4.1 Related Rates

Related rates problems ask you to find how fast one quantity is changing by using its relationship to another quantity whose rate of change you already know. For example, if you know how fast water is filling a cone-shaped tank, you can figure out how fast the water level is rising at any moment.

These problems are one of the most common applications of implicit differentiation and the chain rule. The core idea: when two or more variables are linked by an equation, their rates of change are linked too.

Related Rates

Derivatives in Real-World Scenarios

In related rates problems, every changing quantity is a function of time, even if the original equation doesn't mention time explicitly. A sphere's radius, a ladder's distance from a wall, a shadow's length: if it's changing, it depends on $t$.

• Identify the changing quantities. These become your variables: $V(t)$ for volume, $h(t)$ for height, $r(t)$ for radius, etc.
• Rates of change are derivatives with respect to time. The rate at which volume changes is $\frac{dV}{dt}$, the rate at which height changes is $\frac{dh}{dt}$, and so on.
• You'll typically be given some of these rates and asked to find one you don't know.

Setting Up and Solving Related Rates Problems

The biggest challenge isn't the calculus itself; it's translating the word problem into an equation you can differentiate. Here's a reliable process:

1. Draw a diagram (if the problem is geometric) and label every changing quantity with a variable.
2. Write down what you know and what you need. List the given rates (e.g., $\frac{dV}{dt} = 100 \text{ cm}^3/\text{min}$) and identify the unknown rate (e.g., $\frac{dr}{dt}$).
3. Find an equation that relates the variables. Use geometry, physics, or whatever the problem gives you. For a sphere: $V = \frac{4}{3}\pi r^3$.
4. Differentiate both sides with respect to $t$. This is where the chain rule comes in. Since $r$ depends on $t$, differentiating $V = \frac{4}{3}\pi r^3$ gives:

$\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$

1. Substitute known values and solve. Plug in the given rate and the value of $r$ at the specific instant the problem asks about. If $\frac{dV}{dt} = 100$ and $r = 5$:

$100 = 4\pi (5)^2 \frac{dr}{dt} \implies \frac{dr}{dt} = \frac{100}{100\pi} = \frac{1}{\pi} \text{ cm/min}$

A common mistake: don't plug in the specific values of variables before you differentiate. The variables need to stay as variables during differentiation because they're changing. You substitute only after taking the derivative.

The Chain Rule in Related Rates

Every related rates problem uses the chain rule, whether you notice it or not. When you differentiate $r^3$ with respect to $t$, you're applying the chain rule because $r$ is itself a function of $t$:

$\frac{d}{dt}[r^3] = 3r^2 \cdot \frac{dr}{dt}$

This pattern, $\frac{d}{dt}[f(g(t))] = f'(g(t)) \cdot g'(t)$, shows up every time you differentiate a variable that depends on time.

Pythagorean theorem example: A ladder 10 ft long leans against a wall. The base slides away at 2 ft/s. How fast is the top sliding down when the base is 6 ft from the wall?

The relationship is $x^2 + y^2 = 100$, where $x$ is the distance from the wall and $y$ is the height on the wall. Differentiating both sides with respect to $t$:

$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$

When $x = 6$, you get $y = \sqrt{100 - 36} = 8$. Substituting $x = 6$, $y = 8$, and $\frac{dx}{dt} = 2$:

$2(6)(2) + 2(8)\frac{dy}{dt} = 0 \implies \frac{dy}{dt} = -\frac{12}{8} = -\frac{3}{2} \text{ ft/s}$

The negative sign tells you $y$ is decreasing, which makes sense since the top of the ladder is sliding down.

Advanced Applications and Concepts

• Connection to differential equations. Related rates problems are actually simple differential equations. You're finding $\frac{dy}{dt}$ given a relationship between variables and other known rates. More complex versions of this idea appear throughout Calculus II and beyond.
• Parametric equations. When both $x$ and $y$ depend on $t$, you're already working with parametric curves. The rate $\frac{dy}{dx}$ can be found as $\frac{dy/dt}{dx/dt}$, which connects related rates to parametric differentiation.
• Optimization. Sometimes a related rates result feeds into an optimization question: for instance, finding the moment when a rate of change is at its maximum or minimum.
• Watch for domain restrictions. Make sure the values you substitute are physically reasonable. A radius can't be negative, a height can't exceed the container, and the functions involved need to be continuous at the point you're evaluating.