Question

How do you find the instantaneous rate of change of $g\left( t \right) = 3{t^2} + 6$ at t = 4?

Answer 1

The instantaneous rate of change is the change in the rate at a particular instant, and it is same as the change in the derivative value at a specific point. We can find instantaneous rate of change by finding its derivative with respect to t, i.e., by finding $g'\left( t \right)$ at t=4 we can get the rate of change of the given function.

Complete step by step answer:
The given function is
$g\left( t \right) = 3{t^2} + 6$
The instantaneous rate of change is given by $g'\left( t \right)$, hence find the derivative of the given function with respect to t as
$g'\left( t \right) = 3\left( {2t} \right)$
We get the derivative as
$g'\left( t \right) = 6t$ ……………….. 1
As the value of t is given as t = 4, hence substituting the value in equation 1 we get
$g'{\left( t \right)_{t = 4}} = 6\left( 4 \right)$
Simplifying, we get
$g'{\left( t \right)_{t = 4}} = 24$
Therefore, the instantaneous velocity at t=4 is
$g'\left( 4 \right) = 24$

Hence, the instantaneous rate of change of $g\left( t \right) = 3{t^2} + 6$ at t = 4 is $24$.

Additional information:
For a graph, the instantaneous rate of change at a specific point is the same as the tangent line slope. That is, it is a curve slope. If given the function values before, during, and after the required time, the instantaneous rate of change can be estimated. While estimates of the instantaneous rate of change can be found using values and times, an exact calculation requires using the derivative function. This rate of change is not the same as the average rate of change.

Note: The key point to find the instantaneous rate of change is to find the derivative of g(t), as the instantaneous change is given by $g'\left( t \right)$, and an average velocity between two points on the path in the limit that the time (and therefore the displacement) between the two points approaches zero.