Juan's question at Yahoo! Answers regarding related rates

[MarkFL]

Here is the question:

The length of a rectangle is increasing at a rate of 9 cm/s and its width is increasing at a rate of 5 cm/s...?

The length of a rectangle is increasing at a rate of 9 cm/s and its width is increasing at a rate of 5 cm/s. When the length is 14 cm and the width is 12 cm, how fast is the area of the rectangle increasing?

Using a=lw, we have

(dA/dt) = l * (___?___) + (___?___) * (dl/dt)


___3.9, 2

I have posted a link there to this topic so the OP can see my work.

 

[MarkFL]

Re: Juan's question at Yahoo! Answers regarding releated rates

Hello Juan,

The area of the rectangle is:

\(\displaystyle A=\ell w\)

Implicitly differentiating with respect to time $t$, we find:

\(\displaystyle \frac{dA}{dt}=\ell\frac{dw}{dt}+\frac{d\ell}{dt}w\)

Using the given data:

\(\displaystyle \ell= 14\text{ cm},\,w= 12\text{ cm},\,\frac{d\ell}{dt}= 9\frac{\text{cm}}{\text{s}},\,\frac{dw}{dt}= 5\frac{\text{cm}}{\text{s}}\)

we obtain:

\(\displaystyle \frac{dA}{dt}=\left(14\text{ cm} \right)\left(5\frac{\text{cm}}{\text{s}} \right)+\left(9\frac{\text{cm}}{\text{s}} \right)\left(12\text{ cm} \right)=178\frac{\text{cm}^2}{\text{s}}\)