How to Find the Rates of Change of a Rectangle’s Area, Perimeter, and Diagonal

To find how quickly a rectangle’s area, perimeter, and diagonals change, treat its side lengths as functions of time and differentiate each formula with respect to time. The resulting rates are instantaneous: they describe what is happening at a particular moment, using the side lengths and their rates at that same moment.

A numerical answer cannot be determined unless you know the dimensions and how quickly they are changing. The formulas below give the general answer and show how to substitute values when those data are available.

Define the changing side lengths

Let the rectangle’s side lengths be $x(t)$ and $y(t)$. Their rates of change are $dx/dt$ and $dy/dt$. A positive rate means a side is growing; a negative rate means it is shrinking.

The area and perimeter are

$$A=xy, \qquad P=2(x+y).$$

By the Pythagorean theorem, either diagonal has length

$$d=\sqrt{x^2+y^2}.$$

The two diagonals of a rectangle always have the same length. Consequently, they also have the same rate of change.

Differentiate each formula with respect to time

For the area, use the product rule because both side lengths may change:

$$\frac{dA}{dt}=x\frac{dy}{dt}+y\frac{dx}{dt}.$$

For the perimeter, differentiate the sum of the side lengths:

$$\frac{dP}{dt}=2\left(\frac{dx}{dt}+\frac{dy}{dt}\right).$$

For the diagonal, differentiate $d=(x^2+y^2)^{1/2}$ using the chain rule:

$$\frac{dd}{dt}=\frac{x\,dx/dt+y\,dy/dt}{\sqrt{x^2+y^2}}.$$

This last expression is the rate for each diagonal. The diagonal rate can be positive, zero, or negative, depending on whether the sides’ combined effects make the diagonal grow, stay momentarily constant, or shrink.

Worked example

Suppose that at a particular instant a rectangle is $3$ cm by $4$ cm. At that instant, the 3-cm side is increasing at $2$ cm/min, while the 4-cm side is decreasing at $1$ cm/min. Thus

$$x=3,\quad y=4,\quad \frac{dx}{dt}=2,\quad \frac{dy}{dt}=-1.$$

The area rate is

$$\frac{dA}{dt}=3(-1)+4(2)=5\text{ cm}^2/\text{min}.$$

So the area is increasing at $5$ cm²/min at that instant. The perimeter rate is

$$\frac{dP}{dt}=2(2-1)=2\text{ cm}/\text{min},$$

so the perimeter is increasing at $2$ cm/min. The diagonal is $\sqrt{3^2+4^2}=5$ cm, and its rate is

$$\frac{dd}{dt}=\frac{3(2)+4(-1)}{5}=\frac{2}{5}\text{ cm}/\text{min}.$$

Each diagonal is therefore increasing at $0.4$ cm/min at that instant.

A useful check on the diagonal rate

The diagonal rate’s numerator, $x(dx/dt)+y(dy/dt)$, tells you whether the diagonal is growing or shrinking, since its denominator is positive for a nondegenerate rectangle. In the example, the longer side is shrinking, but the shorter side is growing quickly enough that the diagonal still increases.

Units also provide a quick check: area rate has units of length squared per time, while perimeter and diagonal rates have units of length per time.

Common mistakes

• Treating a changing side as constant. If both dimensions change, include both terms in the area derivative and both rates in the perimeter derivative.
• Dropping a negative sign. A shrinking side has a negative rate, even though its length itself is positive.
• Differentiating the diagonal incorrectly. The diagonal is a square root, so use the chain rule; its rate is not simply the sum of the side rates.
• Using values from different instants. Substitute dimensions and rates that all describe the same moment.
• Giving one diagonal a different rate from the other. A rectangle’s diagonals are equal in length at every instant, so their rates of change are equal too.