Question

How do you find the instantaneous rate of change of `g(t)=3t^2+6` at t=4?

Answer 1

Compute the first derivative and evaluate it at `t = 4`

`g'(4) = 24`

Explanation:

Compute the first derivative:

`g'(t) = 6t`

Evaluate it at `t = 4`

`g'(4) = 24`

Answer 2

It depends on what you have in your mathematical toolbox.

Explanation:

If you have learned the power rule, constant multiple rule and derivative of a constant, you can quickly find the derivative of `g`.

`g'(t) = 3(2x^1)+0 = 6t`.

To find the instantaneous rate of change at a particular value of `t`, evaluate the derivative at that value of `t`.

At `t=4` the instantaneous rate of change is `g'(4) = 6(4) = 24`.

If you are using a definition then it depends on the particular definition you are using.

There are several ways to express the definition.

One way of expressing it is to give:

The rate of change of `g` with respect to `t` at `t=4` is

`lim_(trarr4)(g(t)-g(4))/(t-4)`.

Another is

The rate of change of `g` with respect to `t` at `t=4` is

`lim_(hrarr0)(g(4+h)-g(4))/h`.

Still another is

The rate of change of `g` with respect to `t` at `t` is

`lim_(hrarr0)(g(t+h)-g(t))/h`.

(After we find this, we evaluate at `t=4`.

Here is the work for the first definition above.

`lim_(trarr4)(g(t)-g(4))/(t-4) =lim_(trarr4) ([3t^2+6]-[3(4)^2+6])/(t-4)` (Observe that is we substitute `t = 4`, we get the indeterminate form `0/0`.)

`= lim_(trarr4) (3t^2+6-48-6)/(t-4)` `" "` (Still `0/0`)

`= lim_(trarr4) (3t^2-48)/(t-4)`

`= lim_(trarr4) (3(t^2-16))/(t-4)`

`= lim_(trarr4) (3(t+4)(t-4)))/(t-4)`

`= lim_(trarr4) 3(t+4)`

`= 3(4+4) = 24`