Related Rates: Translating Word Problems into Equations

What “Related Rates” Problems Are Really Asking

In a related rates problem, several quantities change over time and are connected by a relationship (often geometric or physical). You are given one or more rates of change (derivatives with respect to time) and asked to find another rate at a specific instant. The key skill is translating words and a diagram into an equation that links the variables, then differentiating that equation with respect to time.

Typical notation: if x changes with time t, then its rate is dx/dt. Rates always have units “(units of the variable) per (units of time)”, such as m/s, ft/min, cm^2/s, or $/day.

A Repeatable 5-Step Process

Step 1: Identify changing quantities; assign symbols and units

• List the quantities that change (lengths, areas, volumes, angles, costs, revenues, etc.).
• Assign symbols and write units next to each symbol.
• Write down the given rate(s) and the target rate using derivative notation.

Tip: If a quantity is constant, do not assign it a time-dependent symbol (or explicitly note it is constant).

Step 2: Draw a diagram first (when possible) and write a relationship equation

• Sketch the situation and label it with your symbols.
• Write a relationship equation that connects the variables (Pythagorean theorem, similar triangles, area/volume formulas, business formulas, etc.).
• Use a single time variable t for all changing quantities.

Step 3: Differentiate implicitly with respect to time

• Differentiate both sides with respect to t.
• Use the chain rule: if x depends on t, then d/dt(x^2)=2x dx/dt.
• Do not plug in numerical values yet.

Step 4: Substitute known values at the specific instant

• Only after differentiating, substitute the values of the variables and the given rates at the instant described.
• Make sure all values correspond to the same moment in time.

Step 5: Solve for the target rate; interpret sign and units

• Solve algebraically for the requested derivative.
• Attach correct units.
• Interpret the sign: negative often means “decreasing” (e.g., a distance shrinking), positive means “increasing.”

Common Pitfalls (and How to Avoid Them)

• Substituting before differentiating: If you plug numbers into the relationship equation first, you may lose the variable dependence needed for the chain rule. Differentiate first, substitute second.
• Mixing time instants: Values like x=3 and dx/dt=2 must refer to the same instant. If the problem says “when x=3,” then every substituted quantity must be evaluated at that same moment.
• Unit mismatches: Convert units early (e.g., minutes to seconds, inches to feet). Rates like ft/min and lengths in in will cause errors.
• Sign confusion: Decide whether a quantity is increasing or decreasing and assign the sign accordingly (e.g., ladder foot sliding away: dx/dt>0; ladder top descending: dy/dt<0).
• Forgetting chain rule factors: Common misses include d/dt(\pi r^2)=2\pi r dr/dt and d/dt(\sqrt{x})=(1/(2\sqrt{x})) dx/dt.

Diagram-First Example 1: Ladder Sliding Down a Wall

Scenario: A 10 ft ladder leans against a vertical wall. The bottom slides away from the wall at dx/dt = 1.5 ft/s. How fast is the top sliding down the wall when the bottom is 6 ft from the wall?

Step 1: Variables and units

• x(t) = distance of ladder’s bottom from wall (ft)
• y(t) = height of ladder’s top on wall (ft)
• Ladder length is constant: 10 ft
• Given: dx/dt = 1.5 ft/s
• Target: dy/dt (ft/s) at x=6 ft

Step 2: Relationship equation (from the right triangle)

Diagram: right triangle with legs x (horizontal) and y (vertical), hypotenuse 10.

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x^2 + y^2 = 10^2 = 100

Step 3: Differentiate with respect to time

d/dt(x^2 + y^2) = d/dt(100)

2x dx/dt + 2y dy/dt = 0

Solve for dy/dt:

dy/dt = -(x/y) dx/dt

Step 4: Substitute values at the instant x=6

First find y at that instant using the original relationship:

6^2 + y^2 = 100 so y^2 = 64 and y=8 ft (positive height).

Step 5: Compute and interpret

dy/dt = -(6/8)(1.5) = -1.125 ft/s

The negative sign means the top is moving downward at 1.125 ft/s.

Diagram-First Example 2: Filling a Tank (Volume and Height)

Scenario: A cylindrical tank with radius r = 0.5 m is being filled with water at a constant rate dV/dt = 0.03 m^3/s. How fast is the water level rising (dh/dt)?

Step 1: Variables and units

• V(t) = volume of water (m³)
• h(t) = water height (m)
• r is constant: 0.5 m
• Given: dV/dt = 0.03 m^3/s
• Target: dh/dt (m/s)

Step 2: Relationship equation

For a cylinder: V = \pi r^2 h

Since r is constant: V(t) = \pi (0.5)^2 h(t)

Step 3: Differentiate with respect to time

dV/dt = \pi r^2 dh/dt

Step 4: Substitute known values

0.03 = \pi (0.5)^2 dh/dt

Step 5: Solve and interpret

dh/dt = 0.03 / (\pi \cdot 0.25) = 0.12/\pi \approx 0.0382 m/s

The height increases at about 0.0382 m/s.

Diagram-First Example 3: Expanding Circle (Area vs. Radius)

Scenario: A circular oil spill expands so that the radius increases at dr/dt = 0.8 m/min. How fast is the area increasing when r = 20 m?

Step 1: Variables and units

• r(t) = radius (m)
• A(t) = area (m²)
• Given: dr/dt = 0.8 m/min
• Target: dA/dt (m²/min) at r=20 m

Step 2: Relationship equation

A = \pi r^2

Step 3: Differentiate with respect to time

dA/dt = 2\pi r dr/dt

Step 4: Substitute at the instant

dA/dt = 2\pi (20)(0.8)

Step 5: Solve and interpret

dA/dt = 32\pi \approx 100.53 m^2/min

Area is increasing; the positive value matches the expanding radius.

Non-Geometry Example: Cost, Revenue, and Profit Rates

Related rates also apply when quantities are linked by algebraic relationships rather than geometry. A common setup connects quantity produced/sold (q) to cost and revenue.

Example: Profit changing as production changes

Scenario: A company’s revenue and cost (in dollars) depend on the number of units produced q:

• R(q) = 50q
• C(q) = 200 + 20q + 0.01q^2

Production is increasing at dq/dt = 30 units/day. Find the rate of change of profit dP/dt when q = 1000, where P = R - C.

Step 1: Variables and units

• q(t) = units produced (units)
• R(t)=R(q(t)) in dollars
• C(t)=C(q(t)) in dollars
• P(t)=P(q(t)) in dollars
• Given: dq/dt = 30 units/day
• Target: dP/dt ($/day) at q=1000

Step 2: Relationship equation

P(q) = R(q) - C(q) = 50q - (200 + 20q + 0.01q^2)

P(q) = 30q - 200 - 0.01q^2

Step 3: Differentiate with respect to time (chain rule)

dP/dt = (dP/dq)(dq/dt)

Compute dP/dq:

dP/dq = 30 - 0.02q

So:

dP/dt = (30 - 0.02q)(dq/dt)

Step 4: Substitute at q=1000

dP/dt = (30 - 0.02(1000))(30)

dP/dt = (30 - 20)(30) = 300

Step 5: Interpret

dP/dt = 300 $/day. Profit is increasing at that instant because the result is positive.

Technique Spotlight: “Differentiate First, Then Plug In”

Many errors come from substituting too early. Compare these two approaches in the ladder problem:

• Correct: Start with x^2 + y^2 = 100, differentiate to get 2x dx/dt + 2y dy/dt = 0, then plug in x=6, y=8, dx/dt=1.5.
• Incorrect: Plugging x=6 into x^2 + y^2 = 100 gives y=8, then treating 6^2 + 8^2 = 100 as if differentiating it will reveal the rates (it won’t, because it has no variables left).

Quick Checklist for Any Related Rates Problem