# Instantaneous Rate of Change

Imagine a car covering a distance of 50 km in a span of 40 min, and the driver wants to know the rate of change of distance at any instance during those 40 mins. Or imagine the stocks of a company fluctuating every other second and to properly monitor them, the rate of change of the amount of stocks per unit time is to be calculated.

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These are some of the scenarios where there is a need to calculate the instantaneous rate of change of a quantity with respect to another quantity. Applications of this concept are vast in different fields, and sometimes the number of quantities is not limited to two in some complicated situations. But for the scope of this article, we shall restrict our study of the Instantaneous Rate of Change to two quantities only.

## Instantaneous Rate of Change at a point

In the accompanying article on Average Rate of Change, it has been discussed how to calculate the rate of change of a quantity over a considerable span of another quantity, but what happens to the rate of change at a particular instance? In a loose sense, we want to know how a quantity is changing at a particular instance and not over a big part of it.

The Instantaneous Rate of Change at a point on a curve is the rate at which the quantity is changing relative to the other one at the point (or instant).

The Instantaneous Rate of Change at a point is the gradient of the curve at that point. Recall from the definition of the gradient that gradient at a point tells us the rate of change at that point.

The tangent at a point signifying its slope, StudySmarter Originals

In the above diagram, suppose we want to know the Instantaneous Rate of Change at point A, then we need to find the gradient of the curve at point A. And the gradient at that point is the slope at that point.

For any curve, it will vary from point to point and the slope at that point has to be calculated individually.

A special and intriguing case is a straight line, where the gradient at every point would be the same and hence the instantaneous rate of change will be equal to the average rate over any interval.

## Instantaneous Rate of Change using Calculus

An important tool from calculus is really handy here, that of differentiation. Recall that differentiating a function is finding the slope of the curve at that point. But the slope at that point, as we saw, is the instantaneous rate of change at that point. Hence we can use calculus to find the rate of change.

Let $y=f\left(x\right)$ be a function that describes a curve and let us assume that the function is continuous and differentiable over the appropriate interval. Then the slope of the graph at a point $\left({x}_{1},{y}_{1}\right)$ is given by:

$Slope=\frac{dy}{dx}$ $evaluatedat\left({x}_{1},{y}_{1}\right)$

And this is the formula for the instantaneous rate of change of y with respect to x at that point.

The Instantaneous Rate of Change of $y=f\left(x\right)$ with respect to x at a point $\left({x}_{1},{y}_{1}\right)$ is determined by evaluating $\frac{dy}{dx}$ at that point.

The instantaneous rate of change at a point, StudySmarter Originals

We can also calculate the instantaneous rate of change of x with respect to y by calculating $\frac{dx}{dy}$ at that point. Since the majority of functions are given explicitly in terms of x, we can use the following identity to find the instantaneous rate of change of x with respect to y;

$\frac{dx}{dy}=\frac{1}{\frac{dy}{dx}}$

This implies a very crucial property: the instantaneous rate of change of y with respect to x is reciprocal of the instantaneous rate of change of x with respect to y.

## Instantaneous Rate of Change as Limits

We have already seen how we can calculate the instantaneous rate of change using derivatives, but where does that come from? How do we deduce that? We will here explore how limits help in relating the instantaneous rate of change to derivatives.

Recall that the average rate of change is taken over a big interval and is given by $\frac{\Delta y}{\Delta x}$ for a function$y=f\left(x\right)$.

The Average rate of change over two points, StudySmarter Originals

If we bring points A and B closer and closer to each other, they will eventually become a single point. This is the reason we use the limiting condition $\underset{\Delta x\to 0}{\mathrm{lim}}$ and $\underset{\Delta y\to 0}{\mathrm{lim}}$ to avoid both points merging together.

Now notice that as A and B get closer and closer, the distance between them is infinitesimally small and the line we draw through them becomes a tangent line:

Two points infinitesimally close to each other, StudySmarter Originals

As it can be seen, as the distance between the points approaches 0, the secant line becomes a tangent line and the average rate of change becomes an instantaneous rate of change at that point. If we zoom on the two points, we see that the curve becomes a straight line and our tangent proposition is geometrically justified:

Two points infinitely close resulting into a tangent, StudySmarter Originals

We can name the coordinates at A as $\left(x,f\left(x\right)\right)$ and the let the coordinates of point B as $\left(x+a,f\left(x+a\right)\right)$ since the values will vary very slightly as they are extremely close to each other. Hence we have ${x}_{1}=x$, ${x}_{2}=x+a$ and ${y}_{1}=f\left(x\right)$, ${y}_{2}=f\left(x+a\right)$.

Now using the formula of the slope:

$\frac{\Delta y}{\Delta x}=\frac{f\left(x+a\right)-f\left(x\right)}{x+a-x}$$=\frac{f\left(x+a\right)-f\left(x\right)}{a}$

Now since the points are very close, we can use the limiting condition:

$\underset{a\to 0}{\mathrm{lim}}\frac{\Delta y}{\Delta x}$$=\underset{a\to 0}{\mathrm{lim}}\frac{f\left(x+a\right)-f\left(x\right)}{a}$

Now recall from the definition of a derivative that:

$\frac{dy}{dx}=\underset{a\to 0}{\mathrm{lim}}\frac{f\left(x+a\right)-f\left(x\right)}{a}$

This eventually gives us the formula of Instantaneous rate of change at a point, since A and B eventually merge into a single point under the limiting condition.

## Examples of Instantaneous Rate of Change

For the function given by $y=3{x}^{2}-x+4$, find the rate of change of y with respect to x at the point (0,4) and explain what it means.

Solution:

Step 1: Find the derivative of the function with respect to x:

$\frac{dy}{dx}=\frac{d}{dx}\left(3{x}^{2}-x+4\right)$$=6x-1$

Step 2: Evaluate the derivative at the given point, here it is (0,4)

$\frac{dy}{dx}=0-1=-1at\left(0,4\right)$

Hence the rate of change of y with respect to x is -1 at point (0,4).

Thus, the value of the function is decreasing by a unit of 1 with respect to x at the point (0,4).

An object is thrown and it follows a parabolic path and then falls down. The trajectory of the object is given by the equation $y=\frac{{x}^{2}}{2}-x$. Find the rate of change of height attained by the object with respect to the horizontal distance covered when it reaches a distance 4 m and a height of 4 m.

Solution:

Step 1: Convert the given data into mathematical terms:

y = height attained, x = horizontal distance covered, the point where we want to find the rate of change is (4,4).

Step 2: Differentiate the given function $y=\frac{{x}^{2}}{2}-x$:

$\frac{d}{dx}\left(\frac{{x}^{2}}{2}-x\right)=x-1$

Step 3: Evaluate the derivative at the given point (4,4):

$\frac{dy}{dx}=4-1=3$

Hence the rate of change of height with respect to horizontal distance is 3 when the object is at 4 m height and 4 m away.

Find the rate of change of y with respect to x for the function ${y}^{2}=\mathrm{ln}\left|{x}^{3}-4\right|$ at the point (2,1)

Solution:

Step 1: Take the derivative of the function and find $\frac{dy}{dx}$ :

$2y\frac{dy}{dx}=\frac{3{x}^{2}}{{x}^{3}-4}$

$\frac{dy}{dx}=\frac{3{x}^{2}}{2y\left({x}^{3}-4\right)}$

Step 2: Evaluate the derivative at (2,1):

$\frac{dy}{dx}=\frac{3.{2}^{2}}{2.1\left({2}^{3}-4\right)}$$=\frac{3}{2}$

Hence the rate of change of y with respect to x is $\frac{3}{2}$ at the point (2,1).

## Instantaneous Rate of Change - Key takeaways

• The Instantaneous Rate of Change at a point on a curve is the rate at which the quantity is changing relative to the other one at the point (or instant).
• The rate of change at a point on a curve is the slope of the tangent line drawn at that point.
• For a function y=f(x), the rate of change of y with respect to x is at a certain point is reciprocal of the rate of change of x with respect to y at that point.
• The formula for the rate of change at a point $\left({x}_{1},{y}_{1}\right)$ is given by $\frac{dy}{dx}$ evaluated at that point.

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How to find instantaneous rate of change without derivatives?

Another way of finding the instantaneous rate of change at a point is by calculating the tangent at that point.

What is an example of instantaneous rate of change?

Suppose a car has covered 40 km in 20 minutes. To find the rate of change of distance relative to time at any point over the interval of that time would be a typical example.

How to calculate instantaneous rate of change?

By finding the derivative of the curve and evaluating it at that point.

What is instantaneous rate of change?

The Instantaneous Rate of Change at a point on a curve is the rate at which the quantity is changing relative to the other one at the point (or instant).

What is the equation for Instantaneous Rate of Change?

For a function y=f(x), the instantaneous rate of change at a point (x1,y1) is given by dy/dx evaluated at (x1,y1).

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