0
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ im still confused to how to start because I am not sure what dA/dd stands for? $\endgroup$ – user70884 Apr 4 '13 at 5:01
  • $\begingroup$ 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 $dA/dd$. $\endgroup$ – alex.jordan Apr 4 '13 at 6:03
0
$\begingroup$

$D$ 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.

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.