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Solving
Related Rates Problems

Learn to connect changing quantities using the chain rule and implicit differentiation to solve real-world rate problems.

AP Calculus AB

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. Related rates problems ask you to find how fast one quantity is changing given information about how fast another is changing.

The Chain Rule Connection

Related rates problems rely on implicit differentiation with respect to time \( t \). Every variable that changes over time must be differentiated using the chain rule.

Equation linking x, y, z, ... \(\frac{d}{dt}\) Equation linking \(\frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt}\) Example: \( x^2 + y^2 = r^2 \;\to\; 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 2r\frac{dr}{dt} \)

Why Related Rates Matter

Physics
Velocity, acceleration, and changing distances between moving objects
Engineering
Rates of fluid flow, expanding/contracting materials, circuits
Biology
Population growth, rates of reaction, spread of diseases

The Fundamental Principle

\(x\) \(y\) \(z\) \(\frac{dx}{dt}\) \(\frac{dy}{dt}\) \( x^2 + y^2 = z^2 \) \( 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 2z\frac{dz}{dt} \)

Implicit Differentiation Refresher

In related rates, you differentiate every term with respect to \( t \), treating each variable as a function of \( t \). Here are the key differentiation patterns:

Power Rule
\( \frac{d}{dt}[x^n] = nx^{n-1}\frac{dx}{dt} \)
Product Rule
\( \frac{d}{dt}[xy] = x\frac{dy}{dt} + y\frac{dx}{dt} \)
Trig Functions
\( \frac{d}{dt}[\sin\theta] = \cos\theta\cdot\frac{d\theta}{dt} \)

Related Rates vs. Optimization

Related Rates Optimization
Goal Find how fast something changes Find max or min value
Key operation Differentiate with respect to \( t \) Set \( \frac{dy}{dx} = 0 \)
Variables Multiple, all functions of \( t \) Usually one independent variable
Constraint Geometric or physical equation Boundary or domain restriction
Answer type A rate (units/time) A value (max/min)

The 5-Step Strategy

1
Draw and label. Sketch the situation. Assign variables to all changing quantities.
2
Identify knowns and unknowns. List all given rates \( \frac{d(\cdot)}{dt} \) and the target rate you need to find.
3
Write an equation that relates the variables. Use geometry, trigonometry, or physics formulas.
4
Differentiate both sides with respect to \( t \) using the chain rule.
5
Substitute and solve. Plug in all known values at the specific instant, then solve for the unknown rate.

A light-and-camera studio with exact sightlines

A three-part studio scene showing a wall spotlight, overhead camera, and point-light wall shadow.Similar triangles connect a 6 ft person, a wall 20 ft from the light, and the shadow height. Differentiate with respect to time to find how the shadow changes as the person moves.

Point light: a shrinking wall shadow

Similar triangles connect a 6 ft person, a wall 20 ft from the light, and the shadow height. Differentiate with respect to time to find how the shadow changes as the person moves.Similar triangles connect a 6 ft person, a wall 20 ft from the light, and the shadow height. Differentiate with respect to time to find how the shadow changes as the person moves.

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 \( r = 5 \) ft?

\(r = 5\) \(\frac{dr}{dt} = 3\) Solution: \(A = \pi r^2\) \(\frac{dA}{dt} = 2\pi r\frac{dr}{dt}\) \(= 30\pi\) ft²/s

Example 2: The Ladder Problem

A 10 ft ladder leans against a wall. The base slides away at \( \frac{dx}{dt} = 1 \) ft/s. How fast is the top sliding down when the base is 6 ft from the wall?

\(x = 6\) \(y = 8\) \(L=10\) \(\frac{dx}{dt} = 1\) \(\frac{dy}{dt} = ?\) Pythagorean: \(x^2 + y^2 = 100\) \(2x\frac{dx}{dt}+2y\frac{dy}{dt}=0\) \(12(1) + 16\frac{dy}{dt} = 0\) \(\frac{dy}{dt} = -\frac{3}{4}\) ft/s

Example 3: Filling a Cone

Water flows into a conical tank (height 10 ft, radius 5 ft) at \( \frac{dV}{dt} = 12 \) ft³/min. How fast is the water level rising when \( h = 6 \) ft?

\(H = 10\) \(h = 6\) \(R = 5\) \(r = 3\) Similar triangles: \(\frac{r}{h} = \frac{5}{10} \to r = \frac{h}{2}\) Volume formula: \(V = \frac{\pi h^3}{12}\) \(\frac{dV}{dt} = \frac{\pi h^2}{4} \cdot \frac{dh}{dt}\) \(12 = 9\pi\frac{dh}{dt}\) \(\frac{dh}{dt} = \frac{4}{3\pi}\) ft/min

Why Eliminate Variables First?

In the cone example, the volume formula \( V = \frac{1}{3}\pi r^2 h \) has three variables: \( V \), \( r \), and \( h \). Differentiating directly would give:

General rule: Your equation should contain only one unknown rate plus the rate you are solving for. Use constraints to eliminate extra variables before differentiating.

Geometry connects changing radius and depth

A clear inverted conical tank fills from a pipe, with a widening horizontal water surface.As water rises in a conical tank, its surface radius and depth increase together. Similar triangles express r in terms of h before the volume equation is differentiated.

Trigonometric Related Rates

A searchlight rotates at \( \frac{d\theta}{dt} = 2 \) rad/s. The light is 50 m from a straight wall. How fast does the beam move along the wall when \( \theta = \frac{\pi}{6} \)?

Wall Light \(d = 50\) m \(x\) \(\theta\) Setup: \(\tan\theta = \frac{x}{50}\) \(\sec^2\!\theta \cdot \frac{d\theta}{dt} = \frac{1}{50}\frac{dx}{dt}\) \(\frac{dx}{dt} = 50\sec^2\!\theta \cdot \frac{d\theta}{dt}\) \(= \frac{400}{3}\) m/s

Essential Formulas to Know

Circle
\( A = \pi r^2 \)
\( C = 2\pi r \)
Sphere
\( V = \frac{4}{3}\pi r^3 \)
\( S = 4\pi r^2 \)
Cone
\( V = \frac{1}{3}\pi r^2 h \)
Right Triangle
\( a^2 + b^2 = c^2 \)
\( \tan\theta = \frac{\text{opp}}{\text{adj}} \)

Common Mistakes to Avoid

✕ Substituting Too Early
Plugging in numerical values before differentiating eliminates the variable, making it impossible to find its rate of change.
✕ Forgetting the Chain Rule
Writing \( \frac{d}{dt}[x^2] = 2x \) instead of \( 2x\frac{dx}{dt} \). Every variable needs its own \( \frac{d(\cdot)}{dt} \).
✕ Wrong Sign on the Rate
If a quantity is decreasing, its rate must be negative. A shrinking radius means \( \frac{dr}{dt} < 0 \).

Problem Type Summary

Scenario Key Equation After Differentiating
Expanding circle \( A = \pi r^2 \) \( \frac{dA}{dt} = 2\pi r\frac{dr}{dt} \)
Ladder/Pythag. \( x^2 + y^2 = L^2 \) \( 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0 \)
Conical tank \( V = \frac{\pi h^3}{12} \) \( \frac{dV}{dt} = \frac{\pi h^2}{4}\frac{dh}{dt} \)
Rotating light \( x = d\tan\theta \) \( \frac{dx}{dt} = d\sec^2\!\theta\cdot\frac{d\theta}{dt} \)
Expanding sphere \( V = \frac{4}{3}\pi r^3 \) \( \frac{dV}{dt} = 4\pi r^2\frac{dr}{dt} \)

Worked Example — Conical Sand Pile

Sand is poured onto a pile at a rate of 3 ft³/min. The pile maintains the shape of a right circular cone whose height is always equal to twice its base radius. How fast is the height of the pile increasing when the height is 4 ft? Conical Sand Pile Step 1 — Draw and Label Variables We sketch a cone with base radius r and height h. Both r and h are functions of time t. We are given dV/dt = 3 ft³/min and that h = 2r (i.e., r = h/2). We want dh/dt when h = 4. Step 2 — Write the Linking Equation The volume of a cone is V = (1/3)πr²h. Since the cone's proportions are fixed with r = h/2, we eliminate r: V = (1/3)π(h/2)²h = (1/3)π(h²/4)h = πh³/12. V = πh³/12 Step 3 — Differentiate Implicitly with Respect to t Applying the chain rule: dV/dt = (π/12) × 3h² × (dh/dt) = (πh²/4)(dh/dt). dV/dt = (πh²/4)(dh/dt) Step 4 — Substitute Known Values Substitute dV/dt = 3 and h = 4: 3 = (π(4)²/4)(dh/dt) = (16π/4)(dh/dt) = 4π(dh/dt). Step 5 — Solve and Interpret Solving for dh/dt: dh/dt = 3/(4π) ft/min ≈ 0.239 ft/min. Since dh/dt is positive, the height is increasing—consistent with sand being added to the pile. dh/dt = 3/(4π) ft/min ≈ 0.239 ft/min

A cone context photograph beside two idealized cones at a common scale, with heights 2 and 4 and radii 1 and 2. The model uses h=2r and gives dh/dt=3/(4π).
For an idealized cone with h=2 r, adding 3 ft³/min at h=4 ft raises the height at 3/(4π)ft/min. The photo is qualitative; h=2r applies only to the separate idealized cone construction.

Question 1 of 5

A balloon is being inflated so that its radius increases at \( \frac{dr}{dt} = 2 \) cm/s. What is \( \frac{dV}{dt} \) when \( r = 3 \) cm?
(Use \( V = \frac{4}{3}\pi r^3 \))

Question 2 of 5

A 13-foot ladder leans against a wall. The foot slides away at 2 ft/s. How fast is the top sliding down when the foot is 5 ft from the wall?

Question 3 of 5

In a related rates problem, when should you substitute known numerical values for the variables?

Question 4 of 5

If \( xy = 12 \) and \( \frac{dx}{dt} = 3 \), what is \( \frac{dy}{dt} \) when \( x = 4 \)?

Question 5 of 5

Water is poured into a cylindrical tank of radius 4 ft at a rate of \( \frac{dV}{dt} = 10 \) ft³/min. How fast is the water level rising?
(Use \( V = \pi r^2 h \) with \( r \) constant.)