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Variable Rate of Change

Variable Rate of Change

A linear function - such as a straight line on a graph - has a constant rate of change. As `x` changes, so does `y`, and this is constant. The rate of change can be derived from the gradient:

`frac(text(change in )y)(text(change in )x)`.

For other functions, the rate of change varies over the function: in other words, the gradient on a graph becomes steeper or flatter. At any one point on the graph, the rate of change can be derived (by drawing a tangent to the graph at that point). At a different point, a different rate of change can be found by obtaining the gradient at that point.

Example 1

A car covers a distance of 20m in 5s. If the car has a constant rate of acceleration, what is the increase in the velocity of the car beween 5s and 10s?

The kinematic formulae are:

`v = u + at, s=ut+frac(1)(2)at^2` and `v^2=u^2+2as`

where `u` is initial velocity, `v` is final velocity, `s` is distance, `a` is acceleration and `t` is time.

Known values: `s=20, t=5, u=0`

Work out the acceleration `s` `=` `ut + frac(1)(2)at^2`
`(20)` `=` `(0)(5) + frac(1)(2)a(5)^2`
`a` `=` `frac(8)(5)`

Velocity at 5s `v` `=` `u + at`
`v` `=` `(0) + frac(8)(5)(5)`
`v` `=` `8`

Velocity at 10s `v` `=` `u + at`
`v` `=` `(0) + frac(8)(5)(10)`
`v` `=` `16`

Difference in velocity = 16 - 8 = 8ms-1

Answer: 8 ms-1

Example 2

The gradient of `x^3+1` is the same at (-2, -7) as it is at (2, 9). The gradient at (0, 1) is zero.

Jon draws line from (-2, -7) to (2, 9) which passes through (0, 1). Jon states that the rate of change between (-2, -7) and (2, 9) is constant as his line also passes through (0, 1).

Why is Jon wrong?

Draw a tangent at (-1, 0) (or some other point) to find a different gradient.

Graph of cubic and straight line with average rate of change


Jon has drawn the average rate of change between these two points.

The rate of change is constantly varying eg it is zero at (0, 3) and 3 at (-1, 0).