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Introduction to
Related Rates

Learn how to connect rates of change of different quantities using the chain rule and implicit differentiation with respect to time.

AP Calculus BC

What Are Related Rates?

In many real-world situations, two or more quantities change simultaneously. When those quantities are connected by an equation, their rates of change are also connected. Related rates problems ask you to find the rate at which one quantity changes, given information about another.

A Classic Scenario

\( x \) \( y \) L (ladder) \( \frac{dx}{dt} \) \( \frac{dy}{dt} \)

Why Do Related Rates Matter?

Physics
Velocity, acceleration, and the motion of connected parts in a system
Engineering
Filling tanks, expanding materials, changing pressures
Biology
Growth rates of organisms that depend on nutrient flow rates

The Chain Rule Connection

Related rates rely on implicit differentiation with respect to time. The chain rule is the engine that makes it work:

More generally, for any equation involving \( x(t) \) and \( y(t) \), differentiate every term with respect to \( t \), applying the chain rule to each variable.

The Core Equation Setup

Every related rates problem follows the same pattern:

Equation F(x, y) = C Differentiate w.r.t. t (chain rule) Solve for unknown rate

Implicit vs. Explicit Differentiation

Explicit Differentiation Implicit w.r.t. \( t \)
Form \( y = f(x) \) \( F(x, y) = C \), all depend on \( t \)
Derivative \( \frac{dy}{dx} \) \( \frac{dy}{dt}, \frac{dx}{dt} \)
Chain rule Not always needed Always needed on every term
Result Slope of curve Rate of change over time

How to Solve Related Rates

1
Draw a diagram and label all variables that change with time.
2
Write an equation relating the variables (geometric, algebraic, or trigonometric).
3
Differentiate both sides with respect to \( t \), using the chain rule on every variable.
4
Substitute all known values and rates at the given instant.
5
Solve for the unknown rate. Include units and interpret the sign.

Example 1: Expanding Circle

A stone is dropped into a pond, creating a circular ripple. The radius increases at \( \frac{dr}{dt} = 3 \) ft/s. How fast is the area increasing when \( r = 10 \) ft?

\( r \) \( \frac{dr}{dt} = 3 \) \( A = \pi r^2 \)

Example 1: Solution

1
Equation: \( A = \pi r^2 \)
2
Differentiate w.r.t. \( t \): \( \frac{dA}{dt} = 2\pi r \cdot \frac{dr}{dt} \)
3
Substitute: \( r = 10 \), \( \frac{dr}{dt} = 3 \)
4
Solve: \( \frac{dA}{dt} = 2\pi(10)(3) = 60\pi \approx 188.5 \) ft²/s

Example 2: Sliding Ladder

A 13-ft ladder leans against a wall. The base slides away at \( \frac{dx}{dt} = 2 \) ft/s. How fast is the top sliding down when the base is \( x = 5 \) ft from the wall?

\( x = 5 \) ft \( y = ? \) L = 13 ft \( \frac{dx}{dt} = 2 \)

Example 2: Solution

1
Equation: \( x^2 + y^2 = 13^2 = 169 \)
2
Differentiate: \( 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0 \)
3
Find \( y \): \( y = \sqrt{169 - 25} = \sqrt{144} = 12 \) ft
4
Substitute: \( 2(5)(2) + 2(12)\frac{dy}{dt} = 0 \)
5
Solve: \( \frac{dy}{dt} = -\frac{20}{24} = -\frac{5}{6} \) ft/s

Example 3: Filling a Cone

Water pours into a conical tank (height 10 ft, radius 5 ft at the top) at \( \frac{dV}{dt} = 12 \) ft³/min. How fast is the water level rising when \( h = 4 \) ft?

\( h = 4 \) \( r \) H = 10 R = 5 \( \frac{dV}{dt} = 12 \)

Example 3: Solution

1
Volume of cone: \( V = \frac{1}{3}\pi r^2 h \). Use similar triangles: \( \frac{r}{h} = \frac{5}{10} \), so \( r = \frac{h}{2} \).
2
Eliminate \( r \): \( V = \frac{1}{3}\pi \left(\frac{h}{2}\right)^2 h = \frac{\pi h^3}{12} \)
3
Differentiate: \( \frac{dV}{dt} = \frac{\pi}{12} \cdot 3h^2 \cdot \frac{dh}{dt} = \frac{\pi h^2}{4}\frac{dh}{dt} \)
4
Substitute: \( 12 = \frac{\pi(16)}{4}\frac{dh}{dt} = 4\pi\frac{dh}{dt} \)
5
Solve: \( \frac{dh}{dt} = \frac{12}{4\pi} = \frac{3}{\pi} \approx 0.955 \) ft/min

Common Mistakes

✕ Substituting Too Early
Plugging in numerical values before differentiating. You must differentiate first, then substitute the instant's values.
✕ Forgetting the Chain Rule
Writing \( \frac{d}{dt}(r^2) = 2r \) instead of \( 2r\frac{dr}{dt} \). Every variable is a function of \( t \).
✕ Ignoring the Sign
A negative rate means the quantity is decreasing. Always interpret and report the sign correctly.

Summary of Key Scenarios

Scenario Equation After Differentiating
Expanding circle \( A = \pi r^2 \) \( \frac{dA}{dt} = 2\pi r \frac{dr}{dt} \)
Ladder / Pythagorean \( x^2 + y^2 = L^2 \) \( 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0 \)
Filling a cone \( V = \frac{\pi h^3}{12} \) \( \frac{dV}{dt} = \frac{\pi h^2}{4}\frac{dh}{dt} \)
Expanding sphere \( V = \frac{4}{3}\pi r^3 \) \( \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} \)

Question 1 of 5

A circle's radius grows at \( \frac{dr}{dt} = 4 \) cm/s. What is \( \frac{dA}{dt} \) when \( r = 6 \) cm?

Question 2 of 5

Given \( x^2 + y^2 = 25 \), if \( \frac{dx}{dt} = 3 \) and \( x = 3 \), what is \( \frac{dy}{dt} \)?

Question 3 of 5

When solving a related rates problem, which step must come before substituting known values?

Question 4 of 5

A sphere's volume increases at \( \frac{dV}{dt} = 100 \) cm³/s. Find \( \frac{dr}{dt} \) when \( r = 5 \) cm. (Use \( V = \frac{4}{3}\pi r^3 \).)

Question 5 of 5

For a conical tank where \( r = \frac{h}{2} \), the volume is \( V = \frac{\pi h^3}{12} \). If \( \frac{dV}{dt} = 8\pi \) ft³/min, what is \( \frac{dh}{dt} \) when \( h = 4 \) ft?