Related Rates: Fresh Take

Lumen Learning

Related Rates: Fresh Take

Show how quantities change using derivatives and explore how these changes are connected
Apply the chain rule to calculate how one changing quantity affects another

Related-Rates Problem-Solving

The Main Idea

Related rates problems involve finding the rate of change of one quantity when it’s related to another quantity whose rate of change is known.
Use the chain rule to relate the rates of change of different quantities.
Common Applications:
- Geometric problems (e.g., expanding circles, changing triangles)
- Physical scenarios (e.g., water flowing, objects moving)
Important Considerations:
- Draw a diagram if possible
- Don’t substitute values too early (before differentiation)
- Pay attention to units and sign conventions

Problem-Solving Strategy

Assign variables to all relevant quantities
Establish the given information and what needs to be found
Develop an equation relating the variables
Differentiate the equation with respect to time
Substitute known values and solve for the unknown rate

A conical water tank is being filled at a rate of $10 \text{ft}^3/\text{min}$ . The tank has a height of $12 \text{ft}$ and a base radius of $6 \text{ft}$ . At what rate is the water level rising when the water is $4 \text{ft}$ deep?

Assign variables:

$r$ = radius of water surface
$h$ = height of water
$V$ = volume of water

Given information:

$\frac{dV}{dt} = 10$ ft³/min
Tank height = $12$ ft, base radius = $6$ ft
Need to find $\frac{dh}{dt}$ when $h = 4$ ft

Develop equation:

Volume of a cone: $V = \frac{1}{3}\pi r^2h$
Similar triangles: $\frac{r}{h} = \frac{6}{12} = \frac{1}{2}$ , so $r = \frac{h}{2}$
$V = \frac{1}{3}\pi(\frac{h}{2})^2h = \frac{\pi}{12}h^3$

Differentiate with respect to time:

$\frac{dV}{dt} = \frac{\pi}{4}h^2\frac{dh}{dt}$

Substitute and solve:

$10 = \frac{\pi}{4}(4^2)\frac{dh}{dt}$
$\frac{dh}{dt} = \frac{10}{4\pi} \approx 0.796 , \text{ft}/\text{min}$

Therefore, when the water is $4 \text{ft}$ deep, it’s rising at a rate of approximately $0.796 \text{ft}/\text{min}$ .

A street lamp is mounted at the top of a $15$ foot tall pole. A man $6$ feet tall walks away from the pole with a speed of $5 \text{ft}/\text{sec}$ along a straight path. At what rate is the tip of his shadow moving when he is $40 \text{feet}$ from the pole?

Assign variables:

$x$ = distance of man from pole
$s$ = length of shadow
$h$ = height of pole ( $15 , \text{ft}$ )
$H$ = height of man ( $6 , \text{ft}$ )

Given information:

$\frac{dx}{dt} = 5 , \text{ft}/\text{sec}$
$x = 40 , \text{ft}$
Need to find $\frac{ds}{dt}$

Develop equation:

Similar triangles: $\frac{s}{x+s} = \frac{h}{H}$
$\frac{15}{6} = \frac{x+s}{x}$
$5x = 2(x+s)$
$s = \frac{3x}{2} - x = \frac{x}{2}$

Differentiate with respect to time:

$\frac{ds}{dt} = \frac{1}{2}\frac{dx}{dt}$

Substitute and solve:

$\frac{ds}{dt} = \frac{1}{2}(5) = 2.5 , \text{ft}/\text{sec}$

Therefore, when the man is $40 \text{feet}$ from the pole, the tip of his shadow is moving at $2.5 \text{ft}/\text{sec}$ .

A circular oil slick is expanding at a constant rate of $2$ m²/sec. How fast is the radius of the slick increasing when the radius is $10$ meters?

Assign variables:

$A$ = area of oil slick
$r$ = radius of oil slick

Given information:

$\frac{dA}{dt} = 2 , \text{m}^2/\text{sec}$
$r = 10 , \text{m}$
Need to find $\frac{dr}{dt}$

Develop equation:

Area of a circle: $A = \pi r^2$

Differentiate with respect to time:

$\frac{dA}{dt} = 2\pi r\frac{dr}{dt}$

Substitute and solve:

$2 = 2\pi(10)\frac{dr}{dt}$
$\frac{dr}{dt} = \frac{2}{20\pi} = \frac{1}{10\pi} \approx 0.0318 , \text{m}/\text{sec}$

Therefore, when the radius of the oil slick is $10 \text{meters}$ , it’s expanding at a rate of approximately $0.0318 \text{m}/\text{sec}$ or $3.18 \text{cm}/\text{sec}$ .