Answer
a. $14\frac{cm^2}{sec}$ increasing
b. $0\frac{cm}{sec}$ constant
c. $-\frac{14}{13}\frac{cm}{sec}$, decreasing
Work Step by Step
a. The area of a rectangle is given by
$A=L*W$
We then take the derivative of this equation using the chain rule to get:
$dA = (\frac{dL}{dt}*W) + (\frac{dW}{dt}*L)$
Plug in the given values $\frac{dL}{dt} = -2, \frac{dW}{dt} = 2, L = 12, W = 5$
$(-2)(5) + (2)(12) = 14 \frac{cm^2}{sec}$
Since this slope is positive, we can infer that the area is increasing at this point
b. The perimeter of a rectangle is given by
$P = 2L + 2W$
We then take the derivative of this equation using the chain rule to get:
$dP = 2(\frac{dL}{dt}) + 2(\frac{dW}{dt})$
Plug in the given values $\frac{dL}{dt} = -2, \frac{dW}{dt} = 2$
$(2)(-2) + (2)(2) = 0\frac{cm}{sec}$
Since this slope is 0, we can infer that the perimeter is constant at this point.
c. The length of a diagonal is given by
$D = \sqrt{L^2 + W^2}$
We then take the derivative of this equation using the chain rule to get:
$dD = (\frac{L}{\sqrt{L^2 + W^2}})(\frac{dL}{dt}) + (\frac{W}{\sqrt{L^2 + W^2}})(\frac{dW}{dt})$
Plug in the given values $\frac{dL}{dt} = -2, \frac{dW}{dt} = 2, L = 12, W = 5$
$(\frac{12}{13})(-2) + (\frac{5}{13})(2) = -\frac{14}{13}\frac{cm}{sec}$
Since this slope is negative, we can infer that the length of the diagonal is decreasing at this point.
