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Introduction to
Related Rates

Learn how to connect rates of change of two or more quantities that vary with time using the chain rule.

AP Calculus AB

What Are Related Rates?

In many real-world situations, two or more quantities change simultaneously over time. When these 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 the rate at which a related quantity changes. Both rates are derivatives with respect to time \( t \).

A Visual Intuition

Imagine a circular ripple spreading outward when a stone drops into a pond. As the radius \( r \) increases over time, the area \( A \) also increases. How fast is the area growing?

\( r(t) \)
 Ripple at three moments in time expanding

Why Related Rates Matter

Related rates connect calculus to the physical world. Whenever quantities are changing simultaneously, you can use derivatives to answer practical questions.

Physics
How fast does a shadow lengthen as a person walks away from a streetlight?
Engineering
How quickly does water level drop as a conical tank drains?
Medicine
How fast does a tumor's volume grow given the rate of change of its radius?

The Core Rule: Implicit Differentiation

The foundation of every related rates problem is implicit differentiation with respect to time \( t \). If an equation relates variables that all depend on \( t \), differentiate every term using the chain rule.

Equation: \( F(x, y) = C \)
\( \frac{d}{dt} \)
Rate equation with \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \)
 Chain rule links the static equation to the dynamic rates

The Chain Rule Connection

Every time you differentiate a variable with respect to \( t \), the chain rule produces a \( \frac{d(\text{variable})}{dt} \) factor. This is what creates the "related" part.

1
\( \frac{d}{dt}[x^2] = 2x \cdot \frac{dx}{dt} \): power rule times chain rule factor
2
\( \frac{d}{dt}[\sin(\theta)] = \cos(\theta) \cdot \frac{d\theta}{dt} \): trig derivative times chain rule factor
3
\( \frac{d}{dt}[xy] = x\frac{dy}{dt} + y\frac{dx}{dt} \): product rule, each piece gets a chain rule factor

Related Rates vs. Standard Derivatives

Students often confuse these two types of problems. Here is the critical difference:

Related Rates Standard Derivatives
Independent variable Time \( t \) Usually \( x \)
Differentiation type Implicit: \( \frac{d}{dt} \) Explicit: \( \frac{d}{dx} \)
What you find Rate at a specific instant General derivative function
Chain rule usage Every variable gets \( \frac{d(\cdot)}{dt} \) Only for compositions
When to substitute values After differentiating Before or after, depends

The Five-Step Method

1
Draw and label: Sketch the situation. Assign variables to all changing quantities.
2
Write the equation: Find a formula connecting the variables (geometry, trig, physics).
3
Differentiate with respect to \( t \): Apply implicit differentiation using the chain rule on every variable.
4
Substitute known values: Plug in the given rates and the specific values at the instant described.
5
Solve for the unknown rate: Isolate the desired \( \frac{d(\cdot)}{dt} \) and include units.

Example 1: Expanding Circle

A circle's radius increases at \( \frac{dr}{dt} = 3 \text{ cm/s} \). Find \( \frac{dA}{dt} \) when \( r = 5 \text{ cm} \).

\( r = 5 \)
\( \frac{dr}{dt} = 3 \) cm/s
 Step-by-step
1. \( A = \pi r^2 \)
2. \( \frac{dA}{dt} = 2\pi r \cdot \frac{dr}{dt} \)
3. \( = 2\pi(5)(3) \)
4. \( \frac{dA}{dt} = 30\pi \) cm²/s

Example 2: The Ladder Problem

A 10 ft ladder leans against a wall. The base slides away at \( \frac{dx}{dt} = 1 \text{ ft/s} \). Find \( \frac{dy}{dt} \) when the base is 6 ft from the wall.

y
\( x = 6 \) ft
 10 ft
\( \frac{dx}{dt} = 1 \)
\( \frac{dy}{dt} = ? \)

Ladder Problem: Solution

1
Equation: By the Pythagorean theorem: \( x^2 + y^2 = 10^2 = 100 \)
2
Differentiate: \( 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0 \)
3
Find \( y \): When \( x = 6 \): \( y = \sqrt{100 - 36} = 8 \text{ ft} \)
4
Substitute: \( 2(6)(1) + 2(8)\frac{dy}{dt} = 0 \), so \( 12 + 16\frac{dy}{dt} = 0 \)
5
Solve: \( \frac{dy}{dt} = -\frac{12}{16} = -\frac{3}{4} \text{ ft/s} \)

Example 3: Filling a Cone

Water fills a conical tank (height 10 ft, radius 5 ft) at \( \frac{dV}{dt} = 12 \text{ ft}^3\text{/min} \). Find \( \frac{dh}{dt} \) when \( h = 4 \text{ ft} \).

10 ft
\( h = 4 \)
\( R = 5 \)
\( r \)

Cone Solution: Eliminate a Variable

1
Volume formula: \( V = \frac{1}{3}\pi r^2 h \). Two unknowns: we need to eliminate \( r \).
2
Similar triangles: \( \frac{r}{h} = \frac{5}{10} = \frac{1}{2} \), so \( r = \frac{h}{2} \). Substitute: \( V = \frac{1}{3}\pi\left(\frac{h}{2}\right)^2 h = \frac{\pi h^3}{12} \)
3
Differentiate: \( \frac{dV}{dt} = \frac{3\pi h^2}{12}\frac{dh}{dt} = \frac{\pi h^2}{4}\frac{dh}{dt} \)
4
Substitute: \( 12 = \frac{\pi(4)^2}{4}\frac{dh}{dt} = 4\pi\frac{dh}{dt} \)
5
Solve: \( \frac{dh}{dt} = \frac{12}{4\pi} = \frac{3}{\pi} \approx 0.955 \text{ ft/min} \)

A filling cone: connect volume and height

An inverted conical tank fills through an inlet. Its axial diagram marks full height 12 ft and radius 4 ft, water depth 6 ft and radius 2 ft, and the height rate 2/π ft/min.For a tank 12 ft high with top radius 4 ft, r = h/3 and V = πh³/27. An inflow of 8 ft³/min raises the water at 2/π ft/min when h = 6 ft.

A filling cone: connect volume and height

An inverted conical tank fills through an inlet. Its axial diagram marks full height 12 ft and radius 4 ft, water depth 6 ft and radius 2 ft, and the height rate 2/π ft/min.For a tank 12 ft high with top radius 4 ft, r = h/3 and V = πh³/27. An inflow of 8 ft³/min raises the water at 2/π ft/min when h = 6 ft.Open full diagram

Common Formulas in Related Rates

Shape Equation After \( \frac{d}{dt} \)
Circle area \( A = \pi r^2 \) \( \frac{dA}{dt} = 2\pi r\frac{dr}{dt} \)
Sphere volume \( V = \frac{4}{3}\pi r^3 \) \( \frac{dV}{dt} = 4\pi r^2\frac{dr}{dt} \)
Right triangle \( x^2 + y^2 = L^2 \) \( 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0 \)
Cone volume \( V = \frac{1}{3}\pi r^2 h \) Product rule (or eliminate \( r \))
Trigonometric \( \tan\theta = \frac{y}{x} \) \( \sec^2\theta\frac{d\theta}{dt} = \frac{x\frac{dy}{dt} - y\frac{dx}{dt}}{x^2} \)

Common Mistakes to Avoid

✕ Substituting Too Early
Plugging in numbers before differentiating turns variables into constants. Always differentiate first, then substitute.
✕ Forgetting the Chain Rule
Writing \( \frac{d}{dt}[r^2] = 2r \) instead of \( 2r\frac{dr}{dt} \). Every variable needs its \( \frac{d(\cdot)}{dt} \) factor.
✕ Ignoring Signs
A negative rate means a quantity is decreasing. Forgetting to account for direction leads to wrong answers and wrong interpretations.

Question 1 of 5

A circle's radius increases at \( \frac{dr}{dt} = 2 \text{ cm/s} \). What is \( \frac{dA}{dt} \) when \( r = 3 \text{ cm} \)?

Question 2 of 5

Given \( x^2 + y^2 = 25 \), with \( \frac{dx}{dt} = 4 \), what is \( \frac{dy}{dt} \) when \( x = 3 \) and \( y = 4 \)?

Question 3 of 5

A spherical balloon is inflated at \( \frac{dV}{dt} = 100 \text{ cm}^3\text{/s} \). How fast is the radius growing when \( r = 5 \text{ cm} \)?

Question 4 of 5

In a related rates problem, why must you differentiate before substituting known values?

Question 5 of 5

A 13 ft ladder leans against a wall. The base moves away at \( \frac{dx}{dt} = 2 \text{ ft/s} \). When the base is 5 ft from the wall, how fast is the top sliding down?