Mathematics » Differential Calculus » Equation Of A Tangent To A Curve

# Equation of a Tangent to a Curve

## Equation of a Tangent to a Curve

At a given point on a curve, the gradient of the curve is equal to the gradient of the tangent to the curve. The derivative (or gradient function) describes the gradient of a curve at any point on the curve. Similarly, it also describes the gradient of a tangent to a curve at any point on the curve.

### To determine the equation of a tangent to a curve:

1. Find the derivative using the rules of differentiation.

2. Substitute the $$x$$-coordinate of the given point into the derivative to calculate the gradient of the tangent.

3. Substitute the gradient of the tangent and the coordinates of the given point into an appropriate form of the straight line equation.

4. Make $$y$$ the subject of the formula.

The normal to a curve is the line perpendicular to the tangent to the curve at a given point.

$m_{\text{tangent}} \times m_{\text{normal}} = -1$

## Example

### Question

Find the equation of the tangent to the curve $$y=3{x}^{2}$$ at the point $$(1;3)$$. Sketch the curve and the tangent.

### Find the derivative

Use the rules of differentiation:

\begin{align*} y &= 3{x}^{2} \\ & \\ \therefore \cfrac{dy}{dx} &= 3 ( 2x ) \\ &= 6x \end{align*}

### Calculate the gradient of the tangent

To determine the gradient of the tangent at the point $$(1;3)$$, we substitute the $$x$$-value into the equation for the derivative.

\begin{align*} \cfrac{dy}{dx} &= 6x \\ \therefore m &= 6(1) \\ &= 6 \end{align*}

### Determine the equation of the tangent

Substitute the gradient of the tangent and the coordinates of the given point into the gradient-point form of the straight line equation.

\begin{align*} y-{y}_{1} & = m(x-{x}_{1}) \\ y-3 & = 6(x-1) \\ y & = 6x-6+3 \\ y & = 6x-3 \end{align*}

### Sketch the curve and the tangent ## Example

### Question

Given $$g(x)= (x + 2)(2x + 1)^{2}$$, determine the equation of the tangent to the curve at $$x = -1$$ .

### Determine the $$y$$-coordinate of the point

\begin{align*} g(x) &= (x + 2)(2x + 1)^{2} \\ g(-1) &= (-1 + 2)[2(-1) + 1]^{2} \\ &= (1)(-1)^{2} \\ & = 1 \end{align*}

Therefore the tangent to the curve passes through the point $$(-1;1)$$.

### Expand and simplify the given function

\begin{align*} g(x) &= (x + 2)(2x + 1)^{2} \\ &= (x + 2)(4x^{2} + 4x + 1) \\ &= 4x^{3} + 4x^{2} + x + 8x^{2} + 8x + 2 \\ &= 4x^{3} + 12x^{2} + 9x + 2 \end{align*}

### Find the derivative

\begin{align*} g'(x) &= 4(3x^{2}) + 12(2x) + 9 + 0 \\ &= 12x^{2} + 24x + 9 \end{align*}

### Calculate the gradient of the tangent

Substitute $$x = -\text{1}$$ into the equation for $$g'(x)$$:

\begin{align*} g'(-1) &= 12(-1)^{2} + 24(-1) + 9 \\ \therefore m &= 12 – 24 + 9 \\ &= -3 \end{align*}

### Determine the equation of the tangent

Substitute the gradient of the tangent and the coordinates of the point into the gradient-point form of the straight line equation.

\begin{align*} y-{y}_{1} & = m(x-{x}_{1}) \\ y-1 & = -3(x-(-1)) \\ y & = -3x – 3 + 1 \\ y & = -3x – 2 \end{align*}

## Example

### Question

1. Determine the equation of the normal to the curve $$xy = -4$$ at $$(-1;4)$$.
2. Draw a rough sketch.

### Find the derivative

Make $$y$$ the subject of the formula and differentiate with respect to $$x$$:

\begin{align*} y &= -\cfrac{4}{x} \\ &= -4x^{-1} \\ & \\ \therefore \cfrac{dy}{dx} &= -4 ( -1x^{-2} ) \\ &= 4x^{-2} \\ &= \cfrac{4}{x^{2}} \end{align*}

### Calculate the gradient of the normal at $$(-1;4)$$

First determine the gradient of the tangent at the given point:

\begin{align*} \cfrac{dy}{dx} &= \cfrac{4}{(-1)^{2}} \\ \therefore m &= 4 \end{align*}

Use the gradient of the tangent to calculate the gradient of the normal:

\begin{align*} m_{\text{tangent}} \times m_{\text{normal}} &= -1 \\ 4 \times m_{\text{normal}} &= -1 \\ \therefore m_{\text{normal}} &= -\cfrac{1}{4} \end{align*}

### Find the equation of the normal

Substitute the gradient of the normal and the coordinates of the given point into the gradient-point form of the straight line equation.

\begin{align*} y-{y}_{1} & = m(x-{x}_{1}) \\ y-4 & = -\cfrac{1}{4}(x-(-1)) \\ y & = -\cfrac{1}{4}x – \cfrac{1}{4} + 4\\ y & = -\cfrac{1}{4}x + \cfrac{15}{4} \end{align*}

### Draw a rough sketch 