Question

find the rate of change of its area when its length is 65in and its width is 15in

Answer 1

We will have the following:

$A=w\cdot l$

So we use implicit derivations with respect of "t" to obtain:

$\frac{\partial A}{\partial t}=\frac{\partial w}{\partial t}\cdot l+w\cdot\frac{\partial l}{\partial t}$

Now, we replace the values and solve for the rate of change of the area:

$\frac{\partial A}{\partial t}=(65in)(-\frac{2in}{s})+(15in)(\frac{9in}{s})\Rightarrow\frac{\partial A}{\partial t}=-\frac{130in^2}{s}+\frac{135in^2}{s}$$\Rightarrow\frac{\partial A}{\partial t}=\frac{5in^2}{s}$

So, the rate of change of the area is 5 square inches per second.