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# Rate of Change In Calculus, the rate of change equation can easily be obtained from the slope equation. The slope of the equation is also called as rate of change of the equation.

The subject of calculus had its origin mainly in the geometrical problem of determination of the gradient of a curve at a point, thereby resulting in the determination of the tangent at the point. This subject has also rendered possible precise formulations of a large number of physical concepts such as velocity at an instant, acceleration at an instant, curvature at a point, density at a point, specific heat at any temperature. All these appears as local or instantaneous rate of change as against the average rate of change.

Many practical relationships involve independent and dependent quantities. For example, the volume of water in a tank changes when there is change in the depth of water in the tank. Hence, the depth of the tank is the independent quantity and the volume is the dependent quantity. When we through a stone on surface of water there are ripples whose radius increases for few sec. That is the area of the circle increases as the radius increases. Hence, we see that there is increase in area of the circles as there is increase in radius. Hence, radius is the independent variable and area is the dependent variable. These are some of the rate of change examples.

 Related Calculators Average Rate of Change Calculator Calculating Rate of Change Instantaneous Rate of Change Calculator Percent Change

## Rate of Change Definition

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Basically, the ratio of the change in the output value and change in the input value of a function is called as rate of change. Rate of change is the ratio that shows the relationship between the two variables in equation. In general, the coefficient of $x$ is called as the rate of change of an equation. For example, in $y = 3x + 4$, the rate of change is $3$. $3$ is the coefficient of $x$. In this equation, the constant variable is $4$.

### Rate of Change in Position

The displacement in position is known as the rate of change in position. If a particle is initially at A and after time t it is at the point B, then
Rate of change in position = $\frac{\text{Particle at Point B} - \text{Particle at Point A}}{\text{Time Taken for Change in Position}}$.

### Rate of Change in Velocity

Velocity is the speed of a particle or object. The rate of change in velocity is described as the change in position(displacement) divided by the total change in time. Speed is always positive and velocity indicates the direction.

Velocity = $\frac{\text{Displacement}}{\text{Time}}$

Rate of Change in Velocity = $\frac{\text{Final Velocity of the Body - Initial Velocity of the body}}{\text{Time Taken for Change in the Velocity}}$

### Rate of Change in Acceleration

Acceleration is the second order derivative of the displacement.
Rate of change in acceleration = $\frac{\text{Final Acceleration - Initial Acceleration}}{\text{Time Taken for Change}}$

### Rate of Change Examples

Given below are some of the examples on rate of change.

### Solved Examples

Question 1: Find the rate of change of the given points $A(7,3)$ and $B(9,5)$
Solution:
We know that,
Rate of change in point(slope) = $\frac{y_{2} - y_{1}}{x_{2} - x_{1}}$

= $\frac{5 - 3}{9 - 7}$

= $2$

Question 2: A man driving a bus covers $60$ miles in one hour. Over the next two hours, he drives $100$ miles. What is his velocity over the next two hours.
Solution:
Here, we denoted $d$ as the distance and $t$ as the time. We use the points $(1,60)$ and $(3,100)$. Here, 2 additional hours gives us a t value of $3$ and the total number of miles is $d = 60 + 100 = 60$.
So, the velocity is the rate of change of the distance with respect to time. It is given as $\frac{d_{2} - d_{1} }{t_{2} - t_{1}}$

= $\frac{160 - 60}{3 - 1}$

= $\frac{100}{2}$

= $50 \text{mph}$

## Rate of Change Formula

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In calculus, rate of change formula is easily derived from the slope of the equation. The rate of change is equal to slope of the equation.

### Find the Rate of Change:

The formula for slope equation can be written as,

Slope = $\frac{\text{Rise of the equation}}{\text{Run of the equation}}$

Slope = $\frac{y_{2} -y_{1}}{x_{2} - x_{1}}$

The formula for rate of change is given as follows:
Rate of change of $x$ with respect to $y$ is given as $\frac{dx}{dy}$

Rate of change of $y$ = $\frac{dx}{dy}$

This is the relationship between rate of change and slope.

### Solved Example

Question: Find the rate of change of the given points $X(3,7)$ and $Y(4,2)$
Solution:
We know that the rate of change is given by the formula,
Slope = $\frac{y_{2} - y_{1}}{x_{2} - x_{1}}$

Here, $(3,7)$ and $(4,2)$ are $(x_{1},y_{1})$ and $(x_{2},y_{2})$ respectively. Then,

Slope = $\frac{7 - 2}{4 - 3}$

= $5$

## Average Rate of Change

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In geometrical concepts, the term "rate of change" is the slope of the line joining the two points of the line. This is also called as the average rate of change. The average rate of change between the points $(4,3)$ and $(6,7)$ is the change in the y-coordinates over the change in the x-coordinates,which is also called as the slope of the line.

Average rate of change formula is m = $\frac{y_{2} -y_{1}}{x_{2} - x_{1}}$ where, m is called the slope of the line joining the points $(x_{1},y_{1})$ and $(x_{2},y_{2})$.
The slope of the line in the graph = m = $\frac{7 - 3}{6 - 4}$ = $\frac{4}{2}$ = $2$
Hence, the average rate of change of the line joining the points $(4,3)$ and $(6,7)$ is $2$.

### Average Rate of Change Formula

For any two points $(x_{1},y_{1})$ and $(x_{2},y_{2})$ in the co-ordinate plane, if $\delta \text{x}$ is the change in $x$ and if $\delta \text{y}$ is the corresponding change in $y$, then the difference quotient $\frac{\delta y}{\delta x}$ = $\frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ is called the average rate of change of $y$ with respect to $x$ over the interval $(x_{1},y_{1})$. It is also called as the ratio of change in y-coordinates over the change in x-coordinates. We can easily find the average rate of change with the help of the average rate of change formula.

A function is one where there is a relationship between $x$ and $y$ is denoted as $y = f(x)$, where $x$ is an independent quantity and $y$ a dependent quantity.

### Find the Average Rate of Change

To find the average rate of change, we can use formula shown below:

Average change = $\frac{\text{Change in y Value}}{\text{Change in x Value}}$

### Solved Examples

Question 1: Let $y = x^{3} - 2$. Find the average rate of change of $y$ with respect to $x$ over the interval $[3,4]$.
Solution:
We have $y = x^{3} - 2$ ..........................................(1)
For average rate of change, put $x = 3$ in (1)
$f(3) = (3)^{3} - 2$ = $27 - 2$ = $25$

Again, put $x = 4$,

$f(4) = (4)^{3} - 2 = 64 - 2 = 62$

The average rate of change over the interval $[3,4]$ = $\frac{f(4) - f(3)}{4-3}$

= $\frac{62 - 25}{4-3}$

= $37$

Question 2: Area of a circle $A(x) = \pi x^{2}$, where $x$ is the radius. Here, $x$ can take any values.
Therefore, it is called as the independent variable and $y = A(x)$ is a dependent variable which depends on the value of $x$. Find the average rate of which the area of the circle changes with $x$ as the radius which changes from $x = 3$ to $x = 5$.
Solution:
We have, $A(x) = \pi x^{2}$
Put $x = 3$, then $A(3) = \pi (3)^{2} = 9 \pi$
Again, $x = 5$ then $A(5) = \pi (5)^{2} = 25 \pi$

Then, the average rate of change is
$\frac{A(5) - A(3)}{5 -3}$

= $\frac{25 \pi - 9 \pi}{5 - 3}$

= $\frac{16 \pi} {2}$

= $8 \pi$

## Average Rate of Change of a Function

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For a given function $y = f(x)$, the average rate of change from a point $x = a$ to the point $x = b$ is the difference quotient. We can easily find the average rate of change of a function with the help of the formula for the average rate of change of a function.

### Find the Average Rate of Change of the Function:

Average rate of change of a function is denoted as:

$\frac{\delta y}{\delta x}$ = $\frac{f(b) - f(a)}{b - a}$

Here, $(b - a)$ is the change in the input of the function $f$, ${f(b) - f(a)}$ is the change in the function $f$ as the input change from $a$ to $b$ and $\frac{\delta y}{\delta x}$ is the average rate of change function

## Maximum Rate of Change

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The maximum rate of change is obtained when vector $v$ is into the same direction of the gradient $f(x,y)$. The maximum rate of change is the magnitude of the gradient $(x,y)$. Therefore, $grad(x,y)$ is the direction of the maximum rate of change of function $f$ at the point $(x,y)$.

### Solved Example

Question: If $f(t) = xy^{2}z^{2}$, then find the maximum rate of change of the function at $(2,1,3)$
Solution:
The maximum rate of change of the function at $(2,1,3)$ occurs along the direction parallel to $\triangledown f$ at $(2,1,3)$.

For this, $\frac{\partial f}{\partial x}$ = $y^{2}z^{2}$

$\frac{\partial f}{\partial y}$ = $2xyz^{2}$

$\frac{\partial f}{\partial z}$ = $2xy^{2}z$

Then, $\triangledown f$ = $y^{2}z^{2}i + 2xyz^{2}j + 2xy^{2}zk$. Here, $i, j, k$ are the unit vectors in the direction $x, y$ and $z$ respectively. Therefore, at the point $(2,1,3)$, the $grad f = \triangledown f$ at $(2,1,3)$
= $9i + 36j + 12k$
So, the maximum rate of change of $f(t)$ is $|\text{grad} f| = \sqrt{9^{2} + 36^{2}+ 12^{2}}$
=$\sqrt{81 + 144 + 1296}$
= $\sqrt{1521}$

## Instantaneous Rate of Change

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Instantaneous rate of change is the limit of the function that describes the average rate of change. It is used to describe how an object travel in space or around the ground. For example, the value of a function at a point $x = c$ is $f(c)$. If there is a small increment in $x$, which is $\delta x$, then the value of the function at $c + \delta x$ is $f(c + \delta x)$. Hence, the change in the value of the function is $f(c + \delta x) - f(c)$. By the term difference of quotient, we mean the change in the value of the function over the change in the value of $x$ which is $\delta x$.

### Instantaneous Rate of Change Formula:

It is also the difference of quotient = $\lim_{\delta x\rightarrow 0}$ $\frac{\delta y}{\delta x}$ = $\frac{t(a + \delta x) - t(a)}{\delta x}$

### Solved Example

Question: Find the instantaneous rate of change of the function $f(x) = 2x^{2} + x - 3$ at $x = 4$.
Solution:
For the instantaneous rate of change, calculate the first derivative of the given function.
$f'(x) = 4x + 1$
Therefore, for instantaneous rate of change at $x = 4$
$f'(4) = 4 x 4 + 1 = 17$

## Constant Rate of Change

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Objects moving along a straight line are said to move at constant rate of change.

### Constant Rate of Change Definition

The constant rate of change definition states that "If the object moves uniformly with respect to time, then it is said to move at a constant rate of change". For example, observe the following table

 Distance (km) 100 200 300 400 Time taken (hrs) 2 4 6 8

In the above table, as the distance increases the time taken also increases.
Rate of change = $\frac{200 - 100}{4 - 2}$ = $\frac{100}{2}$ = $50$ km/hr, which is also equal to the rate of change between $(4,200)$ and $(8,400)$.
For an object moving with constant rate of change, the motion of the path will be linear. The rate of change will be the slope of the line. The path described by the above a object moving as per the table is,
$y - y_{1} = m(x - x_{1})$ [Point slope formula]

$y - 200 = 50(x - 4)$

$y - 200 = 50x - 200$

$y = 50x - 200 + 200 = 50x$

$y = 50x$ is the equation of the line, which is the path described by the object.

## Rate of Change of a Function

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To find the rate of change of a function we may compare this function with linear function. If we have $y = mx + c$ as a function. A change of $\delta x$ in $x$ produces a change $\delta y$ = $m \delta x$ in $y$, so the rate of change of a function is given by the ratio $\frac{\delta y}{\delta x}$ is equal to $m$, independent of $x$ and $\delta x$.

## Relative Rate of Change

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The relative rate of change of a function $f(x)$ is given by the relation $\frac{f '(x)}{f(x)}$, where $f'(x)$ is the first derivative of the function.

### Solved Example

Question: If $f(x) = 3x + 1$ is the function. Then, find the relative rate of change at $x = 1$.
Solution:
Given that $f(x) = 3x + 1$
$f'(x) = 3$
The relative rate of change at $x = 1$ = $\frac{f '(x)}{f(x)}$
= $\frac{3}{3x +1}$
= $\frac{3}{3 \times 1 + 1}$
= $\frac{3}{4}$

## Rate of Change and Slope

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A rate of change is a ratio that compares the amount of change in a dependent variable to the amount of change in a independent variable. So, we say that
Rate of Change = $\frac{\text{Amount of Change in a Dependent Variable}}{\text{Amount of Change in an Independent Variable}}$

Slope means gradient or incline. By the use of slope, we can easily describe the measurement of the steepness of a straight line. The slope of a line is a rate of change.
Slope = $\frac{\text{Vertical Change}}{\text{Horizontal Change}}$ = $\frac{\text{Rise}}{\text{Run}}$

If all the connected line segment of a line have the same rate of change, then they all have the same steepness and together form a straight line. So, the constant rate of change of a line is called the slope of the line. If we find the slope, then we can easily find the rate of change over the given period.

## Percentage Rate of Change

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The percentage rate of change is $100 \times$ $\frac{f '(x)}{f(x)}$

### Solved Example

Question: If $f(x) = 3x + 1$ is the function, then find the percentage rate of change at $x = 1$.
Solution:
Given that $f(x) = 3x + 1$ then $f'(x) = 3$
The percentage rate of change at $x = 1$ is $100 \times$ $\frac{f '(x)}{f(x)}$

= $100 \times $$\frac{3}{3x + 1} = 100 \times$$\frac{3}{3 x 1 + 1}$

= $100 \times$$\frac{3}{4}$

= $100 \times 0.75$

= $75 %$

## Rate of Change Problems

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Given below are some of the rate of change problems.

### Solved Examples

Question 1: Find the rate of change of the given points $(3,4)$ and $(0,- 2)$
Solution:
Given points $x_{1} = 3$, $x_{2} = 0$ and $y_{1} = 4$, $y_{2} = -2$

Formula:

Rate of change = $\frac{y_{2} - y_{1}}{x_{2} - x_{1}}$

= $\frac{-2 - 4}{0 - 3}$

= $2$.

The rate of change is $2$.

Question 2: Calculate rate of change of the given algebraic equation $y = 4x + 9$
Solution:
Given $y = 4x + 9$

Formula:

Rate of change = $\frac{dy}{dx}$

Therefore,

$y = 4x + 9$

Differentiating with respect to $x$, we get

Rate of change of $y$ = $\frac{dy}{dx}$

$\frac{dy}{dx}$ = 4$The rate of change of$y$is$4$Question 3: Find the rate of change of the given algebraic equation$y = 2x^{2} + 7x - 5$Solution: Given$y = 2x^{2} + 7x - 5$Formula: Rate of change of$y$=$\frac{dy}{dx}$Therefore, differentiating the given equation with respect to$x$, we get$\frac{dy}{dx}$=$4x + 7$The rate of change of$y$is$4x + 7$. ## Rate of Change Word Problems Back to Top Given below are some of the rate of change word problems. ### Solved Examples Question 1: A toy rocket is launched straight up so that its height$s$, in meters, at time$t$, in seconds , is given by$s(t) = -2t^{2} + 30t + 5$. Calculate the instantaneous velocity of the rocket at$t = 3$. Solution: Since$s(t) = -2t^{2} + 30t + 5s(3 + h) = -2(3 + h)^{2} + 30(3 + h) + 5$=$-2(9 + 6h + h^{2}) + 90 + 30h + 5$=$-18 - 12h - 2h^{2} + 90 + 30h + 5$=$-2h^{2} + 18h + 77S(3) = -2(3^{2}) + 30(3) + 5 = -2(9) + 90 + 5 = -18 + 95 = 77$The instantaneous velocity at$t = 3$is,$V(3)$=$\lim_{h \to 0}\frac{s(3 + h) - s(3)}{h}$=$\lim_{h \to 0}\frac{-2h^{2} + 18h + 77 - 77}{h}$=$\lim_{h \to 0}\frac{-2h^{2} + 18h}{h}$=$\lim_{h \to 0}\frac{h(-2h + 18)}{h}$=$\lim_{h \to 0}(-2h + 18)$=$0 + 18 = 18$m/sec Velocity of the toy rocket at$t = 3 sec = 18$m/sec. Since the velocity is positive the rocket will be moving upward. If the velocity at the given instant is negative the rocket will be moving downward. Question 2: A pebble is dropped from a cliff,$50$m high. After$t$sec, the pebble is$s$meters above the ground, where$s(t) = 50 - 2t^{2}$,$0 \leq t \leq 5$. Calculate the average velocity average rate of change of the pebble between$t = 2$sec and$t = 4$sec. Solution: Average velocity =$\frac{\text{Change in distance}}{\text{Change in time}}S(t)= 50 - 2t^{2}S(2) = 50 - 2(2^{2}) - 50 - 4 = 46S(4) = 50 - 2(4^{2}) = 50 - 32 = 18$Average velocity =$\frac{s(4) - s(2)}{4 -2}$=$\frac{18 - 46}{2}$=$\frac{-28}{2}$=$-14$m/sec Hence, the Average rate of change between$t = 2$sec and$t = 4$sec is$-14$m/sec. ### Practice Problems Question 1: An oil tank is to be drained for cleaning. After$t$minutes, there are$V$litres of water in the tank, where$V(t) = 35(20 - t)^{2}$,$0 \leq t \leq 20$(a) Calculate the average rate of change in volume during the first$20$min. (b) Calculate the rate of change in volume at time$t = 20$. Question 2: The average annual salary of a professional baseball player can be modeled by the function$S(x)= 200 + 3x - 6x^{2} + 12x^{3}$, where$S$represents the average annual salary, in thousands of dollars, and$x$is the number of years, since$1982$. Determine the rate at which the average salary was changing in$2005$. ## Rate of Change Graph Back to Top By the use of slope of a line we can easily find the rate of change. ### Solved Example Question: Find the slope of a line passing through the points$(-2,-4)$and$(3,6)$. Solution: We can draw a graph with the help of these points and count the vertical changes and the horizontal changes to use in the formula. Slope =$\frac{\text{Vertical Change}}{\text{Horizontal Change}}$=$\frac{6 + 4}{3 + 2}$=$2\$

If we have a function, then we can easily draw a graph of that and calculate the required rate of change.

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