Finding the gradient of a curve

In a nutshell

Curves don't have a constant gradient, but using tangents can allow you to estimate the gradient of a curve at a particular point on it. This is because the gradient of the curve at some point $A$ is equal to the gradient of the tangent at point $A$ . The gradient of the tangent would be an estimate of the gradient of the curve, since it can be difficult to sketch a perfectly accurate tangent at a point on a curve.

Gradient of a tangent

A tangent is a straight line with equation $y=mx+c$ . The gradient of a tangent can be calculated using:

$\boxed{m=\dfrac{\text{change in }y}{\text{change in }x}}$

Thus, when a tangent is drawn at a point on a curve, the coordinates of two points on that line must be identified and used.

Note: Consider a tangent at a turning point. It would be a horizontal line and thus has a zero gradient. Therefore, it follows that the turning points of curves have gradient of zero.

Example 1

Consider the curve given by the equation $y=x^2$ . Below, this curve is pictured with a tangent at the point $(2,4)$ :

Maths; Differentiation I; KS5 Year 12; Finding the gradient of a curve

Find the gradient of the curve at the point $(2,4)$ .

You need to calculate the gradient of the pictured tangent. You already know one point on the curve: $(2,4)$ , but you need another point to calculate the gradient. The point $(4,12)$ is also on this line:

Maths; Differentiation I; KS5 Year 12; Finding the gradient of a curve

Hence, you can now calculate the gradient using $\frac{\text{change in }y}{\text{change in }x}$ :

$m=\dfrac{12-4}{4-2}=\dfrac{8}{2}=4$

Therefore, the gradient of $y=x^2$ at the point $(2,4)$ is $\underline{4}$ .

Example 2

Below, the graph given by the equation $y=x^3+2x^2-x$ is presented, with a tangent at the point $(-1,2)$ :

Maths; Differentiation I; KS5 Year 12; Finding the gradient of a curve

Note: This tangent does cross the curve, but not at the point $(-1,2)$ . It only touches the curve there. So it is still a tangent, but specifically it is a tangent at $(-1,2)$ and not at $(0,0)$ , where it crosses through the curve.

Using this diagram, calculate the gradient of the curve $y=x^3+2x^2-x$ at the point $(-1,2)$ .

You need a second point on the line to calculate its gradient. The point $(0,0)$ is also on the line.

Note: It is incidental and unimportant that $(0,0)$ is also on the curve itself. Here you are using two points on a line to find the gradient of said line.

Maths; Differentiation I; KS5 Year 12; Finding the gradient of a curve

The gradient of the tangent is:

$m=\dfrac{0-2}{0-(-1)}=\dfrac{-2}1=-2$ 

Therefore, the gradient of $y=x^3+2x^2-x$ at the point $(-1,2)$ is $\underline{-2}$ .