}{4}x + 1$.
4. The normal. Its gradient is $-4$, so $y - 2 = -4(x - 4)$, that is, $y = -4x + 18$.

Practise with new numbers

Try it in Python

The program finds the gradient of a curve from a very short interval, and then writes the tangent and the normal. The product of the two gradients should be $-1$.

Common mistakes

• Using $f(x_0)$ as the gradient. The gradient is $f'(x_0)$, the derivative.
• Putting $x_0$ into the equation of the tangent twice. The point is $(x_0, y_0)$, and $x$ and $y$ stay as letters.
• Forgetting to change the sign for the normal. The gradient is $-\dfrac{1}{m}$, not $\dfrac{1}{m}$ and not $-m$.
• Dividing by zero. A horizontal tangent has a vertical normal $x = x_0$.
• Differentiating an implicit curve and forgetting $\dfrac{dy}{dx}$. The derivative of $y^2$ is $2y\,\dfrac{dy}{dx}$.

Remember these

• Tangent: $y - y_0 = f'(x_0)(x - x_0)$. Normal gradient: $-\dfrac{1}{f'(x_0)}$.
• A horizontal tangent has $f'(x_0) = 0$ and a vertical normal.
• Implicit curve: differentiate both sides. Parametric curve: $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Find the gradient of the tangent to $y = x^2 - 5x$ at $x = 3$.

2. Write the equation of the tangent to $y = x^3$ at the point $(1, 1)$.

3. Find the gradient of the normal to $y = x^3$ at $(1, 1)$.

4. Find the tangent to $y = \sqrt{x}$ at $x = 9$.

5. Find $\dfrac{dy}{dx}$ for $x^2 + y^2 = 25$ at the point $(3, -4)$.

References

• NCERT, Mathematics Textbook for Class XII, Part I, Chapter 6 (Application of Derivatives).
• OpenStax, Calculus Volume 1, Section 3.1 (Defining the Derivative), the part on tangent lines.