In the final exam you may be asked to calculate the average rate of change and then the instantaneous rate of change. So it is important to know both the difference between the two of these and also how to calculate each of them.

## Average Rate of Change

The average rate of change between two points is the slope of the line separating the two points. So the change in y over the change in x for just that range.

## Instantaneous Rate of Change

The instantaneous rate of change is the change at that particular moment or the gradient at that point. This is the value of the derivative at a particular point.

The key difference between the two is that the average rate of change is over a range, while the instantaneous rate of change is applied at a particular point.

## Examples

Question 3c from the 2014 exam 2

##  As we are doing average rate of change we want to be working over a range and we have been given that range in the question. So know we want to work out the gradient of the slope between these two lines, which will be the rise over the run.

So we get the equation: $average.rate.of.change=\frac{c(3)-c(\frac{2}{3})}{(3-\frac{2}{3})} \newline=\frac{(\frac{5}{2} \times 3 \times e^{-\frac{3}{2} \times 3} -\frac{5}{2}\times\frac{2}{3}\times(e)^{-\frac{3}{2}\times\frac{2}{3}})}{3-\frac{2}{3}}\newline =\frac{\frac{15}{2}e^{-\frac{9}{2}}-\frac{5}{3}e^{-1}}{\frac{7}{3}}\newline =-0.23mg/L/h$

Remembering to include the negative and the units.

### Instantaneous Rate of Change As we are now dealing with the instantaneous rate of change we need to calculate the derivative.
Here we have two different functions multiplied together so that we need to use the product rule.

So we need to set one part of the function to u and the other to v. $u=\frac{5}{2}t \newline v=e^{\frac{-3}{2}t}$

Then we need to calculate their derivatives. $\frac{du}{dx}=\frac{5}{2} \newline \frac{du}{dx}=-\frac{3}{2}e^{\frac{-3}{2}t}$

If any of these steps don’t make sense, feel free to have a look at our power or exponential differentiation pages

Then we use the formula from the exam one formula sheet that $\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}$

And sub in our calculated values to get $\frac{d}{dx}(\frac{5}{2}e^{\frac{-3}{2}t} )= \frac{5}{2}t \times -\frac{3}{2}e^{\frac{-3}{2}t}+e^{\frac{-3}{2}t}\times \frac{5}{2} \newline c'(t)=\frac{5}{2}e^{\frac{-3}{2}t}(\frac{-3}{2}t+1)$

We know that the instantaneous range must equal the previously calculated average rate of change. $c'(t)=\frac{5}{2}e^{\frac{-3}{2}t}(\frac{-3}{2}t+1)=-0.227$

This is too complex to be solved by hand and should be calculated using your calculator to get

t=0.90, 2.12

You may get a slightly different answer depending on how many decimal places you used to calculate the average rate of change.