4.4 Introduction to Related Rates

Topic 4.3 used a derivative to describe how one quantity responds to another. In many applied problems several quantities change at once, all as time passes, and a known rate for one of them determines the rate for another. These are related-rates problems, and the chain rule is what links the rates.

Related rates: several quantities, one independent variable

A related-rates problem involves two or more quantities that change together because an equation connects them. Although time is the most common independent variable, every changing quantity must be differentiated with respect to the same independent variable. In applications, that variable is usually $t$.

Imagine a circular chemical spill whose radius $r$ changes as the spill spreads. Its area is

$$ A=\pi r^2. $$

If the radius grows at $\dfrac{dr}{dt}=0.6$ meters per minute when $r=10$ meters, differentiate both sides with respect to $t$:

$$ \frac{dA}{dt}=2\pi r\frac{dr}{dt}. $$

Substitute the known values:

$$ \frac{dA}{dt}=2\pi(10)(0.6)=12\pi\ \text{m}^2/\text{min}. $$

Thus, at that instant, the contaminated area is increasing at $12\pi$ square meters per minute. The factor $\dfrac{dr}{dt}$ appears because $r$ itself depends on $t$; this is the chain rule in action.

Topic 4.4 Introduction to Related Rates is aligned with Learning Objective CHA-3.D: Calculate related rates in applied contexts and Essential Knowledge CHA-3.D.1: The chain rule is the basis for differentiating variables in a related rates problem with respect to the same independent variable. The related-rates workflow is:

1. Name the variables and their units.
2. Write an equation relating the quantities.
3. Differentiate the entire equation with respect to the common independent variable, usually $t$.
4. Substitute values only after differentiating.
5. Interpret the resulting rate, including its sign and units.

Implicit differentiation in an applied relationship

Sometimes the relationship between quantities is not solved explicitly for one variable. Consider a chemical process in which concentration variables $x$ and $y$ satisfy

$$ 6xy=2+y^3. $$

Suppose both concentrations change as time $t$ passes. Differentiate implicitly with respect to $t$:

$$ 6\left(x\frac{dy}{dt}+y\frac{dx}{dt}\right) =3y^2\frac{dy}{dt}. $$

Collect the terms containing $\dfrac{dy}{dt}$:

$$ 6x\frac{dy}{dt}-3y^2\frac{dy}{dt} =-6y\frac{dx}{dt}, $$

so

$$ \frac{dy}{dt}

\frac{-6y,\dfrac{dx}{dt}}{6x-3y^2}. $$

If $x=1$, $y=2$, and $\dfrac{dx}{dt}=0.05$ concentration units per minute, then

$$ \frac{dy}{dt}

\frac{-6(2)(0.05)}{6(1)-3(2)^2}

\frac{-0.6}{-6}

0.1. $$

Therefore, $y$ is increasing at $0.1$ concentration units per minute at that instant. The sign matters: the equation forces $y$ to rise while $x$ rises at the stated point.

If the same equation is viewed geometrically rather than chemically, differentiating with respect to $x$ gives

$$ 6\left(x\frac{dy}{dx}+y\right)=3y^2\frac{dy}{dx}, $$

and therefore

$$ \frac{dy}{dx}=\frac{-6y}{6x-3y^2}. $$

This derivative is the slope of the curve at the point $(x,y)$. Replacing $\dfrac{dy}{dx}$ with $\dfrac{dy/dt}{dx/dt}$ produces the same relationship when both variables change with time.

Rules and reasoning that prevent errors

The chain rule is not the only rule that may be required. Essential Knowledge CHA-3.D.2: Other differentiation rules, such as the product rule and the quotient rule, may also be necessary to differentiate all variables with respect to the same independent variable. For instance, if

$$ Q=\frac{x^2}{y}, $$

then differentiating with respect to $t$ requires the quotient rule:

$$ \frac{dQ}{dt}

\frac{y\left(2x\frac{dx}{dt}\right)-x^2\frac{dy}{dt}}{y^2}. $$

This expression relates the rates of all three quantities through the same variable $t$.

The associated Implementing Mathematical Processes skill 1.E: Apply appropriate mathematical rules or procedures, with and without technology is assessed when you select and execute the correct differentiation rule, preserve factors such as $\dfrac{dx}{dt}$, and substitute data accurately.

Common misconception — differentiating with mismatched inputs: Writing $\dfrac{dA}{dr}=2\pi r$ when the problem asks for $\dfrac{dA}{dt}$ gives a valid derivative but answers a different question. Convert all rates to the same independent variable before interpreting them.

Retrieval check: A tank’s volume $V$ depends on its radius $r$, and both change with time. If $V=\pi r^2h$ with constant height $h$, which derivative represents the requested rate of volume change: $\dfrac{dV}{dr}$ or $\dfrac{dV}{dt}$? The answer is $\dfrac{dV}{dt}$, found by differentiating with respect to $t$ and using $\dfrac{dr}{dt}$.