Introduction
A circular region can grow or shrink as its radius changes. Knowing how quickly the radius is changing does not, by itself, tell us the area rate: the same change in radius has a different effect on area for a small circle than for a large one.
The relationship follows by differentiating the circle-area formula with respect to time. The result links three quantities: the area rate, the current radius, and the radius rate. This tutorial derives that relationship and applies it to expanding and shrinking circular regions.
Area and Radius as Time-Dependent Quantities
The area of a circle with radius \(r\) is \(A=\pi r^2\). In a changing-circle problem, the radius is a function of time. Write it as \(r(t)\), and let \(A(t)\) denote the corresponding area at time \(t\). Then the area formula becomes
The radius rate \(\frac{dr}{dt}\) measures how quickly the radius changes with time. The area rate \(\frac{dA}{dt}\) measures how quickly the area changes with time. Both are instantaneous rates at a particular time, so each depends on the state of the circle at that time.
A positive value of \(\frac{dr}{dt}\) means the radius is increasing, and a negative value means it is decreasing. Since \(2\pi r\) is nonnegative, the area rate has the same sign as the radius rate when \(r>0\). Thus an expanding circle has positive area rate, while a shrinking circle has negative area rate.
Why the Rate Formula Works
The area depends on time through the radius: first \(r\) changes, and then its square determines \(A\). The chain rule accounts for this dependence. Differentiating the outer expression \(\pi u^2\) with respect to \(u\) gives \(2\pi u\); multiplying by the rate at which \(u=r(t)\) changes gives the time derivative.
This also explains why the current radius must appear in the rate equation. The derivative of area with respect to radius is \(2\pi r\), so at a larger radius, a small change in radius produces a larger change in area. The time rate is that area-per-radius change multiplied by the radius-per-time change.
For example, if the radius grows at the same rate at two different times, the area will generally grow faster at the time when the circle is larger. The radius rate alone is not enough to determine the area rate; the current radius is also needed.
Applying the Relationship
In a related-rates calculation, identify the radius at the time of interest, the given rate, and the rate being requested. Then use the formula and keep the units attached to the rates. If the radius is measured in centimeters and time in seconds, then \(\frac{dr}{dt}\) has units of centimeters per second and \(\frac{dA}{dt}\) has units of square centimeters per second.
Use \(\frac{dA}{dt}=2\pi r\frac{dr}{dt}\), which follows by differentiating \(A=\pi r^2\) with respect to time.
Use the radius and rate values that apply at that same time, including the sign of a shrinking radius rate.
Substitute the values, simplify, verify the units, and substitute the result back into the relationship.
Worked Examples
Worked Example: Area Rate for an Expanding Circle
A circular region has radius \(3\) centimeters at a particular instant. Its radius is increasing at \(2\) centimeters per second. Find the area rate at that instant.
Start with the rate relationship:
At the specified instant, \(r=3\) centimeters and \(\frac{dr}{dt}=2\) centimeters per second. Substitute both values:
The area is increasing at \(12\pi\) square centimeters per second. To check, consider a radius \(r(t)=3+2t\) near \(t=0\), with \(t\) measured in seconds. Then \(A(t)=\pi(3+2t)^2\), so differentiation gives \(A'(t)=4\pi(3+2t)\). At \(t=0\), \(A'(0)=4\pi(3)=12\pi\) square centimeters per second, as required.
Worked Example: Area Rate for a Shrinking Circle
A circle has radius \(8\) centimeters at a particular instant, and its radius is decreasing at \(0.5\) centimeter per minute. Find its area rate.
Because the radius is decreasing, its signed rate is \(\frac{dr}{dt}=-0.5\) centimeter per minute. Substitute this and the current radius into the relationship:
The negative sign means the area is decreasing. To check, use \(r(t)=8-0.5t\) near \(t=0\), with \(t\) in minutes. Then \(A(t)=\pi(8-0.5t)^2\), and differentiation gives \(A'(t)=2\pi(8-0.5t)(-0.5)\). At \(t=0\), this is \(2\pi(8)(-0.5)=-8\pi\) square centimeters per minute, matching the result.
Worked Example: Finding the Radius Rate from the Area Rate
A circular region has radius \(6\) meters, and its area is increasing at \(15\pi\) square meters per minute. Find the radius rate.
Use the same relationship, but solve for \(\frac{dr}{dt}\). First substitute the known area rate and radius:
Divide both sides by \(12\pi\):
Thus the radius is increasing at \(\frac{5}{4}\) meter per minute. Check by substituting this rate into the area-rate equation:
This recovers the given area rate.
Common Pitfall: Losing the Factor \(2r\)
A frequent error is to differentiate \(A=\pi r^2\) as though the area rate were just \(\pi\frac{dr}{dt}\). That omits the derivative of \(r^2\). The chain rule requires differentiating the square as well as accounting for the time dependence of the radius.
For instance, at \(r=3\) centimeters with \(\frac{dr}{dt}=2\) centimeters per second, the incorrect calculation would be
This is wrong because \(\frac{d}{dr}(\pi r^2)=2\pi r\), not \(\pi\). The correct calculation is
The direct check is to differentiate \(A(t)=\pi(3+2t)^2\): its derivative at \(t=0\) is \(12\pi\), confirming the correct result.
Units and Interpretation
The factor \(2\pi r\) has units of length, while \(\frac{dr}{dt}\) has units of length per time. Their product therefore has units of area per time, as required for an area rate. A units check can reveal an error: an answer in centimeters per second cannot be an area rate when the area is measured in square centimeters.
The sign also carries information. A positive area rate means the circle’s area is increasing; a negative area rate means it is decreasing. If a problem asks “how fast” in the sense of a nonnegative speed, report the magnitude and state the direction of change. For example, an area rate of \(-8\pi\ \text{cm}^2/\text{min}\) means the area decreases at \(8\pi\) square centimeters per minute.
Check Your Understanding
Use the area-rate relationship to answer these questions. Include units and interpret the sign where appropriate.
- A circle has radius \(4\) centimeters and is expanding at \(3\) centimeters per second. What is its area rate?
- A circle’s radius is \(5\) meters and decreases at \(0.2\) meter per minute. What is its signed area rate?
- At radius \(7\) centimeters, the area increases at \(28\pi\) square centimeters per second. What is the radius rate?
- Why can two circles with the same radius rate have different area rates?
- What factor is missing from the incorrect formula \(\frac{dA}{dt}=\pi\frac{dr}{dt}\)?
