Calculus Applications of differentiation

Related rates

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Related Rates

California Calculus • Standard 12.0 • Applications of Differentiation

Related Rates is a topic in Applications of Differentiation in the California Calculus Standards. It is aligned to Standard 12.0, which requires students to use differentiation to solve related-rate problems in a variety of applied contexts.

A related-rates problem links two changing quantities through an equation and finds how fast one changes given the other, by differentiating the relationship with respect to time.

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Theory

A related-rates problem connects two changing quantities through an equation and asks how fast one changes given the other. Differentiate the relationship with respect to time, then solve for the unknown rate.

When two quantities are linked by an equation and both change over time, their rates are linked too. Related rates finds one rate from another by differentiating the equation with respect to \(t\).

Every variable is a function of time, so the chain rule attaches a rate to each term — for example \(\dfrac{d}{dt}(r^2)=2r\dfrac{dr}{dt}\).

After differentiating, substitute the known values and solve for the unknown rate. Watch the sign: a shrinking quantity has a negative rate.

Key idea: substitute numbers only after differentiating — plugging in too early freezes a quantity that is actually changing.

An expanding circle: \(\dfrac{dA}{dt}=2\pi r\dfrac{dr}{dt}\).

A sliding ladder: \(x^2+y^2=L^2\) links the two rates.

Differentiate the geometric relation with respect to time:

\[A=\pi r^2\ \Rightarrow\ \dfrac{dA}{dt}=2\pi r\,\dfrac{dr}{dt}\]

\[x^2+y^2=L^2\ \Rightarrow\ x\dfrac{dx}{dt}+y\dfrac{dy}{dt}=0\]

differentiate area or the Pythagorean relation with respect to time

Units and sign: a rate carries a unit (ft/s, sq ft/s, cubic ft/s); a decreasing quantity gets a negative rate.

How to solve a related-rates problem

Write the relation among the quantities (often geometry).
Differentiate both sides with respect to \(t\).
Substitute the known values and rates, then solve for the unknown rate — keeping the units and sign.

Example 1 — Expanding circle

A circle's radius grows at \(2\) ft/s. How fast is the area changing when \(r=5\) ft?

Solution

Differentiate \(A=\pi r^2\) with respect to \(t\).

\(\dfrac{dA}{dt}\)	\(=\)	\(2\pi r\dfrac{dr}{dt}\)
	\(=\)	\(2\pi(5)(2)\)
	\(=\)	\(20\pi\)

The area grows at \(20\pi\) sq ft/s.

area increases at 20 pi square feet per second

Example 2 — Growing square

A square's side grows at \(3\) ft/s. How fast is the area changing when the side is \(4\) ft?

Solution

Differentiate \(A=s^2\).

\(\dfrac{dA}{dt}\)	\(=\)	\(2s\dfrac{ds}{dt}\)
	\(=\)	\(2(4)(3)=24\)

The area grows at \(24\) sq ft/s.

area increases at 24 square feet per second

Example 3 — Sliding ladder

A \(13\) ft ladder leans on a wall; its base slides out at \(2\) ft/s. How fast does the top slide when the base is \(5\) ft out?

Solution

With \(x^2+y^2=169\), differentiate: \(2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}=0\). At \(x=5\), \(y=12\).

\(5(2)+12\dfrac{dy}{dt}\)	\(=\)	\(0\)
\(\dfrac{dy}{dt}\)	\(=\)	\(-\dfrac{5}{6}\)

The top slides down at \(\dfrac{5}{6}\) ft/s (negative).

top slides down at five sixths feet per second

Example 4 — Expanding sphere

A balloon's radius grows at \(1\) ft/s. How fast is its volume changing when \(r=3\) ft?

Solution

Differentiate \(V=\dfrac{4}{3}\pi r^3\).

\(\dfrac{dV}{dt}\)	\(=\)	\(4\pi r^2\dfrac{dr}{dt}\)
	\(=\)	\(4\pi(9)(1)=36\pi\)

The volume grows at \(36\pi\) cubic ft/s.

volume increases at 36 pi cubic feet per second

Common pitfalls

Do not substitute too early. Differentiate the general relation first; only plug in the instant's values afterward.

Every term gets a time rate. Differentiating \(r^2\) gives \(2r\dfrac{dr}{dt}\), not \(2r\).

Mind the sign. A shrinking length or falling height has a negative rate; report it with the correct sign and unit.

Frequently asked questions

What is a related-rates problem?

A problem where two quantities change together through an equation, and you find one rate of change from another by differentiating with respect to time.

Why do you differentiate with respect to time?

Because every quantity depends on time. The chain rule then produces a rate \(\dfrac{d}{dt}\) for each variable, linking the rates together.

When should you plug in the numbers?

After differentiating the general relationship. Substituting the instant's values too early treats a changing quantity as constant.

What does a negative rate mean?

The quantity is decreasing. For a sliding ladder, a negative \(\dfrac{dy}{dt}\) means the top is moving down the wall.

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