From a Context to a Related-Rate Response
A related-rates problem describes quantities that change over time and asks how quickly one of them is changing when the quantities have a specified relationship. A complete response must connect the context to the calculus: identify the changing quantities, state the relationship among them, differentiate that relationship with respect to time, and interpret the rate that the question requests.
The order matters. In particular, differentiating before stating the relationship can hide where the equation came from, while substituting numerical values too early can turn a changing quantity into a constant in the calculation. A clear written solution keeps the quantities variable until after differentiation.
Representing Changing Quantities
Use a variable for each quantity that changes. When helpful, make the dependence on time explicit: \(x(t)\) might represent a distance at time \(t\), and \(A(t)\) might represent an area at time \(t\). A rate such as \(\frac{dx}{dt}\) describes how that quantity changes per unit of time.
Units help keep the response tied to the situation. If \(x\) is measured in feet and \(t\) in seconds, then \(\frac{dx}{dt}\) has units of feet per second. If \(A\) is measured in square feet, then \(\frac{dA}{dt}\) has units of square feet per second. These units are different because the quantities being measured are different.
The relationship is usually an equation supplied by the context or by geometry. For example, the area of a circle of radius \(r\) is \(A=\pi r^2\), and the legs \(x\) and \(y\) of a right triangle with fixed hypotenuse \(L\) satisfy \(x^2+y^2=L^2\). In a related-rates response, state the relevant relationship before taking its derivative.
Why Differentiation Connects the Rates
The central calculus idea is the chain rule. If \(y=F(x)\) and \(x\) changes with time, then \(y\) also changes with time, and
This equation explains why a rate of change in one quantity can determine a rate of change in another. The derivative \(F'(x)\) describes how the quantities are related at the current value of \(x\); multiplying by \(\frac{dx}{dt}\) accounts for how quickly \(x\) itself changes.
For a relationship involving two changing quantities, the same idea applies to each one. If \(x(t)^2+y(t)^2=L^2\), where \(L\) is constant, differentiating both sides with respect to time gives
Each squared quantity contributes a factor of its derivative with respect to time. The constant \(L^2\) has derivative zero. This is the reason related-rate equations include terms such as \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\): the variables describe changing quantities, not fixed numbers.
A Reliable Written Structure
Name each changing quantity, state what it measures, and include its units. Identify which rate is given and which rate is requested.
Write an equation connecting the quantities. Briefly identify its source, such as a geometric formula or a fixed length in the situation.
Use the chain rule for every changing quantity. A quantity stated to be constant has time derivative zero.
Use the values at the specified instant, along with the given rate, to find the requested rate.
Include units and explain whether a positive or negative rate means the quantity is increasing or decreasing.
Worked Examples
Worked Example: Area of an Expanding Circle
The radius of a circular ripple is increasing at \(3\) centimeters per second. How quickly is its area increasing when the radius is \(4\) centimeters?
Define the quantities. Let \(r(t)\) be the ripple's radius in centimeters and \(A(t)\) its area in square centimeters. The given rate is \(\frac{dr}{dt}=3\) centimeters per second. We seek \(\frac{dA}{dt}\) when \(r=4\) centimeters.
State the relationship. The area of a circle is \(A=\pi r^2\). Both \(A\) and \(r\) vary with time, so this is a relationship between changing quantities.
Differentiate with respect to time.
Substitute the values.
Check the rate equation. Substituting \(r=4\) and \(\frac{dr}{dt}=3\) into the differentiated relationship gives \(2\pi(4)(3)=24\pi\), as found. The units also match: centimeters times centimeters per second gives square centimeters per second.
Interpret the result. When the radius is \(4\) centimeters, the area is increasing at \(24\pi\) square centimeters per second.
Worked Example: A Ladder Sliding Down a Wall
A \(10\)-foot ladder rests against a vertical wall. Its foot moves away from the wall at \(2\) feet per second. At the instant when the foot is \(6\) feet from the wall, how quickly is the top moving vertically?
Define the quantities. Let \(x(t)\) be the horizontal distance, in feet, from the wall to the ladder's foot, and let \(y(t)\) be the vertical height, in feet, of the ladder's top. The ladder length is fixed at \(10\) feet. We are given \(\frac{dx}{dt}=2\) feet per second at \(x=6\), and we seek \(\frac{dy}{dt}\).
State the relationship. The wall and ground form a right angle, and the ladder is the hypotenuse. By the Pythagorean theorem, \(x^2+y^2=10^2=100\). At the stated instant, find \(y\) from this relationship:
The positive value \(y=8\) is appropriate because \(y\) is a height.
Differentiate with respect to time. The ladder length is constant, so the derivative of \(100\) is zero:
Substitute and solve.
Check the rate equation. Substitution gives \(2(6)(2)+2(8)(-\frac{3}{2})=24-24=0\), so the differentiated relationship is satisfied.
Interpret the result. At that instant, the top of the ladder is moving downward at \(\frac{3}{2}\) feet per second. The negative value of \(\frac{dy}{dt}\) indicates that the height is decreasing; the rate's units are feet per second.
Worked Example: Water Filling a Conical Tank
A conical tank is \(12\) meters high and has a top radius of \(4\) meters. Water is poured in so that its depth is increasing at \(2\) meters per minute. How quickly is the water's volume increasing when the depth is \(6\) meters?
Define the quantities. Let \(h(t)\) be the water depth in meters, \(r(t)\) the radius of the water's surface in meters, and \(V(t)\) the water volume in cubic meters. We are given \(\frac{dh}{dt}=2\) meters per minute and seek \(\frac{dV}{dt}\) when \(h=6\) meters.
State the geometric relationships. A horizontal cross-section of the water forms a smaller cone similar to the entire tank. The radius-to-height ratio is therefore the same for both cones:
The volume of a cone is \(V=\frac{1}{3}\pi r^2h\). Substituting \(r=\frac{h}{3}\) gives a relationship involving only the changing quantities \(V\) and \(h\):
Differentiate with respect to time.
Substitute and solve.
Check the rate equation. Differentiating \(V=\frac{\pi h^3}{27}\) gives \(\frac{dV}{dt}=\frac{3\pi h^2}{27}\frac{dh}{dt}=\frac{\pi h^2}{9}\frac{dh}{dt}\), the same rate equation used above. Substituting \(h=6\) and \(\frac{dh}{dt}=2\) yields \(8\pi\).
Interpret the result. When the water depth is \(6\) meters, the water's volume is increasing at \(8\pi\) cubic meters per minute.
A Common Error: Dropping a Rate Factor
In the ladder example, a tempting but incorrect step is to differentiate \(x^2\) with respect to time as if \(x\) were an ordinary fixed input:
This calculation omits how quickly \(x\) and \(y\) are changing. Since both depend on time, the chain rule gives \(\frac{d}{dt}(x^2)=2x\frac{dx}{dt}\) and \(\frac{d}{dt}(y^2)=2y\frac{dy}{dt}\). The correct equation is therefore
The units reveal the problem with the incorrect calculation: \(x\) and \(y\) have units of feet, so \(2x+2y\) has units of feet and cannot represent a sum of rates in feet per second. The correct terms have units of feet squared per second, as required when differentiating \(x^2+y^2\) with respect to time.
A second frequent error is to report only the numerical value and omit what its sign means. If a requested rate is negative, state that the quantity is decreasing; if it is positive, state that it is increasing. Include the units in either case.
What Makes the Response Complete?
A related-rates answer is not just a number produced by differentiation. Its written structure lets a reader see why the equation applies, which values belong to the specified instant, and how the calculated rate answers the question.
- Each changing quantity has a definition and suitable units.
- The relationship between the quantities is stated before differentiating.
- The differentiation is with respect to time, with chain-rule factors included.
- Instantaneous values are substituted into the rate equation after differentiation.
- The final rate includes units and a statement of what its sign means in context.
This structure is useful across many settings. A changing radius can determine an area rate, a fixed ladder length can connect horizontal and vertical rates, and similar shapes can connect a fluid's depth to its volume. The details differ, but in each case the explanation follows the same chain from context to equation to derivative to interpretation.
Check Your Understanding
For each question, focus on making the relationship between the context and the rate explicit.
- A circle's radius \(r(t)\) is changing. Write its area relationship and the rate equation obtained by differentiating with respect to time.
- A right triangle has a fixed hypotenuse and changing legs \(x(t)\) and \(y(t)\). Write the equation that relates \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) after differentiating.
- Why should a related-rates response generally substitute the values at the specified instant after differentiating the relationship?
- If a calculated height rate is \(-2\) feet per second, what does the sign mean, and what units should accompany the rate?
- In a cone problem, explain why a geometric relationship between the water-surface radius and depth may be needed before differentiating the volume formula.
