## Lecture 6 | Gradient of a curve

## The Gradient of a Curve

The gradient of a curve changes from point to point. It can be easily calculated by differentiation.

**Example:**

Let the equation of the curve be, y = x² - x + 3

dy/dx = 2x - 1

This gives the gradient at any point; all you have to do now is to substitute the x value into the equation.

If x = 1 => dy/dx = 1

If x = 0 => dy/dx = -1

If x = 3 => dy/dx = 5

You can check these values interactively using the following applet. In addition, you can find the gradient of the curve any point you wish - just by moving the slider.

In addition, you can find the gradient of the tangents, drawn to the same points, to verify the values.

**Question:**

Find the gradient of the same curve at the following points, using the differentiation. Then verify the values with the applet:

- x = -2
- x = -3
- x = 5

## Problem Solving with Curves

**Example 1**

The equation of a curve is y = x² -x -6. Find the gradient of the curve at x = -2 and x = 3. Find the coordinates of the point where the gradient is zero.

y = x² -x -6

dy/dx = 2x - 1

When x = -2

dy/dx = 2 x -2 - 1 = -5

When x = 3

dy/dx = 2 x 3 - 1 = 5

When the gradient is zero, dy/dx = 0

2x - 1 = 0

x = 0.5

Sub into y = x² -x -6,

y = 0.5² - 0.5 - 6 = - 6.25

The coordinates are, (0.5, -6.25)

**Example 2**

The equation of a curve is y = 4x - x². Find the gradient of the curve at x = 1. Find the coordinates of the point where the gradient is -4.

y = 4x - x²

dy/dx = 4 - 2x

When x = 1

dy/dx =4 - 2 x 1 = 2

When the gradient is -4, dy/dx = -4

4 - 2x = -4

2x = 8

x = 4

Sub into y = 4x - x²

y = 4 x 4 - 16 = 0

The coordinates are, (4, 0)

**Example 3**

The equation of a curve is y = x³ - 4x. Find the coordinates of the point/s where the gradient is 8. Comment on the answer.

y = x³ - 4x

dy/dx = 3x² - 4

When the gradient is 8, dy/dx = 8

3x² - 4 = 8

3x² = 12

x² = 4

x = ± 2

There are two points on the curve with the same gradient, x = 2 and x = -2

Sub into y = x³ - 4x

The coordinates of the two points are, (-2, 0) and (2,0).

The following graph shows how it happens:

## Additional Problems

- At what points on the curve, y = x
^{3}- 12x + 5, is the gradient of the curve equal to 10? - At what points on the curve, y = (1/3)x
^{3}- x^{2}+ 1/3, are the tangents parallel to the x-axis? - At what point on the curve, y = x
^{3}+ 1, can the tangent to the curve be y = 3x - 1? - At what points on the curve, y = (2/3)x
^{3}- (1/2)x^{2}+ 1, is the gradient equal to 2? - At what points on the curve, y = 4/x, does the tangent to the curve become, y = -3x + 7 ?

- x = -3, x = 3
- x = 0, x= 2
- x = 1
- x= 1, x = -1
- x = 1