Problem Overview:

We are solving a calculus problem involving rates of change for a rectangle. Here is what we know:

• The base $b$ is decreasing at $\frac{db}{dt} = -2 \, \text{cm/sec}$.
• The height $h$ is increasing at $\frac{dh}{dt} = +2 \, \text{cm/sec}$.
• The dimensions at the moment of interest are $b = 12 \, \text{cm}$ and $h = 5 \, \text{cm}$.

To solve:

1. The rate of change of the area ($\frac{dA}{dt}$).
2. The rate of change of the perimeter ($\frac{dP}{dt}$).
3. The rate of change of the diagonal ($\frac{dD}{dt}$).

(a) The rate of change of the area ($\frac{dA}{dt}$):

The formula for the area of a rectangle is: $A = b \cdot h$ Differentiating both sides with respect to time $t$: $\frac{dA}{dt} = \frac{db}{dt} \cdot h + b \cdot \frac{dh}{dt}$

Substituting values:

• $\frac{db}{dt} = -2 \, \text{cm/sec}$,
• $\frac{dh}{dt} = 2 \, \text{cm/sec}$,
• $b = 12 \, \text{cm}$,
• $h = 5 \, \text{cm}$.

$\frac{dA}{dt} = (-2)(5) + (12)(2)$ $\frac{dA}{dt} = -10 + 24 = 14 \, \text{cm}^2/\text{sec}.$

Answer:

The area of the rectangle is increasing at a rate of 14 cm²/sec.

(b) The rate of change of the perimeter ($\frac{dP}{dt}$):

The formula for the perimeter of a rectangle is: $P = 2b + 2h$ Differentiating both sides with respect to $t$: $\frac{dP}{dt} = 2\frac{db}{dt} + 2\frac{dh}{dt}$

Substituting values:

• $\frac{db}{dt} = -2 \, \text{cm/sec}$,
• $\frac{dh}{dt} = 2 \, \text{cm/sec}$.

$\frac{dP}{dt} = 2(-2) + 2(2)$ [ \frac{dP}{dt} = -4 + 4 =