# related rates, how fast the area enclosed by the square growing at the instant the area is 100 square meters?

The diagonal length of a square is growing at a rate of 20 meters per second. How fast is the area enclosed by the square growing at the instant the area is 100 square meters? I am lost i don't know how to start, i need help

## 2 Answers

Both the diagonal length $d$ and the area $A$ depend on $t$. You have been asked to find $\frac{dA}{dt}$ at a particular moment. The chain rule says $$\frac{dA}{dt}=\frac{dA}{dd}\frac{dd}{dt}$$ and this applies all the time - no matter how early or late it is, no matter how small or big the square is.

Now at the moment in question, you can find $\frac{dA}{dd}$ if you find a way to express $A$ as a function of $d$. And $\frac{dd}{dt}$ was provided directly in the setup.

• im still confused to how to start because I am not sure what dA/dd stands for? – user70884 Apr 4 '13 at 5:01
• The derivative of $A$ with respect to $d$. Draw a picture and find the relationship $A=\text{formula with$d$in it}$. Then you can use derivative rules to compute $dA/dd$. – alex.jordan Apr 4 '13 at 6:03

$D$ is diagonal then $\frac{dD}{dt}$ is rate of change of diagonal with respect to time.

In question given that $\frac{dD}{dt}=20\;m/s$

Now area $A$ and perimeter $D$ are related in square.

If $s$ is one side of square then area of square is $A=s^2$ If $s$ is one side of square then diagonal of square is $D=\sqrt{2}s \Rightarrow s=\frac{D}{\sqrt{2}}$ putting the value of $s$ into area equation $A=\frac{D^2}{2}$ let's call this area Diagonal relation equation.

Differentiating both sides with respect to $t$

$\frac{dA}{dt}=\frac{2D}{2}\frac{dD}{dt}$

At $t=0$, Area $A=100\;m^2\Rightarrow s=10\;m\Rightarrow D=\sqrt{2}*10\;m$

Put the value of $D$ and $\frac{dD}{dt}$ into differential form of area diagonal equation.

$\frac{dA}{dt}=\sqrt{2}*200\;m/s$