91Ó°ÊÓ

The area of a rectangle is determined by the formula \( A = l \times w \), where \( l \) represents the length and \( w \) the width. For related rates, we need to consider how changes in these dimensions affect the area over time.
When solving these types of problems, remember:

• Rate of change in length is given as \( \frac{dl}{dt} \)
• Rate of change in width is given as \( \frac{dw}{dt} \)

Differentiating the area formula with respect to time provides us: \[ \frac{dA}{dt} = w \frac{dl}{dt} + l \frac{dw}{dt} \]This formula expresses how quickly the area changes concerning both length and width. For example, in the exercise above, when you plug in the values \( l = 12 \text{ cm} \), \( w = 5 \text{ cm} \), \( \frac{dl}{dt} = -2 \text{ cm/sec} \), and \( \frac{dw}{dt} = 2 \text{ cm/sec} \), the area change rate is \( 14 \text{ cm}^2/\text{sec} \). This positive change signifies that the area is increasing over time, despite the length decreasing, because the width is increasing faster.

The perimeter of a rectangle is calculated using the formula \( P = 2(l + w) \). This accounts for the total distance around the rectangle.
To find how the perimeter changes over time, differentiate the expression:

• \( \frac{dP}{dt} = 2 \left( \frac{dl}{dt} + \frac{dw}{dt} \right) \)

Applying this formula in related rates problems is straightforward. In the exercise:

• Given \( \frac{dl}{dt} = -2 \text{ cm/sec} \) and \( \frac{dw}{dt} = 2 \text{ cm/sec} \)
• The rate of change of the perimeter is \( 2(0) = 0 \text{ cm/sec} \)

This result tells us that the perimeter remains constant over time. Even as length decreases and width increases, they do so in exactly the right way to balance each other out for no net change in perimeter.

The diagonal of a rectangle adds an interesting dynamic to changes in dimensions. It is found using the Pythagorean theorem: \( d = \sqrt{l^2 + w^2} \).
For the diagonal's rate of change, differentiate using the chain rule:

• \( \frac{dd}{dt} = \frac{1}{\sqrt{l^2 + w^2}}(l \frac{dl}{dt} + w \frac{dw}{dt}) \)

When you substitute the known values from the exercise: \( l = 12 \text{ cm} \), \( w = 5 \text{ cm} \), \( \frac{dl}{dt} = -2 \text{ cm/sec} \), and \( \frac{dw}{dt} = 2 \text{ cm/sec} \), you get \( \frac{-14}{13} = -1.08 \text{ cm/sec} \).
This negative value indicates that the diagonal is decreasing, which is intuitive since the rectangle's length is shrinking more significantly than the width's increase can counterbalance. Thus, as you trace across the rectangle diagonally, you cover less distance than before.