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Solving Related Rates
Problems

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

AP Calculus BC

What Are Related Rates?

In many real-world scenarios, two or more quantities change simultaneously over time. When these quantities are linked by an equation, their rates of change are also linked. A related rates problem asks you to find how fast one quantity is changing given information about the other.

The Chain Rule Connection

Related rates problems are powered by implicit differentiation with respect to time. Every variable is treated as a function of \(t\), so the chain rule applies automatically.

\(x^2 + y^2 = r^2\)
\(\frac{d}{dt}\)
\(2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 2r\frac{dr}{dt}\)
 Original Equation After Implicit Differentiation
with respect to \(t\)

Why Do Related Rates Matter?

Physics
Tracking how distance, velocity, and acceleration relate as an object moves along a path.
Engineering
Determining how fast water rises in a tank as it fills at a known flow rate.
Biology
Modeling how a tumor's radius growth rate affects its volume growth rate.

The Core Framework

Every related rates problem follows a single principle: differentiate both sides of an equation with respect to time \(t\), then solve for the unknown rate.

Geometric / Physical Relationship
\(\frac{d}{dt}\)
 Equation Relating All Rates substitute Solve for Unknown Rate Example: Expanding Circle
\(A = \pi r^2\)
 differentiate
\(\frac{dA}{dt} = 2\pi r \frac{dr}{dt}\)
 substitute known values
\(\frac{dA}{dt} = ?\)

Related Rates vs. Optimization

Related Rates Optimization
Goal Find an unknown rate \(\frac{dy}{dt}\) Find a max or min value
Variable Everything changes with \(t\) One independent variable
Tool Implicit diff. w.r.t. \(t\) Set \(f'(x) = 0\)
Given Rates and snapshot values Constraints and a quantity to optimize
Output A rate (units/time) A quantity (units)

The 5-Step Method

1
Draw a diagram. Label all variables that change with time. Do not plug in numbers yet.
2
Write an equation relating the variables. Use geometry, trigonometry, or a given formula.
3
Differentiate implicitly with respect to \(t\). Apply the chain rule to every term.
4
Substitute known values (snapshot values and given rates) into the differentiated equation.
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 \(\frac{dr}{dt} = 3 \text{ ft/s}\). How fast is the area increasing when \(r = 10\) ft?

\(r = 10\)
\(\frac{dr}{dt} = 3\)
 Solution
\(A = \pi r^2\)
\(\frac{dA}{dt} = 2\pi r \frac{dr}{dt}\)
\(= 2\pi(10)(3)\)
\(= 60\pi \text{ ft}^2/\text{s}\)

Example 2: Sliding Ladder

A 10 ft ladder leans against a wall. The base slides away at \(\frac{dx}{dt} = 2 \text{ ft/s}\). How fast is the top sliding down when the base is 6 ft from the wall?

\(y\)
\(x = 6\)
 10 ft
\(\frac{dx}{dt}=2\)
\(\frac{dy}{dt}=?\)
 Solution
\(x^2 + y^2 = 100\)
\(2x\frac{dx}{dt}+2y\frac{dy}{dt}=0\)
\(y = \sqrt{100-36} = 8\)
\(2(6)(2)+2(8)\frac{dy}{dt}=0\)
\(24+16\frac{dy}{dt}=0\)
\(\frac{dy}{dt}=-\frac{3}{2} \text{ ft/s}\)

Ladder Problem: Key Takeaways

1
Pythagorean Theorem connects the sides: \(x^2 + y^2 = L^2\), where \(L\) is constant.
2
Constant ladder length means \(\frac{dL}{dt} = 0\), so the right side vanishes after differentiation.
3
Negative sign on \(\frac{dy}{dt}\) confirms the top is sliding down (decreasing \(y\)).

Example 3: Filling a Cone

Water flows into a conical tank at \(\frac{dV}{dt} = 8 \text{ ft}^3/\text{min}\). The tank has height 12 ft and top radius 4 ft. How fast is the water level rising when \(h = 6\) ft?

12
\(h\)
\(r\)
 4 ft Solution
Similar triangles: \(\frac{r}{h}=\frac{4}{12}\)
\(r = \frac{h}{3}\)
\(V = \frac{1}{3}\pi r^2 h = \frac{\pi h^3}{27}\)
\(\frac{dV}{dt}=\frac{\pi h^2}{9}\frac{dh}{dt}\)
\(8=\frac{\pi(36)}{9}\frac{dh}{dt}\)
\(8=4\pi\frac{dh}{dt}\)
\(\frac{dh}{dt}=\frac{2}{\pi} \text{ ft/min}\)

Cone Problem: Critical Step

✕ Wrong approach
Differentiating \(V = \frac{1}{3}\pi r^2 h\) with product rule keeping both \(r\) and \(h\), then having two unknown rates.
✓ Right approach
Substituting \(r = \frac{h}{3}\) first to get \(V = \frac{\pi h^3}{27}\), then differentiating with only one variable.

Example 4: Trigonometric Rates

A searchlight rotates at \(\frac{d\theta}{dt} = 2 \text{ rad/s}\). The light is 40 m from a straight wall. How fast does the beam move along the wall when \(\theta = \frac{\pi}{6}\)?

Wall Light 40 m
\(\theta\)
\(x\)
\(\frac{dx}{dt}=?\)
\(\tan\theta = \frac{x}{40}\)
\(\sec^2\!\theta\;\frac{d\theta}{dt}=\frac{1}{40}\frac{dx}{dt}\)
\(\frac{4}{3}\cdot 2 = \frac{1}{40}\frac{dx}{dt}\)
\(\frac{dx}{dt}=\frac{320}{3} \text{ m/s}\)

Common Mistakes

✕ Substituting Too Early
Plugging in snapshot values before differentiating. Always differentiate first, then substitute.
✕ Forgetting the Chain Rule
Writing \(\frac{d}{dt}[x^2] = 2x\) instead of \(2x \cdot \frac{dx}{dt}\). Every variable depends on \(t\).
✕ Wrong Sign
Ignoring whether a quantity is increasing (positive rate) or decreasing (negative rate). The sign matters.

Summary of Key Formulas

Scenario Equation Differentiated Form
Circle area \(A = \pi r^2\) \(\frac{dA}{dt} = 2\pi r \frac{dr}{dt}\)
Sphere volume \(V = \frac{4}{3}\pi r^3\) \(\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}\)
Pythagorean \(x^2 + y^2 = L^2\) \(2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0\)
Cone volume \(V = \frac{1}{3}\pi r^2 h\) Eliminate \(r\) first, then differentiate
Right triangle trig \(\tan\theta = \frac{y}{x}\) \(\sec^2\!\theta\;\frac{d\theta}{dt} = \frac{x\frac{dy}{dt} - y\frac{dx}{dt}}{x^2}\)

Question 1 of 5

The radius of a sphere is increasing at \(\frac{dr}{dt} = 4 \text{ cm/s}\). How fast is the volume increasing when \(r = 3\) cm?

Question 2 of 5

A 13 ft ladder slides along a wall. If the base moves away at \(\frac{dx}{dt} = 3 \text{ ft/s}\), how fast does the top slide down when \(x = 5\) ft?

Question 3 of 5

Water flows into a conical tank at \(\frac{dV}{dt} = 10 \text{ m}^3/\text{min}\). The cone has height 20 m and top radius 5 m. What is \(\frac{dh}{dt}\) when \(h = 8\) m?

Question 4 of 5

When solving a related rates problem, at what point should you substitute known numerical values?

Question 5 of 5

A light is 30 m from a wall. An object between the light and wall moves toward the light at \(2 \text{ m/s}\), casting a shadow on the wall. If \(x\) is the distance from the object to the light and the shadow length on the wall is \(s\), what relationship connects \(s\) and \(x\)?