Why Track the Area Rate?
A rectangle’s area depends on both its length and its width. If both sides change over time, the area’s rate of change depends on how quickly each side moves and on the current size of the other side. A side’s rate alone does not determine how quickly the area changes.
For example, increasing the width by one centimetre adds more area when the rectangle is long than when it is short. The calculation must account for the changing sides together.
Let \(x(t)\) and \(y(t)\) denote the side lengths at time \(t\), measured in a consistent unit. The area is \(A(t)=x(t)y(t)\). The goal is to find the instantaneous rate of change of \(A\) when the values of the sides and their rates are known.
The Area-Rate Formula
In this formula, \(x\) and \(y\) are the side lengths at the instant of interest, while \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) are their signed rates at that same instant. If lengths are measured in metres and time in seconds, for example, the side rates have units of metres per second and the area rate has units of square metres per second.
The two terms have a useful interpretation. The term \(x\frac{dy}{dt}\) is the contribution to the area rate from the changing \(y\)-side, scaled by the current \(x\)-side. The term \(y\frac{dx}{dt}\) is the contribution from the changing \(x\)-side, scaled by the current \(y\)-side. Both contributions must be included.
Why the Formula Works
The formula follows from how the area changes over a short time interval. Suppose the dimensions at time \(t\) are \(x(t)\) and \(y(t)\). After a small time increment \(h\), write the changes in the sides as \(\Delta x=x(t+h)-x(t)\) and \(\Delta y=y(t+h)-y(t)\). The new area is \((x+\Delta x)(y+\Delta y)\), so the change in area is
Divide by \(h\). The values \(x\) and \(y\) in this calculation are the dimensions at time \(t\). Thus
As \(h\) approaches zero, differentiability gives \(\frac{\Delta x}{h}\to\frac{dx}{dt}\) and \(\frac{\Delta y}{h}\to\frac{dy}{dt}\). Also, \(\Delta y\to 0\), so the final term tends to zero: \(\frac{\Delta x}{h}\) approaches the finite value \(\frac{dx}{dt}\), while \(\Delta y\) approaches zero. The left side approaches \(\frac{dA}{dt}\). Therefore,
This argument also shows why multiplying the two side rates is not the right method. The area changes through the contributions of each side, with each contribution scaled by the other side’s current length.
A Reliable Calculation Sequence
For a related-rates problem about a rectangle, keep the relationship and the changing quantities clear before inserting values.
Use \(A=xy\), with both side lengths understood to vary with time.
Apply the product rule to obtain \(\frac{dA}{dt}=x\frac{dy}{dt}+y\frac{dx}{dt}\).
Use both side lengths and both side rates at the specified instant. Keep shrinking rates negative.
Combine terms, include the area-rate units, and verify the result in the differentiated equation.
The formula requires the side lengths to be differentiable at the instant being studied. The dimensions are usually positive for a nondegenerate rectangle. Rates are signed: an increasing side has a positive rate, and a decreasing side has a negative rate.
Worked Examples
Worked Example: Both Sides Are Increasing
A rectangle is \(8\) metres long and \(5\) metres wide at a particular instant. At that instant, the length is increasing at \(0.3\) metres per second and the width is increasing at \(0.2\) metres per second. Find the rate at which its area is changing.
Let \(x=8\) metres, \(y=5\) metres, \(\frac{dx}{dt}=0.3\) metres per second, and \(\frac{dy}{dt}=0.2\) metres per second. Apply the area-rate formula:
The area is increasing at \(3.1\) square metres per second. To check, substitute the given values into the two contributions: the width’s contribution is \(8(0.2)=1.6\ \mathrm{m^2/s}\), and the length’s contribution is \(5(0.3)=1.5\ \mathrm{m^2/s}\). Their sum is \(3.1\ \mathrm{m^2/s}\), as calculated.
Worked Example: One Side Is Shrinking
A rectangle has side lengths \(12\) centimetres and \(7\) centimetres at an instant. The \(12\)-centimetre side is shrinking at \(0.2\) centimetres per minute, while the \(7\)-centimetre side is increasing at \(0.5\) centimetres per minute. Find the area rate.
Let \(x=12\) centimetres and \(y=7\) centimetres. Since \(x\) is shrinking, \(\frac{dx}{dt}=-0.2\) centimetres per minute; since \(y\) is increasing, \(\frac{dy}{dt}=0.5\) centimetres per minute. Therefore,
The area is increasing at \(4.6\) square centimetres per minute at this instant. The positive contribution from the growing \(y\)-side is larger than the negative contribution from the shrinking \(x\)-side. Checking the signed terms gives \(6+(-1.4)=4.6\ \mathrm{cm^2/min}\), which confirms the result.
Worked Example: Find an Unknown Side Rate
At a particular instant, a rectangle is \(6\) feet by \(10\) feet. The \(6\)-foot side is increasing at \(0.5\) feet per second. If the area is increasing at \(7\) square feet per second, how quickly is the \(10\)-foot side changing?
Set \(x=6\) feet, \(y=10\) feet, \(\frac{dx}{dt}=0.5\) feet per second, and \(\frac{dA}{dt}=7\) square feet per second. Let \(\frac{dy}{dt}\) be the unknown. The formula gives
The \(10\)-foot side is increasing at \(\frac{1}{3}\) feet per second. To check, substitute this rate into the area formula: \(6\left(\frac{1}{3}\right)+10(0.5)=2+5=7\ \mathrm{ft^2/s}\), matching the stated area rate.
Signs, Units, and Interpretation
A positive value of \(\frac{dA}{dt}\) means that the area is increasing at the specified instant. A negative value means the area is decreasing. If one side grows while the other shrinks, the signs of the two terms show how their effects compete. The result depends on their sizes as well as their rates.
For instance, a shrinking side does not necessarily mean the area is shrinking: a sufficiently rapid increase in the other side can make the total area increase. Conversely, even if both sides are changing, their contributions could balance so that the instantaneous area rate is zero. The formula gives the net rate; it does not require that each side move in the same direction.
Units provide a useful check. Each term combines a length with a length per time, so each must have units of area per time. If a calculation produces units of length per time or area per time squared, something has gone wrong in the setup or arithmetic.
Common Pitfall: Omitting One Contribution
A tempting but incorrect shortcut is to use only the rate of one side. For the first example, using only the width’s contribution would give \(8(0.2)=1.6\ \mathrm{m^2/s}\). That omits the area change caused by the increasing length, whose contribution is \(5(0.3)=1.5\ \mathrm{m^2/s}\). The correct total is \(1.6+1.5=3.1\ \mathrm{m^2/s}\).
Another error is to multiply the side rates together. In that same example, \((0.3)(0.2)=0.06\) has units of square metres per square second, not square metres per second, and it is not the area rate. The correct calculation uses each side rate multiplied by the other side’s current length.
Check Your Understanding
Use the area-rate formula and keep units and signs explicit.
- A rectangle has sides \(9\) metres and \(4\) metres. They are increasing at \(0.2\) metres per second and \(0.5\) metres per second, respectively. What is the area rate?
- A rectangle has sides \(10\) centimetres and \(6\) centimetres. The first side is shrinking at \(0.3\) centimetres per minute, and the second is increasing at \(0.4\) centimetres per minute. Is the area increasing or decreasing, and at what rate?
- In the formula for \(\frac{dA}{dt}\), why is each side rate multiplied by the other side length?
- A rectangle is \(5\) feet by \(8\) feet. The \(5\)-foot side increases at \(0.4\) feet per second, and the area increases at \(6\) square feet per second. Find the rate of change of the \(8\)-foot side.
- What does a negative value of \(\frac{dA}{dt}\) say about the rectangle at that instant?
