Example 4: Rectangle with shrinking length and growing width: perimeter and area rates

Basics you need

Part (a): the perimeter

1. Turn the words into signed rates (used in both parts)

   At time $t$ minutes let the length be $x$ cm and the width $y$ cm. Then

   $$\frac{dx}{dt} = -3\ \text{cm/min}, \qquad \frac{dy}{dt} = 2\ \text{cm/min}.$$
   "Decreasing at the rate of 3 cm/minute" tells us the size of the rate (3) and its direction (down). A derivative carries both in one number, so $\frac{dx}{dt} = -3$⁠. Forgetting the minus sign is the most common error in this example.
   Basics: rate of change as a derivative

2. Write the perimeter and differentiate

   $$P = 2(x + y) .$$

   By the constant multiple and sum rules:

   $$\begin{aligned} \frac{dP}{dt} &= 2\left(\frac{dx}{dt} + \frac{dy}{dt}\right) \\ &= 2(-3 + 2) = -2 . \end{aligned}$$
   The perimeter is a sum, and the derivative of a sum is the sum of the derivatives. The length loses 3 cm per minute and the width gains 2, so the half-perimeter $x + y$ loses 1 cm per minute, and the full perimeter loses 2.
   Here we did not even need $x = 10$ and $y = 6$⁠: $\frac{dP}{dt} = -2$ at every instant while the rates stay as given.

3. State the answer with its meaning

   $\frac{dP}{dt} = -2$ cm/min. The negative sign means the perimeter is decreasing at 2 cm/min.

   The units are those of $P$ (cm) per unit of time (minute). The sign is part of the answer: it tells the reader the direction of change.
   Lesson: rate of change of quantities

Part (b): the area

1. Write the area and differentiate with the product rule

   $$A = x\,y .$$

   Both $x$ and $y$ depend on $t$⁠, with the signed rates of Part (a), Step 1. By the product rule:

   $$\frac{dA}{dt} = \frac{dx}{dt}\cdot y + x\cdot\frac{dy}{dt} .$$
   The area is a product of two changing quantities. Each term of the product rule measures one effect: $\frac{dx}{dt}\cdot y$ is the area lost because the length shrinks (a strip of height $y$⁠), and $x\cdot\frac{dy}{dt}$ is the area gained because the width grows (a strip of length $x$⁠).
   $(uv)' = u'v + uv'$⁠, Section 5.3. Writing $\frac{dA}{dt} = \frac{dx}{dt}\cdot\frac{dy}{dt}$ would be wrong.

2. Substitute the instant

   Now, and only now, put $x = 10$⁠, $y = 6$⁠, $\frac{dx}{dt} = -3$⁠, $\frac{dy}{dt} = 2$⁠:

   $$\begin{aligned} \frac{dA}{dt} &= (-3)(6) + (10)(2) \\ &= -18 + 20 = 2 . \end{aligned}$$

   So the area is increasing at 2 cm$^2$/min at that instant.

   The shrinking length removes 18 cm$^2$ per minute; the growing width adds 20 cm$^2$ per minute. The gain wins by 2. Unlike part (a), this rate depends on the instant, because the strips have lengths $y$ and $x$⁠, which keep changing.
   Units: cm$^2$ per minute. Positive sign: increasing.

3. A rate describes one instant

   Suppose the two rates stayed exactly as given. Then $x = 10 - 3t$ and $y = 6 + 2t$ (with $t = 0$ at our instant), so

   $$A(t) = (10 - 3t)(6 + 2t) = 60 + 2t - 6t^2 .$$

   Indeed $A'(0) = 2$⁠. But $A'(t) = 2 - 12t$ becomes negative after $t = \frac{1}{6}$ min: the area rises only to $A(\tfrac16) = \tfrac{361}{6} \approx 60.17$ cm$^2$ and then falls. After one minute the rectangle is 7 cm by 8 cm, with area 56 cm$^2$⁠, less than the 60 cm$^2$ we started with.

   "The area is increasing at 2 cm$^2$/min" is a statement about one instant (when $x = 10$⁠, $y = 6$⁠), not a promise about the future. This is why the given instant goes in only after differentiating.

Every fact used, and where it comes from