Daniel's question at Yahoo Answers regarding related rates

[MarkFL]

Here is the question:

Question on Rate of Change?

The breadth of a rectangle is 1/4 of its length. Calculate the rate of increase of the area of the rectangle when its length is increasing at the rate of 0.4 cm s^-1, at the instant the length is 8 cm.

I came out with this formula : da/dt = da/dl x dl/dt
but i didn't know how to subsitute the values into the formula. Can anyone help me with this.

Answer given : 1.6 cm2 s^-1

Here is a link to the question:

Question on Rate of Change? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

 

[MarkFL]

Hello Daniel,

Let's let $W$ represent the breadth (width) of the rectangle and $L$ represent the length. We are told the breadth is 1/4 the length, hence we may state:

$\displaystyle W=\frac{L}{4}$

Now, we are asked to find the rate of change of the area with respect to time, so a good place to begin is with the formula for the area of a rectangle:

$\displaystyle A=WL$

Since we are given information on the time rate of change of the length, we want to express the area as a function of the length alone, so we may substitute for the width as follows:

$\displaystyle A=\frac{L}{4}\cdot L=\left(\frac{L}{2} \right)^2$

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

$\displaystyle \frac{dA}{dt}=2\cdot\frac{L}{2}\cdot\frac{1}{2} \cdot\frac{dL}{dt}=\frac{L}{2}\cdot\frac{dL}{dt}$

Now, using the given data $\displaystyle \frac{dL}{dt}=0.4\,\frac{\text{cm}}{s},\,L=8\text{ cm}$, we have:

$\displaystyle \frac{dA}{dt}=\frac{\left(8\text{ cm} \right)}{2}\cdot\left(0.4\,\frac{\text{cm}}{s} \right)=1.6\,\frac{\text{cm}^2}{s}$