# Solve Rate of Change Problems in Calculus

Rate of change calculus problems and their detailed solutions are presented.

## Problem 1

A rectangular water tank (see figure below) is being filled at the constant rate of $$20$$ liters / second. The base of the tank has dimensions $$w = 1$$ meter and $$L = 2$$ meters. What is the rate of change of the height of water in the tank? (express the answer in cm / sec).

Solution to Problem 1:
The volume $$V$$ of water in the tank is given by.
$$V = w \times L \times H$$
We know the rate of change of the volume $$\dfrac{{dV}}{{dt}} = 20$$ liter /sec.
We need to find the rate of change of the height $$H$$ of water $$\dfrac{{dH}}{{dt}}$$. $$V$$ and $$H$$ are functions of time.
Differentiate both sides of the above volume formula to obtain
$$\dfrac{{dV}}{{dt}} = w \times L \times \dfrac{{dH}}{{dt}}$$
Note that $$w$$ and $$L$$ do not change with time and are therefore considered as constants in the above operation of differentiation.
We now find a formula for $$\dfrac{{dH}}{{dt}}$$ as follows.
$$\dfrac{{dH}}{{dt}} = \dfrac{{\dfrac{{dV}}{{dt}}}}{{w \times L}}$$
We need to convert liters into cubic cm and meters into cm as follows
$$1$$ liter = $$1$$ cubic decimeter
$$= 1000$$ cubic centimeters
$$= 1000$$ cm^3
and $$1$$ meter = $$100$$ centimeter.
We now evaluate the rate of change of the height $$H$$ of water.
$$\dfrac{{dH}}{{dt}} = \dfrac{{\dfrac{{dV}}{{dt}}}}{{w \times L}}$$
$$= \dfrac{{(20 \times 1000 \, \text{cm}^3/\text{sec})}}{{(100 \, \text{cm} \times 200 \, \text{cm})}}$$
$$= 1$$ cm / sec.

## Problem 2

An airplane is flying in a straight direction and at a constant height of $$5000$$ meters (see figure below). The angle of elevation of the airplane from a fixed point of observation is $$a$$. The speed of the airplane is $$500$$ km / hr. What is the rate of change of angle $$a$$ when it is $$25$$ degrees? (Express the answer in degrees / second and round to one decimal place).

Solution to Problem 2:
The airplane is flying horizontally at the rate of $$\dfrac{{dx}}{{dt}} = 500$$ km/hr. We need a relationship between angle $$a$$ and distance $$x$$. From trigonometry, we can write
$$\tan \, a = \dfrac{{h}}{{x}}$$
angle $$a$$ and distance $$x$$ are both functions of time $$t$$. Differentiate both sides of the above formula with respect to $$t$$.
$$\dfrac{{d( \tan \, a)}}{{dt}} = \dfrac{{d(\dfrac{{h}}{{x}})}}{{dt}}$$
We now use the chain rule to further expand the terms in the above formula
$$\dfrac{{d( \tan \, a)}}{{dt}} = (sec^2 \, a) \dfrac{{da}}{{dt}}$$
$$\dfrac{{d(\dfrac{{h}}{{x}})}}{{dt}} = h \left( \dfrac{{-1}}{{x^2}} \right) \dfrac{{dx}}{{dt}}$$.
(note: height $$h$$ is constant)
Substitute the above into the original formula to obtain
$$( \sec^2 \, a) \dfrac{{da}}{{dt}} = h \left( \dfrac{{-1}}{{x^2}} \right) \dfrac{{dx}}{{dt}}$$
The above can be written as
$$\dfrac{{da}}{{dt}} = \left[ h \left( \dfrac{{-1}}{{x^2}} \right) \dfrac{{dx}}{{dt}} \right] / ( \sec^2 \, a)$$
We now use the first formula to find $$x$$ in terms of $$a$$ and $$h$$ follows
$$x = \dfrac{{h}}{{ \tan \, a}}$$
Substitute the above into the formula for $$\dfrac{{da}}{{dt}}$$ and simplify
$$\dfrac{{da}}{{dt}} = \left[ h \left( \dfrac{{- \tan^2 a}}{{h^2}} \right) \dfrac{{dx}}{{dt}} \right] / ( \sec^2 \, a)$$
$$= \left[ \left( \dfrac{{- \tan^2 a}}{{h}} \right) \dfrac{{dx}}{{dt}} \right] / (sec^2 \, a)$$ Use the values for $$a$$, $$h$$ and $$\dfrac{{dx}}{{dt}}$$ to approximate $$\dfrac{{da}}{{dt}}$$ with the right conversion of units: $$1 \, \text{km} = 1000 \, \text{m}$$ and $$1 \, \text{hour} = 3600 \, \text{sec}$$.
$$\dfrac{{da}}{{dt}} = \left[ - sin^2(25 \, \text{deg}) \dfrac{{500 \, \text{000} \, \text{m}}}{{3600 \, \text{sec}}} \right]$$
$$= -0.005 \, \text{radians/sec}$$
$$= -0.005 \times \left[ \dfrac{{180 \, \text{degrees}}}{{\pi \, \text{radians}}} \right] / \text{sec}$$
$$= -0.3 \, \text{degrees/sec}$$

## Problem 3

If two resistors with resistances $$R_1$$ and $$R_2$$ are connected in parallel as shown in the figure below, their electrical behavior is equivalent to a resistor of resistance $$R$$ such that
$$\dfrac{{1}}{{R}} = \dfrac{{1}}{{R_1}} + \dfrac{{1}}{{R_2}}$$

If $$R_1$$ changes with time at a rate $$r = \dfrac{{dR_1}}{{dt}}$$ and $$R_2$$ is constant, express the rate of change $$\dfrac{{dR}}{{dt}}$$ of the resistance of $$R$$ in terms of $$\dfrac{{dR_1}}{{dt}}$$, $$R_1$$ and $$R_2$$.

Solution to Problem 3:
We start by differentiating, with respect to time, both sides of the given formula for resistance $$R$$, noting that $$R_2$$ is constant and $$\dfrac{{d(1/R_2)}}{{dt}} = 0$$
$$\left( \dfrac{{-1}}{{R^2}} \right) \dfrac{{dR}}{{dt}} = \left( \dfrac{{-1}}{{R_1^2}} \right) \dfrac{{dR_1}}{{dt}}$$
Arrange the above to obtain
$$\dfrac{{dR}}{{dt}} = \left( \dfrac{{R}}{{R_1}} \right)^2 \dfrac{{dR_1}}{{dt}}$$
From the formula $$\dfrac{{1}}{{R}} = \dfrac{{1}}{{R_1}} + \dfrac{{1}}{{R_2}}$$, we can write
$$R = \dfrac{{R_1 \times R_2}}{{R_1 + R_2}}$$
Substitute $$R$$ in the formula for $$\dfrac{{dR}}{{dt}}$$ and simplify
$$\dfrac{{dR}}{{dt}} = \left( \dfrac{{R_1 \times R_2}}{{R_1 \times (R_1 + R_2)}} \right)^2 \dfrac{{dR_1}}{{dt}}$$
$$= \left( \dfrac{{R_2}}{{R_1 + R_ 2}} \right)^2 \dfrac{{dR_1}}{{dt}}$$

## Exercises

1 - Find a formula for the rate of change $$\dfrac{{dV}}{{dt}}$$ of the volume of a balloon being inflated such that it radius $$R$$ increases at a rate equal to $$\dfrac{{dR}}{{dt}}$$.
2 - Find a formula for the rate of change $$\dfrac{{dA}}{{dt}}$$ of the area $$A$$ of a square whose side $$x$$ centimeters changes at a rate equal to $$2$$ cm/sec.
3 - Two cars start moving from the same point in two directions that makes $$90$$ degrees at the constant speeds of $$s_1$$ and $$s_2$$. Find a formula for the rate of change of the distance $$D$$ between the two cars.

### Solutions to the Above Exercises

1 -    $$\dfrac{{dV}}{{dt}} = 4 \times \pi \times R^2 \times \dfrac{{dR}}{{dt}}$$
2 -    $$\dfrac{{dA}}{{dt}} = 4x \, \text{cm}^2/\text{sec}$$
3 -    $$\dfrac{{dD}}{{dt}} = \sqrt{{s_1^2 + s_2^2}}$$

## References and Links

calculus problems