The gradient is a measure of how steep a line is. It is calculated by finding the rate of change in the $$y$$-direction with respect to the positive $$x$$-direction.

Estimating the gradient of a curve

Maths

The work on gradient in the prior unit is expanded further in this unit by considering gradient as the rate of change and interpreting this in context. The gradient of the tangent is introduced as a way of estimating the gradient of a curve.

Real-life graphs

The gradient of the hypotenuse of the triangle is not the exact gradient of the curve between $$x=2$$ and $$x=6$$, it is only __________.

Estimate the gradient of this curve between $$x=2$$ and $$x=6$$

Estimate the gradient of this curve between $$x=-6$$ and $$x=-4$$

Nothing. It is quick and accurate to estimate this gradient this way.

The distance between the hypotenuse and the curve means it lacks accuracy.

The changes in $$x$$ and $$y$$ are not both $$+11$$

What is wrong with using the coordinate pairs (-7, 12) and (-2, 1) to estimate the gradient of this curve?

Match the intervals to the estimated gradient of the curve in those intervals.

The interval $$x=0$$ to $$x=$$__________ would be poor because it would give you an estimated gradient of $$0$$

rate of change in $$y$$-direction with respect to positive $$x$$-direction

rate of change in $$y$$-direction with respect to negative $$x$$-direction

What is the gradient of the line going through coordinate pairs $$(18,73)$$ and $$(50,233)$$?

What is the gradient of the line going through coordinate pairs $$\left(-{{1}\over{4}},{{5}\over{3}}\right)$$ and $$\left({{7}\over{3}},-{{16}\over{5}}\right)$$?

This lesson will be removed by end of Summer Term 2025.

Estimating the gradient of a curve

Lesson outcome

This lesson will be removed by end of Summer Term 2025.