Related Rates

Some common related rates problems that you can expect to see on an exam are:

1. Sphere volume
2. Cone volume
3. Shadow length
4. Wall ladder
5. Triangle area

Sphere Volume

A ballon is being inflated such that its volume increases at a rate of $100{\text{ cm}}^3/\text{min}$. How fast is the radius of the balloon increasing when the radius is $5\text{ cm}$?

Information that we know:

• The formula for volume: $V=\frac{4}{3}\pi r^3$
• The rate of change of the volume: $\frac{dV}{dt}=100$
• The radius: $r=5$

Information that we want:

• The rate of change of the radius: $\frac{dr}{dt}$

Steps to solve:

• Differentiate with respect to time $t$
  • $\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$
• Isolate $\frac{dr}{dt}$
  • $\frac{1}{4\pi r^2}\cdot \frac{dV}{dt}=\frac{dr}{dt}$
• Substitute our known values:
  • $\frac{1}{4\pi (5{)}^2}\cdot 100=\frac{dr}{dt}{|}_{r=5}$
• Simplify the expression:
  • $\frac{\cancel{100}}{\cancel{100}\pi }=\frac{dr}{dt}$
  • $\frac{1}{\pi }=\frac{dr}{dt}$

Cone Volume

Water is being poured into a conical tank at a rate of $50{\text{ cm}}^3/\text{min}$. If the tank has a height of $10\text{ cm}$ and a base radius of $5\text{ cm}$, how fast is the water level rising when the water is $4\text{ cm}$ deep?

Information that we know:

• The rate of change of the volume: $\frac{dV}{dt}=50$
• The formula for volume: $V=\frac{1}{3}\pi r^{2}h$
• The height of the tank: $10$
• The base radius of the tank: $5$

Information that we want:

• The rate of change of the height: $\frac{dh}{dt}$

Steps:

• Express radius $r$ in terms of height $h$
  • $\frac{r}{h}=\frac{5}{10}⟹r=\frac{1}{2}h$
• Rewrite the volume with our new $r$
  • $V=\frac{1}{3}\pi (\frac{1}{2}h{)}^{2}h$
  • $V=\frac{\pi }{12}{h}^3$
• Differentiate with respect to $h$
  • $\frac{dV}{dt}=\frac{3\pi }{12}{h}^2\frac{dh}{dt}$
• Isolate $\frac{dh}{dt}$
  • $\frac{12}{3\pi h^2}\cdot \frac{dV}{dt}=\frac{dh}{dt}$
• Substitute known values
  • $\frac{dh}{dt}{|}_{h=4}=\frac{12}{3\pi (4{)}^2}\cdot 50$
  • $\frac{dh}{dt}=\frac{600}{48\pi }$
  • $\frac{dh}{dt}=\frac{25}{2\pi }$

Shadow Length

A 2-meter tall person is walking away from a 5-meter tall light source of $1.5\text{m}/\text{s}$. How fast is the length of their shadow increasing when they are 6 meters away from the base of the light?

Know:

• The light is 5 meters tall
• The person is 2 meters tall
• The person is moving away from the light at $1.5\text{ m}/\text{s}$

Want:

• The rate of change of the length of the person’s shadow

Steps:

• Set up a right triangle
  • The bottom side of the triangle can be expressed as the sum of $A$ and $B$, where
    • $A$ is the distance from the base of the light to the person
    • $B$ is the distance from the person to the tip of their shadow
  • The tall side of the triangle is the 5-meter tall light
  • We know $\frac{dA}{dt}$ and want $\frac{dB}{dt}$
• Recognize that these triangles have similar proportions
  • $\frac{5}{A+B}=\frac{2}{B}$
• Rearrange to isolate $B$
  • $5B=2A+2B$
  • $3B=2A$
  • $B=\frac{2}{3}A$
• Differentiate with respect to $t$
  • $\frac{dB}{dt}=\frac{2}{3}\cdot \frac{dA}{dt}$
• Substitute known values
  • $\frac{dB}{dt}=\frac{2}{3}\cdot 1.5$
  • $\frac{dB}{dt}=1$

Wall Ladder

A 10-meter ladder is sliding down a wall. If the bottom of the ladder is moving away from the wall at a rate of $1\text{ m/s}$, how fast is the top of the ladder moving down the wall when the bottom is 6 meters away from the wall?

Know:

• Length of the ladder (hypotenuse) is 10 meters
• Horizontal rate of change $\frac{dx}{dt}$ is 1 m/s
• Horizontal length is 6 meters

Want:

• Vertical rate of change $\frac{dy}{dt}$

Steps:

• Set up a triangle using the Pythagorean theorem
  • $x^2+y^2=z^2⟹x^2+y^2=10^2=100$
• Solve for $y$ when $x=6$
  • $6^2+y^2=100⟹y=8$
• Differentiate with respect to $t$
  • $2x\frac{dx}{dt}+2y\frac{dy}{dt}=\frac{d}{dt}100=0$
• Isolate $\frac{dy}{dt}$
  • $2y\frac{dy}{dt}=-2x\frac{dx}{dt}$
  • $\frac{dy}{dt}=-\frac{x}{y}\cdot \frac{dx}{dt}$
• Substitute known values
  • $\frac{dy}{dt}=-\frac{6}{8}\cdot 1$
  • $\frac{dy}{dt}=-\frac{3}{4}$

Triangle Area

The base of a right triangle is increasing at a rate of $2\text{ cm/min}$ and the height is increasing at a rate of $3\text{ cm/min}$. How fast is the area of the triangle increasing when the base is $10\text{ cm}$ and the height is $6\text{ cm}$?

Know:

• The base length $b$ = 10 cm
  • The rate of change of the base $\frac{db}{dt}$ = 2 cm
• The height length $h$ = 6 cm
  • The rate of change of the height $\frac{dh}{dt}$ = 3 cm
• The area formula $A=\frac{1}{2}bh$

Want:

• The rate of change of the area $\frac{dA}{dt}$

Steps:

• Differentiate with respect to $t$
  • $\frac{dA}{dt}=\frac{1}{2}(h\frac{db}{dt}+b\frac{dh}{dt})$
• Substitute know values
  • $\frac{dA}{dt}=\frac{1}{2}(6(2)+10(3))$
  • $\frac{dA}{dt}=\frac{1}{2}(12+30)$
  • $\frac{dA}{dt}=21$