Understanding Related Rates

In calculus, related rates problems involve finding the rate at which a quantity changes by relating that quantity to other quantities whose rates of change are known. Because these quantities usually change over time, we differentiate their relating equations with respect to time, $t$.

The General Strategy

To solve any related rates problem, follow these standard steps:

1. Draw a picture and label all changing quantities with variables (like $x$, $y$, $V$) and all constant quantities with their given numerical values.
2. Identify what you know and what you need to find, expressing rates as derivatives with respect to time (e.g., $\frac{dx}{dt}$, $\frac{dV}{dt}$).
3. Write an equation that relates the variables in the problem.
4. Differentiate both sides of the equation with respect to time $t$ using the Chain Rule.
5. Substitute all known values and known rates into your derivative equation.
6. Solve for the unknown rate.

Note: Never substitute changing values into your equation before differentiating! Only constants can be plugged in at the beginning.

Example 1: The Sliding Ladder

Problem: A ladder $10 \text{ ft}$ long rests against a vertical wall. The bottom of the ladder slides away from the wall at a rate of $2 \text{ ft/s}$. How fast is the top of the ladder sliding down the wall when the bottom is $6 \text{ ft}$ from the wall?

Solution:

1. Let $x$ be the distance from the bottom of the ladder to the wall, and $y$ be the height of the top of the ladder on the wall. The length of the ladder is a constant $10$.
2. We know $\frac{dx}{dt} = 2 \text{ ft/s}$. We want to find $\frac{dy}{dt}$ when $x = 6$.
3. The Pythagorean theorem relates the variables: $x^2 + y^2 = 10^2 = 100$
4. Differentiate both sides with respect to $t$: $2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$
5. Before substituting, we need the value of $y$ when $x = 6$. Using the original equation: $6^2 + y^2 = 100 \implies 36 + y^2 = 100 \implies y = 8$.
6. Substitute the knowns into the derivative equation: $2(6)(2) + 2(8)\frac{dy}{dt} = 0$ $24 + 16\frac{dy}{dt} = 0$ $\frac{dy}{dt} = -\frac{24}{16} = -1.5 \text{ ft/s}$

The top of the ladder is sliding down the wall at a rate of $1.5 \text{ ft/s}$.

Example 2: The Conical Tank

Problem: Water is pouring into a conical tank (radius $r=3 \text{ ft}$, height $h=6 \text{ ft}$) at a rate of $2 \text{ ft}^3\text{/min}$. How fast is the water level rising when the water is $4 \text{ ft}$ deep?

Solution:

1. Let $V$ be the volume of the water, $r$ be the radius of the water's surface, and $h$ be the depth of the water.
2. We know $\frac{dV}{dt} = 2$. We want to find $\frac{dh}{dt}$ when $h = 4$.
3. The volume of a cone is: $V = \frac{1}{3}\pi r^2 h$ Because we only want to deal with $h$, we can eliminate $r$ using similar triangles formed by the cross-section of the cone: $\frac{r}{h} = \frac{3}{6} \implies r = \frac{1}{2}h$. Substitute this into the volume equation: $V = \frac{1}{3}\pi \left(\frac{1}{2}h\right)^2 h = \frac{\pi}{12}h^3$
4. Differentiate with respect to $t$: $\frac{dV}{dt} = \frac{\pi}{12} \cdot 3h^2 \frac{dh}{dt} = \frac{\pi}{4}h^2 \frac{dh}{dt}$
5. Substitute the knowns ($h = 4$, $\frac{dV}{dt} = 2$): $2 = \frac{\pi}{4}(4)^2 \frac{dh}{dt}$ $2 = 4\pi \frac{dh}{dt}$
6. Solve for $\frac{dh}{dt}$: $\frac{dh}{dt} = \frac{2}{4\pi} = \frac{1}{2\pi} \text{ ft/min}$

The water level is rising at a rate of $\frac{1}{2\pi} \text{ ft/min}$.