Rate of change calculus problems and their detailed solutions are presented below.
In what follows, \( C \) represents the constant of integration where applicable.
Worked Examples
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).
Show 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 volume \( \dfrac{dV}{dt} = 20 \) liters/sec. We need to find the rate of change of the height \( H \), which is \( \dfrac{dH}{dt} \). Since \( V \) and \( H \) are functions of time \( t \), we differentiate both sides of the volume formula with respect to \( t \):
\[ \frac{dV}{dt} = w \times L \times \frac{dH}{dt} \]Note that \( w \) and \( L \) do not change with time and are therefore treated as constants during differentiation.
Solving for \( \dfrac{dH}{dt} \):
\[ \frac{dH}{dt} = \frac{\dfrac{dV}{dt}}{w \times L} \]We convert units to centimeters and cubic centimeters:
- \( 1 \) liter = \( 1 \) cubic decimeter = \( 1000 \; \text{cm}^3 \)
- \( 1 \) meter = \( 100 \; \text{cm} \)
Substituting the values:
\[ \frac{dH}{dt} = \frac{20 \times 1000 \; \text{cm}^3/\text{sec}}{100 \; \text{cm} \times 200 \; \text{cm}} = \frac{20,000}{20,000} = 1 \; \text{cm/sec} \]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 \( 25^\circ \)? (Express the answer in degrees/second and round to one decimal place).
Show Solution to Problem 2
The airplane is flying horizontally at the rate of \( \dfrac{dx}{dt} = 500 \; \text{km/hr} \). From trigonometry, we relate angle \( a \), height \( h \), and horizontal distance \( x \):
\[ \tan a = \frac{h}{x} \]Since \( a \) and \( x \) are functions of time \( t \), we differentiate both sides with respect to \( t \):
\[ \frac{d(\tan a)}{dt} = \frac{d\left(\frac{h}{x}\right)}{dt} \]Using the chain rule:
\[ \frac{d(\tan a)}{dt} = (\sec^2 a) \frac{da}{dt} \] \[ \frac{d\left(\frac{h}{x}\right)}{dt} = h \left(-\frac{1}{x^2}\right) \frac{dx}{dt} \quad (\text{since } h \text{ is constant}) \]Equating both derivatives:
\[ (\sec^2 a) \frac{da}{dt} = -\frac{h}{x^2} \frac{dx}{dt} \implies \frac{da}{dt} = \frac{-\frac{h}{x^2} \frac{dx}{dt}}{\sec^2 a} \]From \( \tan a = \dfrac{h}{x} \), we have \( x = \dfrac{h}{\tan a} \). Substituting \( x^2 = \dfrac{h^2}{\tan^2 a} \):
\[ \frac{da}{dt} = \frac{-\frac{h \tan^2 a}{h^2} \frac{dx}{dt}}{\sec^2 a} = -\frac{\tan^2 a}{h} \frac{dx}{dt} \cdot \cos^2 a = -\frac{\sin^2 a}{h} \frac{dx}{dt} \]Using unit conversions (\( 1 \; \text{km} = 1000 \; \text{m} \), \( 1 \; \text{hour} = 3600 \; \text{sec} \)), speed \( \dfrac{dx}{dt} = 500 \; \text{km/hr} = \dfrac{500,000 \; \text{m}}{3600 \; \text{sec}} \), height \( h = 5000 \; \text{m} \), and \( a = 25^\circ \):
\[ \frac{da}{dt} = -\frac{\sin^2(25^\circ)}{5000} \left(\frac{500,000}{3600}\right) \approx -0.005 \; \text{radians/sec} \]Converting radians to degrees (\( 180/\pi \)):
\[ \frac{da}{dt} = -0.005 \times \frac{180}{\pi} \approx -0.2865 \approx -0.3 \; \text{degrees/sec} \]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:
\[ \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} \]If \( R_1 \) changes with time at a rate \( \dfrac{dR_1}{dt} \) and \( R_2 \) is constant, express the rate of change \( \dfrac{dR}{dt} \) in terms of \( \dfrac{dR_1}{dt} \), \( R_1 \), and \( R_2 \).
Show Solution to Problem 3
Differentiating both sides of the parallel resistance equation with respect to time \( t \) (noting \( R_2 \) is constant):
\[ -\frac{1}{R^2} \frac{dR}{dt} = -\frac{1}{R_1^2} \frac{dR_1}{dt} \]Rearranging for \( \dfrac{dR}{dt} \):
\[ \frac{dR}{dt} = \left(\frac{R}{R_1}\right)^2 \frac{dR_1}{dt} \]From \( \dfrac{1}{R} = \dfrac{1}{R_1} + \dfrac{1}{R_2} \), we can express \( R \) explicitly:
\[ R = \frac{R_1 R_2}{R_1 + R_2} \]Substituting \( R \) into the derivative equation:
\[ \frac{dR}{dt} = \left(\frac{\frac{R_1 R_2}{R_1 + R_2}}{R_1}\right)^2 \frac{dR_1}{dt} = \left(\frac{R_2}{R_1 + R_2}\right)^2 \frac{dR_1}{dt} \]Exercises & Solutions
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Find a formula for the rate of change \( \dfrac{dV}{dt} \) of the volume of a balloon being inflated such that its radius \( R \) increases at a rate equal to \( \dfrac{dR}{dt} \).
Show Solution to Exercise 1
\[ V = \frac{4}{3}\pi R^3 \implies \frac{dV}{dt} = \frac{4}{3}\pi (3R^2) \frac{dR}{dt} = 4\pi R^2 \frac{dR}{dt} \] -
Find a formula for the rate of change \( \dfrac{dA}{dt} \) of the area \( A \) of a square whose side length \( x \) centimeters changes at a rate equal to \( 2 \; \text{cm/sec} \).
Show Solution to Exercise 2
\[ A = x^2 \implies \frac{dA}{dt} = 2x \frac{dx}{dt} = 2x(2) = 4x \; \text{cm}^2/\text{sec} \] -
Two cars start moving from the same point in two directions making a \( 90^\circ \) angle at constant speeds \( s_1 \) and \( s_2 \). Find a formula for the rate of change of the distance \( D \) between the two cars.
Show Solution to Exercise 3
\[ D^2 = x^2 + y^2 \] Differentiating implicitly with respect to \( t \): \[ 2D \frac{dD}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt} \implies \frac{dD}{dt} = \frac{x s_1 + y s_2}{D} \] Since \( x = s_1 t \), \( y = s_2 t \), and \( D = \sqrt{(s_1 t)^2 + (s_2 t)^2} = t\sqrt{s_1^2 + s_2^2} \): \[ \frac{dD}{dt} = \sqrt{s_1^2 + s_2^2} \]