Related Rates

Calculus

When two quantities are connected by an equation, changing one forces the other to change too. Related rates problems ask: if I know how fast one quantity is changing, how fast is the other one changing?

The tool is implicit differentiation with respect to time.

1. The Core Idea

Imagine a circle whose radius is growing. As the radius r increases, the area A = pi * r^2 increases too. But the area does not grow at a constant rate — it grows faster and faster because the circle is getting bigger.

A = \pi r^2 \implies \frac00 = 2\pi r \cdot \frac00

If the radius grows at a steady rate (dr/dt is constant), the area’s rate of change dA/dt depends on how big r already is.

r (current radius)1

0.55

r = 1, \quad \frac00 = 1, \quad \frac00 = 2\pi \cdot 1 \cdot 1 \approx 6.28 \cdot 1

Try This

Try this: As r increases from 1 to 5, the red curve (rate of area change) grows linearly. A small circle with r = 1 gains about 6.28 square units per second, but a bigger circle with r = 5 gains about 31.4 square units per second — five times more, even though the radius grows at the same speed.

2. The Ladder Problem

A 10-foot ladder leans against a wall. The base slides away from the wall at 2 ft/sec. How fast is the top sliding down?

If x is the distance from the wall to the base, and y is the height of the top:

x^2 + y^2 = 100 \implies 2x\frac00 + 2y\frac00 = 0

\frac00 = -\frac00 \cdot \frac00

x (base distance from wall)3

0.59.5

x = 3, \quad y = \sqrt{100 - 3^2}

\frac00 = -\frac{{3}}{\sqrt{100 - 3^2}} \cdot 2

Connection

When x is small (ladder nearly vertical), the top barely moves. But as the base gets far from the wall, the top accelerates downward. Near x = 10, the rate becomes enormous — the top crashes down. This makes physical sense: the same horizontal push has a bigger effect when the ladder is nearly flat.

3. The Balloon Problem

A spherical balloon is being inflated at a constant rate of dV/dt = 100 cm^3/sec. How fast is the radius changing?

V = \frac00\pi r^3 \implies \frac00 = 4\pi r^2 \cdot \frac00

\frac00 = \frac{dV/dt}{4\pi r^2} = \frac0{4\pi r^2}

r (balloon radius, cm)3

110

r = 3 \text{ cm}, \quad \frac00 = \frac0{4\pi \cdot 3^2} \text{ cm/sec}

Try This

Try this: When the balloon is tiny (r = 1), the radius grows fast — about 8 cm/sec. But when it is big (r = 10), the radius barely changes — about 0.08 cm/sec. Even though air enters at the same rate, a bigger balloon needs more air to increase its radius by the same amount. The relationship is an inverse square.

4. Rates Depend on the Current State

Here is the pattern in every related rates problem: the rate of change of one quantity depends on the current values of the quantities, not just on the other rates.

dx/dt (how fast x changes)2

0.55

x (current value)1

0.55

For y = x^3, the related rate is:

\frac00 = 3x^2 \cdot \frac00 = 3 \cdot 1^2 \cdot 2

Challenge

Challenge: Two cars leave an intersection at the same time. Car A drives north at 30 mph and Car B drives east at 40 mph. How fast is the distance between them increasing after 2 hours? Hint: the distance is d = sqrt(x^2 + y^2), where x and y are the positions of the cars. Differentiate with respect to time.

The Big Idea

When quantities are linked by an equation, their rates of change are linked too. Differentiate the equation with respect to time to find the connection.

The strategy is always the same: write an equation relating the quantities, differentiate both sides with respect to t (using the chain rule!), plug in the known values and rates, and solve for the unknown rate. The chain rule is what makes it all work — related rates is really just the chain rule applied to real-world connections.

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