Exercise 6.1 · Q7: Rectangle with $x$ shrinking, $y$ growing: rates of perimeter and area

Basics you need

Part (a): the rate of change of the perimeter

1. Name the quantities and fix the signs of the rates

   At time $t$ minutes let the length be $x$ cm and the width $y$ cm. Both are functions of $t$⁠. From the question,

   $$\frac{dx}{dt} = -5, \qquad \frac{dy}{dt} = +4 .$$
   The book's Note after Example 3 ties the sign of a rate to the direction of change. The length is decreasing, so its rate is negative; the width is increasing, so its rate is positive. Dropping the minus sign is the most common way to get this question wrong.
   "Decreasing at 5 cm/min" gives the size of the rate (5) and its direction (down), and together these say $\frac{dx}{dt} = -5$⁠. Basics: rate of change as a derivative

2. Write the perimeter and differentiate it

   $$P = 2(x + y) \quad\Longrightarrow\quad \frac{dP}{dt} = 2\left(\frac{dx}{dt} + \frac{dy}{dt}\right).$$
   The perimeter goes once round all four sides: two lengths and two widths. Its rate is found with the sum rule and the constant multiple rule; no product appears, so no product rule is needed.
   Sum rule $(u+v)' = u' + v'$ and constant multiple rule, Section 5.3. Lesson: rate of change of quantities

3. Substitute the rates

   $$\frac{dP}{dt} = 2(-5 + 4) = 2(-1) = -2 .$$

   So the perimeter is changing at $-2$ cm/min: it is decreasing at 2 cm per minute.

   The negative sign means the perimeter is getting smaller: the length loses 5 cm/min while the width gains only 4 cm/min, and each counts twice. Notice the values $x = 8$ and $y = 6$ were not needed: the perimeter's rate depends only on the rates of the sides.
   A negative rate means the quantity is decreasing (Note after Example 3, Section 6.2). Basics: rate of change as a derivative

Part (b): the rate of change of the area

1. Write the area and differentiate by the product rule

   $$A = x\,y \quad\Longrightarrow\quad \frac{dA}{dt} = \frac{dx}{dt}\,y + x\,\frac{dy}{dt} .$$
   Both factors change with time, so the area's rate has two contributions: the change in length acting on the current width, and the change in width acting on the current length. Multiplying the two rates together would be wrong.
   Product (Leibnitz) rule $(uv)' = u'v + uv'$⁠, rule (2) of the algebra of derivatives, Section 5.3, with $u = x(t)$ and $v = y(t)$⁠. Basics: related rates (chain rule)

2. Substitute the instant

   Use $\frac{dx}{dt} = -5$⁠, $\frac{dy}{dt} = 4$ (Part (a), Step 1) and $x = 8$⁠, $y = 6$⁠:

   $$\begin{aligned} \frac{dA}{dt} &= (-5)(6) + (8)(4) \\ &= -30 + 32 = 2 . \end{aligned}$$

   So the area is changing at $2$ cm$^2$/min: it is increasing at 2 cm$^2$ per minute.

   Unlike the perimeter, the area's rate depends on the current sides, so the instant is needed here. The shrinking length removes area at $5 \times 6 = 30$ cm$^2$/min (a strip of width 6 losing 5 cm per minute); the growing width adds area at $8 \times 4 = 32$ cm$^2$/min. The gain wins by 2.
   Units: area (cm$^2$⁠) per time (min), so cm$^2$/min. Lesson: rate of change of quantities

3. Interpret the two answers together

   At this instant the perimeter is decreasing while the area is increasing. There is no contradiction: the rectangle is becoming less elongated (8 by 6 moving towards a square), and for a fixed perimeter a squarer rectangle encloses more area.

   It is tempting to assume that perimeter and area always move together. Signs computed honestly from the rates show they need not.
   The sign of each rate says the direction of change; compare signs, not just sizes. Basics: rate of change as a derivative

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