Instantaneous Rate of Change Formula

The instantaneous rate of change is the change in the concentration of rate that occurs at a particular instant of time. The variation in the derivative values at a specific point also denotes the instantaneous rate of change.  The instantaneous rate of change at a point is equal to the derivative function evaluated at that point. 

Further, the average and instantaneous rate of change at a specific point can map in the graph as the tangent slope line, which shows like a curve slope. The value of the instantaneous rate of change is also equal to the slope of the tangent at a point of a curve.

The instantaneous rate of change formula can also be defined with the differential quotient and limits. The average rate of $y$ shift with respect to $x$ is the quotient of difference. The instantaneous rate of change formula represents with limit exists in,

$f’(a)$ = \[\lim_{\Delta x \rightarrow 0}\] \[\frac{\Delta y}{\Delta x}\] = \[\lim_{x \rightarrow 0}\] \[\frac{t(a+h)-(t(a))}{h}\]

How to Calculate Instantaneous Rate of Change in Graph?

The tangent straight line at a point can be drawn, which touches a curve at the point without crossing over the curve. Or else, the tangent should touch only one point of the curve. The slope of the tangent line can be calculated at a point, which represents the instantaneous rate of change, or derivative of the particular point.

The tangent line can be drawn in the graph to calculate the instantaneous rate of change as shown in the above image. Further, the slope of the tangent line can calculate using the formulas;

\[\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\] 

Or 

f’(\[x_{0}\]) = \[\lim_{(x_{0} \rightarrow x_{1})}\] \[\frac{f(x_{1})-f(x_{0})}{x_{1}-x_{0}}\] = \[\lim_{\Delta x \rightarrow 0}\]  \[\frac{f(x_{0}-\Delta x)-f(x_{0})}{\Delta x}\] 

Problems Related to Instantaneous Rate 

Problem 1: Calculate the Instantaneous rate of change of the function $f(x) = 4x^2 + 24$ at $x = 5$?

Answer: Give that, 

$y = f(x) = 4x^2 + 24$

$f'(x) = 4(2x) + 0$

$f'(x) =8x$

As per the given date, we need to calculate the instantaneous rate of change at the value $x = 5$.

$f'(5) = 8(5)$

$f'(5) = 40$

So, the instantaneous rate of change for the given function at $x = 5$ is $40$. 

Problem 2: Derive the Instantaneous rate of change for the function $f(x) = 7x^3 – 3x^2 + 2x + 5$ at $x = 3$?

Answer: Given that, 

$y = f(x) = 7x^3 – 3x^2 + 2x + 5$

$f'(x) = 7(3x^2) – 3(2x) + 2 + 0$

$f'(x) = 21x^2 – 6x + 2$

As per the given date, we need to calculate the instantaneous rate of change at the value $x = 3$.

$f'(3) = 21(3)2 –  6(3) + 2 =  189 – 18 + 2 = 173$

$f'(3) = 173$

So, the instantaneous rate of change for the given function at $x = 3$ is $173$. 

Conclusion

In this post, we have studied about the instantaneous rate of change of formula and discussed how it will be derived. We also come to know that the variance in derivative values at a given position also represents the instantaneous pace of change. We have also discussed the solved problem so that the application of this formula should be understood properly.