B. Derivatives and Their Applications · Applications of Derivatives: Motion, Rates and Optimization

Related Rates: Linking Rates with the Chain Rule

B2.2 B2.3About 70 minutes

Learning goals

I can use the chain rule to connect the rates of change of related quantities.
I can follow a strategy: diagram, known and wanted rates, equation, differentiate, substitute.
I can use similar triangles and the Pythagorean theorem to link variables.
I can solve related-rates problems involving areas, volumes, flow and sliding objects.

Words to know

related rates
Rates of change of two or more quantities that are linked by an equation.

chain rule (in time)
dQ/dt = (dQ/dx)(dx/dt) when Q depends on x and x depends on t.

similar triangles
Triangles with equal angles, so their sides are in proportion; used to link r and h in cones.

instant
One particular moment in time, at which the variables have particular values.

When one rate drives another

Drop a pebble in a still pond. The circular ripple's radius grows at a steady rate, but its area grows faster and faster. Pump air into a weather balloon at a steady rate, and its radius grows more and more slowly. In each case, two quantities are linked by an equation, so their rates are linked too.

Problems like these are called related rates problems. The Ontario expectations do not name them, but they are a natural use of flow rates (B2.2) and instantaneous rates (B2.3), and they are a standard topic in first-year university calculus. All you need is the chain rule from Derivatives and Differentiation Rules.

The key idea: the chain rule in time

Suppose Q depends on x, and x changes with time t. Then Q is a composite function of t, and the chain rule says

dQ/dt = (dQ/dx) × (dx/dt).

For example, a circle's area is A = πr². If the radius is changing, then dA/dt = 2πr × dr/dt.

A strategy

Draw a diagram and label the quantities that change with letters, and constants with numbers.
List the rate you know and the rate you want, with units (for example, dr/dt = 0.5 m/s; want dA/dt).
Find an equation linking the variables (area, volume, Pythagoras, similar triangles). If it has extra variables, use geometry to remove them, so the quantity you want is a function of one variable.
Differentiate with respect to t, using the chain rule.
Substitute the values at the instant you care about, only now, and solve.
Answer in a sentence, with units and a sign that makes sense.

Substitute last

Do not plug in the instant's values (like r = 3) before you differentiate. A quantity that is changing is not a constant, and r = 3 is true for only one instant. Differentiate first, then substitute.

Worked example

Example 1: A ripple on a pond

A circular ripple's radius grows at 0.5 m/s. How fast is the area inside the ripple growing when the radius is 3 m?

Know: dr/dt = 0.5 m/s. Want: dA/dt when r = 3.
Equation: A = πr².
Differentiate with respect to t: dA/dt = 2πr × dr/dt.
Substitute: dA/dt = 2π(3)(0.5) = 3π ≈ 9.42 m²/s.

The area is growing at 3π ≈ 9.42 m²/s at that instant. (When r = 6 m, it will be growing twice as fast.)

Worked example

Example 2: A weather balloon

Helium is pumped into a spherical weather balloon at 0.2 m³/min. How fast is the radius increasing when the radius is 1 m?

Know: dV/dt = 0.2 m³/min. Want: dr/dt when r = 1.
Equation: V = (4/3)πr³.
Differentiate: dV/dt = 4πr² × dr/dt.
Substitute: 0.2 = 4π(1)² × dr/dt, so dr/dt = 0.2/(4π) ≈ 0.0159 m/min.

The radius is growing at about 0.016 m/min (1.6 cm/min). Because 4πr² grows with r, the radius grows more slowly as the balloon gets bigger.

Worked example

Example 3: Filling a conical tank

Water flows at 0.5 m³/min into a tank shaped like an upside-down cone, 4 m deep with a top radius of 2 m. How fast is the water level rising when the water is 2 m deep?

Know: dV/dt = 0.5 m³/min. Want: dh/dt when h = 2.
Equation: V = (1/3)πr²h, where r is the radius of the water's surface. Two variables, so use similar triangles: r/h = 2/4, so r = h/2.
Substitute r = h/2 before differentiating (this is geometry true at every instant, not a single instant's value): V = (1/3)π(h²/4)h = πh³/12.
Differentiate: dV/dt = (3πh²/12) dh/dt = (πh²/4) dh/dt.
At h = 2: 0.5 = (π × 4/4) dh/dt = π dh/dt, so dh/dt = 0.5/π ≈ 0.159 m/min.

The level is rising at about 0.16 m/min. As the cone widens, the same inflow raises the level more slowly.

The conical tank in Example 3, with water 2 m deep. The water forms a smaller cone similar to the tank, so r = h/2 at every instant.

Worked example

Example 4: A sliding ladder

A 5 m ladder leans against a wall. The bottom slides away from the wall at 0.3 m/s. How fast is the top sliding down the wall when the bottom is 3 m from the wall?

Let x = distance from the wall to the bottom, y = height of the top. Know dx/dt = 0.3 m/s; want dy/dt when x = 3.
Pythagoras: x² + y² = 25, so y = √(25 − x²) = (25 − x²)^(1/2).
Chain rule: dy/dt = (1/2)(25 − x²)^(−1/2) × (−2x) × dx/dt = (−x/√(25 − x²)) dx/dt.
At x = 3: √(25 − 9) = 4, so dy/dt = (−3/4)(0.3) = −0.225 m/s.

The top slides down at 0.225 m/s at that instant (dy/dt is negative because the height is decreasing).

The sliding ladder of Example 4 at the instant x = 3 m and y = 4 m. The length stays 5 m, so when x grows, y must shrink.

Just for fun

Why do related-rates problems always involve ladders sliding down walls? Because in calculus textbooks, nobody ever holds the ladder. (In real life, always have someone hold it.)

Hands-on activity

Fill a cone and a cylinder

You need: a cone-shaped paper cup or a funnel with its spout plugged with a little modelling clay, a straight-sided glass, a ruler, a measuring spoon (15 mL), and water. Work over the sink.

Stand the ruler in the glass. Add water one tablespoon (15 mL) at a time, and record the height after each spoonful for 8 spoonfuls.
Do the same for the cone (hold it point-down, or set it in a narrow glass so it stands up).
Make a table of the height rise per spoonful for each container.

What to look for: in the straight glass, each spoonful raises the level by about the same amount (dh/dt is constant when you pour at a steady rate, because V = πr²h with r fixed). In the cone, the first spoonfuls raise the level a lot and later ones less and less, exactly as dh/dt = (dV/dt) ÷ (πh²/4) predicts for Example 3: the bigger h is, the smaller dh/dt.

Your turn

The side of a square grows at 2 cm/s. How fast is its area growing when the side is 10 cm?

A circle's radius grows at 3 cm/s. How fast is its circumference growing?

A sphere's radius grows at 0.1 cm/s. How fast is its volume growing when r = 5 cm?

A cylindrical tank of radius 1 m is filled at 0.3 m³/min. How fast does the level rise?

A 10 m ladder's bottom is pulled from a wall at 0.5 m/s. How fast is the top sliding down when the bottom is 6 m from the wall?

The area of a circular oil spill grows at 10 m²/s. How fast is the radius growing when r = 4 m?

The edges of a cube of ice grow at 0.5 cm/min as water freezes onto it. How fast is its volume growing when the edge is 4 cm?

More practice: warm-up

Differentiate x² with respect to t (x depends on t).

Differentiate r³ with respect to t.

y = 3x and dx/dt = 4. Find dy/dt.

V is in m³ and t in minutes. What are the units of dV/dt?

A = πr². Find dA/dr.

More practice: on your own

For the conical tank of Example 3 (inflow 0.5 m³/min, r = h/2), how fast is the level rising when h = 3 m?

A spherical balloon loses air at 50 cm³/s. How fast is the radius changing when r = 10 cm?

A plane flies horizontally at 600 km/h at an altitude of 3 km, toward a point directly above a radar station and then past it. How fast is its distance from the station increasing when it is 4 km past that point (horizontally)?

A spherical snowball melts so that its volume decreases at 2 cm³/min. How fast is its radius shrinking when r = 5 cm?

For the same snowball, how fast is its surface area S = 4πr² changing when r = 5 cm?

Challenge

A 1.8 m tall person walks away from a 6 m lamppost at 1.2 m/s. How fast is the length of their shadow increasing?

Water drains from a conical paper cup (point down) that is 10 cm deep with a top radius of 4 cm, at 2 cm³/s. How fast is the water level falling when the water is 5 cm deep?

For the ladder in Example 4, dy/dt = (−x/y)(0.3). What happens to dy/dt as the top gets close to the ground, and what does this tell you about the model?

Check your understanding

Quiz

A circle's radius grows at 2 cm/s. How fast is the area growing when r = 5 cm?

In a related-rates problem, when should you substitute the values at the given instant?

V = (4/3)πr³. What is dV/dt?

A cone-shaped tank (point down) fills at a constant rate. What happens to the rate at which the water level rises?

A ladder slides with x² + y² = 100 (m). When x = 8 and dx/dt = 1 m/s, what is dy/dt?

Which rule of differentiation is at the heart of related rates?

For parents and teachers

Big idea (B2.2, B2.3): when quantities are linked by an equation, their rates are linked by the chain rule. Related rates are not named in the Ontario expectations, but they apply flow rates and instantaneous rates and are a standard first-year university topic.

How to help: insist on a labelled diagram and a list of "know" and "want" rates before any algebra. Ask, "Is that number true for all time or only at this instant?"

Common difficulties:

Substituting the instant's values before differentiating.
Forgetting the dx/dt factor from the chain rule.
Not removing an extra variable (using similar triangles for cones).
Sign errors: a decreasing quantity has a negative rate.