# 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

The tangent to a curve at a point $$P$$ along its length is the line which passes through point $$P$$ and has same gradient as the curve 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: $f'(a)$ 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.

## Method: Equation of the Tangent to a Curve

Given a function $$f(x)$$, described by a curve $$y=f(x)$$, we find the equation of the tangent to the curve at a point $$P\begin{pmatrix}a,b\end{pmatrix}$$ along its length, in two steps:

• Step 1: find the gradient of the curve at point $$P\begin{pmatrix}a,b\end{pmatrix}$$, this equals to the gradient of the tangent
.
The gradient of the curve at $$P\begin{pmatrix}a,b\end{pmatrix}$$ equals to the derivative, $$f'(x)$$, evaluated at $$x = a$$: $f'(a)$ By definition, the tangent's gradient $$m$$ at $$P$$ equals to the curve's gradient at $$P\begin{pmatrix}a,b\end{pmatrix}$$, so: $m = f'(a)$
• Step 2: find the tangent's equation in the form $$y=mx+c$$ by making $$y$$ the subject in the formula: $y-b = m \begin{pmatrix}x - a \end{pmatrix}$ where $$a$$ and $$b$$ are the $$x$$ and $$y$$ coordinates of point $$P$$ and $$m$$ is the gradient of the curve at point $$P$$ (found in step 1).

## 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 $$P\begin{pmatrix}2,3\end{pmatrix}$$.
Keep pen and paper handy and try and follow the working.

## 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

• 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 $$\begin{pmatrix}3,5 \end{pmatrix}$$: $y = 6x - 13$

## Exercise 1

1. Given the function defined by: $f(x) = 2x^2 - 2$
1. Find $$f'(x)$$.
2. Calculate $$f'(2)$$.
3. Find the equation of the tangent to this function's graph $$y=2x^2-2$$ at the point along its length with coordinates $$\begin{pmatrix}2,6 \end{pmatrix}$$

2. Consider the function defined by: $f(x) = 5x+\frac{3}{x}$
1. Calculate the $$y$$-coordinate of the curve $$y=f(x)$$ when $$x = 1$$. State the coordinates of the point on the curve with $$x$$-coordinate $$x = 1$$.
2. Find an expression for $$f'(x)$$.
3. Find the equation of the tangent to the curve $$y=f(x)$$ at the point along its length with $$x$$-coordinate $$x = 1$$.

1. Given the function defined by: $f(x) = 2x^2 - 2$
1. For $$f'(x)$$, we find: $f'(x) = 4x$
2. We find: $f'(2)= 8$
3. We find the equation of the tangent to be: $y=8x-10$

1. The $$y$$-coordinate when $$x=1$$ is $$f(1)$$. We find: $y = 8$ So the point along the curve with $$x$$-coordinate $$x = 1$$ is: $\begin{pmatrix}1,8\end{pmatrix}$
2. We find: $f'(x) = 5 - \frac{3}{x^2}$
3. The tangent to the curve at the point $$\begin{pmatrix}1,8\end{pmatrix}$$ has equation: $y = 2x+6$

## Method: Equation of the Normal to a Curve

Given a function $$f(x)$$, described by a curve $$y=f(x)$$, we find the equation of the normal to the curve at a point $$P\begin{pmatrix}a,b\end{pmatrix}$$ along its length, in three steps:

• Step 1: find the gradient of the curve at point $$P\begin{pmatrix}a,b\end{pmatrix}$$.

By definition this equals to the derivative, $$f'(x)$$, evaluated at $$x = a$$: $f'(a)$
• Step 2: find the gradient of the normal to the curve at $$P\begin{pmatrix}a,b\end{pmatrix}$$.

The normal's gradient $$m$$ is the negative reciprocal of the gradient of the curve at the point $$P\begin{pmatrix}a,b\end{pmatrix}$$, that's: $m = -\frac{1}{f'(a)}$
• Step 3: find the normal's equation in the form $$y=mx+c$$ by making $$y$$ the subject in the formula: $y-b = m \begin{pmatrix}x - a \end{pmatrix}$ where $$a$$ and $$b$$ are the $$x$$ and $$y$$ coordinates of point $$P$$ and $$m$$ is the gradient of the curve at point $$P$$ (found in step 2).

## 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

• 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 $$\begin{pmatrix}3,2 \end{pmatrix}$$: $y = -\frac{x}{3} + 3$

## Exercise 2

1. Given the function defined by: $f(x)= x^2-1$
1. Find the $$y$$-coordinate of the point along the curve $$y=f(x)$$, with $$x$$-coordinate $$x = 2$$. Call this point $$P$$.
2. Find an expression for $$f'(x)$$.
3. Find the gradient of the curve $$y=f(x)$$ at point $$P$$.
4. State the value of the gradient of the normal to the curve $$y=f(x)$$ at point $$P$$.
5. Find the equation of the normal to the curve $$y=f(x)$$ at point $$P$$.

2. Given the function defined as: $y = 4x + \frac{9}{x}$
1. Calculate the $$y$$-coordinate of the point along the function's curve with $$x$$-coordinate $$x = 3$$. Call this point $$P$$.
2. Find $$\frac{dy}{dx}$$.
3. Find the gradient of the curve $$y = 4x + \frac{9}{x}$$ at point $$P$$.
4. State the value of the gradient of the normal to the curve at point $$P$$.
5. Find the equation of the normal to the curve at point $$P$$.

1. The $$y$$ coordinate is equal to $$f(2)$$. We find: $y = 3$ So $$P$$ has coordinates: $P\begin{pmatrix}2,3\end{pmatrix}$
2. We find: $f'(x)=2x$
3. The gradient of the curve at $$P$$ equals to $$f'(2)$$. We find: $f'(2)=4$
4. The normal to the curve at $$P$$ has a gradient equal to the negative reciprocal of the gradient of the curve at $$P$$. That's: $m = - \frac{1}{4}$
5. The equation of the normal to the curve at point $$P$$ is: $y = - \frac{x}{4}+\frac{7}{2}$
1. When $$x=3$$, we find $$y=15$$. So point $$P$$ has coordinates: $P\begin{pmatrix}3,15\end{pmatrix}$
2. We find: $\frac{dy}{dx} = 4 - \frac{9}{x^2}$
3. The gradient of the curve $$y = 4x + \frac{9}{x}$$ at point $$P$$ equals to the derivative $$\frac{dy}{dx}$$ when $$x = 3$$. We find: $\frac{dy}{dx} = 3$
4. The gradient $$m$$ of the normal to the curve at point $$P$$ is the negative reciprocal of the gradient of the curve at $$P$$, that's: $m = -\frac{1}{3}$
5. The equation of the normal to the curve at point $$P$$ is: $y = - \frac{x}{3}+16$