Answer
(a) When the side length is $15~mm$, the instantaneous rate of change of the area is $30~mm^2/mm$
(b) The rate of change of the area with respect to the side length is half the perimeter.
Work Step by Step
(a) $A(x) = x^2$
$A'(x) = 2x$
$A'(15) = 2(15) = 30$
When the side length is $15~mm$, the instantaneous rate of change of the area is $30~mm^2/mm$
(b) The perimeter of a square is $4x$
Since $A'(x) = 2x$, the rate of change of the area with respect to the side length is half the perimeter.
Suppose the original area of a square is $x^2$
After we increase the sides by $\Delta x$, the new area is $(x+\Delta x)^2$
We can find the change in area:
$\Delta A = (x+\Delta x)^2-x^2$
$\Delta A = x^2+2x\Delta x+(\Delta x)^2-x^2$
$\Delta A = 2x\Delta x+(\Delta x)^2$
$\Delta A \approx 2x\Delta x$
Note that we can say this because $(\Delta x)^2 \approx 0$
Then:
$\Delta A \approx 2x\Delta x$
$\frac{\Delta A}{\Delta x} \approx 2x$
