Question

How do you estimate instantaneous rate of change at a point?

Answer 1

Hint: The derivative tells us the rate of change of one quantity compared to another at a particular instant or point. That is what we called "instantaneous rate of change". We have \[\dfrac{{dy}}{{dx}}\] meaning that change in ‘y’ compared to change in ‘x’ at a precise value of ‘x’. We have a formula for instantaneous rate of change of a point and we give one example for it.

Complete step-by-step answer:
We know that, the instantaneous rate of change of the function \[y = f(x)\] at the point \[{x_0}\] in its domain is \[f'({x_0}) = \mathop {\lim }\limits_{x \to {x_0}} \dfrac{{\Delta y}}{{\Delta x}} = \mathop {\lim }\limits_{x \to {x_0}} \dfrac{{f({x_0}) - f(x)}}{{{x_0} - x}}\] , provided limit exists.
Now let’s take an example.
Let \[f(x) = \dfrac{1}{x}\] and let’s find the instantaneous rate of change of \[f\] at \[{x_0} = 2\] .
The first step is to compute the average rate of change over some interval \[{x_0} = 2\] to ‘x’.
 \[\dfrac{{\Delta y}}{{\Delta x}} = \dfrac{{f(2) - f(x)}}{{2 - x}}\]
Since we have \[f(x) = \dfrac{1}{x}\] , we have:
 \[ \Rightarrow = \dfrac{{\left( {\dfrac{1}{2} - \dfrac{1}{x}} \right)}}{{2 - x}}\]
Taking L.C.M and substituting we have,
 \[ = \dfrac{{\left( {\dfrac{{x - 2}}{{2x}}} \right)}}{{2 - x}}\]
 \[ = \dfrac{{x - 2}}{{2x\left( {2 - x} \right)}}\]
Taking negative sign in the denominator, we have
 \[ = \dfrac{{\left( {x - 2} \right)}}{{2x \times - (x - 2)}}\]
Cancelling the terms we have,
 \[ = - \dfrac{1}{{2x}}\]
Thus the instantaneous rate of change at \[{x_0} = 2\] , is
 \[\mathop {\lim }\limits_{x \to 2} \dfrac{{\Delta y}}{{\Delta x}} = \mathop {\lim }\limits_{x \to 2} \dfrac{{ - 1}}{{2x}}\] \[ = - \dfrac{1}{4}\] .

Note: The average rate of change between two input values is the total change of the function values (output values) divided by the change in the input values.