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Rate of Change Word Problems

Rate of change refers to the rate at which a function changes when its independent variable changes at a specific rate. It is an application of derivatives. In terms of a graph, the rate of change of a function is represented by the slope of the function at a given point.
Rate of Change Word Problems
 The rate of change of the green curve at the point $P$ is equal to the slope of the line that is tangential to it at the point $P$.

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Formula

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There are primarily two concepts pertaining to rate of change:

Average Rate of Change: The formula for average rate of change of a function between two points $x$ = $a$ and $x$ = $b$ can be given by:

Avg rate of change = $\frac{f(b) - f(a)}{b - a}$

Instantaneous Rate of Change: The instantaneous rate of change of a function can be found using the following formula:

Instantaneous rate of change = $f'(x)$ = $lim_{x \rightarrow h}$⁡ $\frac{f(x + h)\ -\ f(x)}{h}$

The $f'(x)$ can also be denoted by $\frac{dy}{dx}$. The above formula is nothing but the limit definition of derivative. We can use the various derivative formulas also to find the instantaneous rate of change.

Examples

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Example 1: 

Suzy is a writer. She is able to write the same number of pages every hour. If she has written $40$ pages by the end of $2^nd$ hour and $100$ pages by the end of the $5^th$ hour, then what is the rate at which she writes?

Solution:

From the question we see that $a$ = $2$ and $b$ = $5$. Also from the question we see that $f(b)$ = $100$ and $f(a)$ = $40$. Thus the average rate of change would be:

Avg.rate of change = $\frac{f(b)\ -\ f(a)}{b - a}$

Avg.rate of change = $\frac{100 - 40}{5 - 2}$

Avg.rate of change = $\frac{60}{3}$ = $20 \leftarrow\ Answer$
Example 2: 

A wrestler weighted $190$ lbs in $1996$ and $205$ lbs in $2005$. Find the average rate of change of weight of the wrestler.

Solution:

The formula for average rate of change is:

Avg.rate of change = $\frac{f(b)\ -\ f(a)}{b - a}$

For this problem, the function $f(x)$ refers to the weight and the variable $x$ is the year. Thus we have:

$f(1996)$ = $190$ lbs

And

$f(2005)$ = $205$ lbs

Plugging in the values into the above formula we have:

= $\frac{f(2005)\ -\ f(1996)}{2005 - 1996}$

Replacing the functions with their corresponding values we have:

= $\frac{205 - 190}{2005 - 1996}$

Simplifying and evaluating that gives:

= $\frac{15}{9}$ = $\frac{5}{3}$ = $1$ $\frac{2}{3}$ lbs per year.
Example 3: 

Two cars, a Beetle and a sedan are initially $600$ miles apart. The Beetle is to the west of the sedan. Beetle is moving towards east at a speed of $40$ miles per hour and at the same time the sedan is moving south at a speed of $60$ miles per hour. What is the rate of change of distance between the two cars at the end of $4$ ours of driving?

Solution:

For such problems, it is best to first make a picture of the situation described in the question. It would probably look something like this:

Word Problem on Rate of Change

From the figure we see that:

$x$ = Distance between final position of beetle and sedan's initial position

$y$ = Distance moved by sedan

$D$ = Distance between the two cars.

The distance travelled by the Beetle in $3$ hours at the speed of $40$ miles per hour would be:

= $40 \times 3$ = $120$ miles

Thus, the value of $x$ would be:

$x$ = $600 - 120$ = $480$ miles

The distance travelled by the sedan in $3$ hours at $60$ miles per hour would be:

= $y$ = $60 \times 3$ = $180$ miles

So now we can find the distance between the two cars, $d$ using the Pythagorean theorem:

$D^2$ = $x^2 + y^2$

$D^2$ = $480^2 + 180^2$

$D^2$ = $230400 + 32400$

$D^2$ = $262800$

$D$ = $\sqrt{262800}$

$D$ = $512.64$ miles

However we are not interested in this distance $D$. What we are actually interested in is the rate of change of this $D$ with respect to time $t$. That can be represented by $\frac{dD}{dt}$.

We know from the question that:

$\frac{dx}{dt}$ = $-40$ miles per hour

And

$\frac{dy}{dt}$ = $60$ miles per hour

The $\frac{dx}{dt}$ is negative since the distance $x$ is decreasing with time. 

From the Pythagorean theorem we have:

$D^2$ = $x^2 + y^2$

Differentiating this equation implicitly with respect to time $t$ we have:

$2D$ $\frac{dD}{dt}$ = $2x$ $\frac{dx}{dt}$ + $2y$ $\frac{dy}{dt}$

Dividing the entire equation by $2$ gives:

$D$ $\frac{dD}{dt}$ = $x$ $\frac{dx}{dt}$ + $y$ $\frac{dy}{dt}$
 
Substituting the values of $d, x, y$, $\frac{dx}{dt}$ and $\frac{dy}{dt}$ into the above equation we have:

$512.64$ $\frac{dD}{dt}$ = $480(-40) + 180(60)$

Simplifying the right side we have:

$512.64$ $\frac{dD}{dt}$ = $-19200 + 10800$

$512.64$ $\frac{dD}{dt}$ = $-8400$

Dividing both the sides by $512.64$ we have:

$\frac{dD}{dt}$ = $\frac{-8400}{512.64}$ = $-16.39$ miles per hour

Thus the distance between the two cars is DECREASING at a rate of $16.39$ miles per hour.
Example 4:

Water is being filled into a rectangular water tank at the rate of $30$ litres per second. The dimensions of the base of the tank are $2$ meters by $3$ meters. At what rate is the height of the water increasing in the tank?

Solution:

The volume of a rectangle tank is given by:

$V$ = $L \times W \times h$

From the question we know that:

$L$ = $3$ meters

$W$ = $2$ meters

So the volume equation would now look like this:

$V$ = $3 \times 2 \times h$

$V$ = $6h$

The rate of change of volume can be found by taking derivative of this function with respect to time $t$. Thus we have:

$\frac{dV}{dt}$ = $6$ $\frac{dh}{dt}$

However, we are given in the question that the rate of change of volume is $30$ liters per second. Replacing the $\frac{dV}{dt}$ in the above equation by that we have:

$30$ = $6$ $\frac{dh}{dt}$

Solving that for $\frac{dh}{dt}$ we have:

$\frac{30}{6}$ = $\frac{dh}{dt}$

$\frac{dh}{dt}$ = $5$ litres per seconds $\leftarrow\ Answer!$

Thus the rate at which the height is increasing in the tank is $5$ litres per second.
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