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:

`(\partial A)/(\partial t)=(\partial w)/(\partial t)\cdot l+w\cdot(\partial l)/(\partial t)`

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

`(\partial A)/(\partial t)=(65in)(-(2in)/(s))+(15in)((9in)/(s))\Rightarrow(\partial A)/(\partial t)=-(130in^2)/(s)+(135in^2)/(s)`
`\Rightarrow(\partial A)/(\partial t)=(5in^2)/(s)`

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