Tangents & Normals

(equation of the tangent and the normal to the curve at point \(P\))

In this section we learn how to find the equation of the tangent and the normal to a curve at a point along its length.

For each we learn a two-step method as well as view a tutorial and work our way through exercises to consolidate our knowlede.

Tangent to a Curve

Say the curve has equation \(y = f(x)\), then its gradient at a point \(P\begin{pmatrix}a,b\end{pmatrix}\) along its length is equal to: \[f'(a)\] Where \(f'(a)\) is the derivative of \(f(x)\) evaluated at \(x = a\).

So the tangent will have gradient: \[m = f'(a)\]

Normal to a Curve

The normal to a curve at a point \(P\) along its length is the line which passes through point \(P\) and is perpendicular to the tangent at \(P\).

Say the curve has equation \(y = f(x)\), then its gradient at a point \(P\begin{pmatrix}a,b\end{pmatrix}\) along its length is equal to: \[f'(a)\] Since the normal is perpendicular to the tangent, its gradient is the negative reciprocal of the gradient of the tangent. That's: \[m = -\frac{1}{f'(a)}\]

How to find the Equation of a Tangent & a Normal

A tangent to a curve as well as a normal to a curve are both lines. They therefore have an equation of the form: \[y = mx+c\] The methods we learn here therefore consist of finding the tangent's (or normal's) gradient and then finding the value of the \(y\)-intercept \(c\) (like for any line).

We start by learning how to find the equation of a tangent to a curve. Further-down, we learn how to find the equation of a normal.

Tutorial

In the following tutorial we illustrate how to use this two-step method with two examples, in which we find:

• the tangent to the curve of the function \(f(x) = x^2 - 1\) at the point \(P\begin{pmatrix}2,3\end{pmatrix}\).
• the tangent to the curve of the function \(f(x) = x^2 - 1\) at the point \(P\begin{pmatrix}2,3\end{pmatrix}\).

Example

Find the equation of the tangent to the curve \[y = x^2 - 4\] at the point along its length with coordinates \(\begin{pmatrix}3,5\end{pmatrix}\).

Solution

We follow our two-step method:

• Step 1: find the gradient of the tangent.
  
  At the point \(\begin{pmatrix}3,5\end{pmatrix}\) the \(x\) coordinate is \(x = 3\) so the gradient of the tangent is: \[m = f'(3)\] We therefore start by finding \(f'(x)\). Using the power rule for differentiation we find this function's derivative: \[f'(x) = 2x\] Evaluating it when at \(\begin{pmatrix}3,5\end{pmatrix}\) that's when \(x = 3\), we find: \[\begin{aligned} f'(3) & = 2\times 3 \\ f'(x) & =6 \end{aligned} \] So the tangent's gradient is: \[m = 6\]
• Step 2: find the equation of the tangent to the curve at the point \(\begin{pmatrix}a,b \end{pmatrix}\) by making \(y\) the subject in: \[y - b = m \begin{pmatrix}x - a \end{pmatrix}\] Since the point we're working with is \(\begin{pmatrix}3,5 \end{pmatrix}\) and the gradient, we found in step 1, is \(m = 6\), we this becomes: \[y - 5 = 6 \begin{pmatrix}x - 3 \end{pmatrix}\] Distributing the \(6\) on the right hand side: \[y - 5 = 6x - 18 \] Adding \(5\) to each side: \[y = 6x - 18 +5\] Finally we find the equation of the tangent to the curve at the point \(\begin{pmatrix}3,5 \end{pmatrix}\): \[y = 6x - 13\]

Example

Find the equation of the normal to the curve \[y = \frac{x^2}{2}-\frac{5}{2}\] at the point along its length with coordinates \(\begin{pmatrix}3,2\end{pmatrix}\).

Solution

We follow our three-step method:

• Step 1: find the gradient of the curve at the point \(\begin{pmatrix}3,2\end{pmatrix}\).
  
  We start by finding this function's derivative. Using the power rule for differentiation, we find: \[\begin{aligned} \frac{dy}{dx} & = 2\times \frac{x^{2-1}}{2} - 0 \\ & = \frac{2x^1}{2} \\ \frac{dy}{dx} & = x \end{aligned}\] The gradient of the curve at \(\begin{pmatrix}3,2\end{pmatrix}\) equals to the value of \(\frac{dy}{dx}\) when \(x = 3\); that's: \[\frac{dy}{dx} = 3\]
• Step 2: find the gradient of the normal.
  
  The normal's gradient equals to the negative reciprocal of the gradient of the curve. Since the gradient of the curve at the point is \(3\), we find the normal's gradient: \[m = -\frac{1}{3}\]
• Step 3: find the equation of the normal to the curve at the point \(\begin{pmatrix}a,b \end{pmatrix}\) by making \(y\) the subject in: \[y - b = m \begin{pmatrix}x - a \end{pmatrix}\] Since the point we're working with is \(\begin{pmatrix}3,2 \end{pmatrix}\) and the gradient, we found in step 2, is \(m = -\frac{1}{3}\), this becomes: \[y - 2 = -\frac{1}{3} \begin{pmatrix}x - 3 \end{pmatrix}\] Distributing the \(-\frac{1}{3}\) on the right hand side: \[y - 2 = -\frac{1}{3}.x - \begin{pmatrix} -\frac{1}{3}\end{pmatrix}3 \] Leading to: \[y - 2 = -\frac{x}{3} - \begin{pmatrix} -\frac{3}{3}\end{pmatrix} \] That's \[y - 2 = -\frac{x}{3} - \begin{pmatrix} -1\end{pmatrix} \] \[y - 2 = -\frac{x}{3} + 1 \] We now add \(2\) to each side of this equation, which leads to: \[y = -\frac{x}{3} + 1 +2\] Finally we find the equation of the normal to the curve at the point \(\begin{pmatrix}3,2 \end{pmatrix}\): \[y = -\frac{x}{3} + 3\]