# Rate of increase in the area of a square

I really do not understand how to do these problems, so many weird math tricks and rules and I am getting caught up on at least a dozen in this problem. Anyways I am supposed to find:

Each side of a square is increasing at a rate of $6 \text{ cm/s}$. At what rate is the area of the square increasing when the area of the square is $16 \text{ cm}^2$?

I think what I need to do is set it equal to 16 or 4, but I am not sure which so the problem will look like $4=s(36)$ but I am not sure what to do with that.

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You are familiar with opening sentences like the one you wrote here. These bring nothing to your questions and can only alienate people participating to this site, as was already explained to you. It was also already explained to you that maths is NOT (I repeat, NOT) a collection of weird tricks. Why you persist in this vein is a mystery to me. // Regarding your question: what is the side of the initial square? What will be the side of the square at time $t$? Hence? – Did Oct 2 '11 at 16:10
I don't know what the initial square is, I don't think there is a way to find that out. I am not too sure how to find the side of a square, but I do know that the area at time t would be a=t(36) – user138246 Oct 2 '11 at 16:13
The area of the initial square is 16 cm${}^2$ but you do not know the length of the side? – Did Oct 2 '11 at 16:14
Oh yeah I knew that it is 4. – user138246 Oct 2 '11 at 16:16
Yes it is. Next step: what is the length of the side of the square at time $t$? – Did Oct 2 '11 at 16:16

We may define rate as $\frac{dP}{dt}$ , so let's find first derivative of the area formula $P=a^2$

$\frac{dP}{dt}=2a\frac{da}{dt}$ since $P=a^2 \Rightarrow a=\sqrt{P}$

$\frac{dP}{dt}=2\sqrt{P}\frac{da}{dt} \Rightarrow \frac{dP}{dt}=2*4*6 \frac{cm^2}{s} \Rightarrow \frac{dP}{dt}=48\frac{cm^2}{s}$

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I don't really know what the purpose of all those symbols are, I know what they mean but there are too many varying and changing letters for me to keep track of. – user138246 Oct 2 '11 at 16:26
@Jordan, keep telling yourself that, and you will certainly never learn it. – Henning Makholm Oct 2 '11 at 16:35
@jordan there really are only three essential symbols in pedja's answer: $P$ , $a$ , and $t$. $P$ is the area of the square, $a$ is the side length, and $t$ is just your time variable. $dP$, for example, is just shorthand to say, "the change in P" However, that's pretty meaningless without knowing what P is changing relative to. So, we write $\frac{dP}{dt}$, meaning the change in P relative to the change in t. $\frac{da}{dt}$ follows a similar pattern Pedja's first step, then, is implicit differentiation of the area formula $P = a^2$. From there, everything is replaced by givens. – Drew Christianson Oct 2 '11 at 16:36
I am getting confused by dp/dt and da/dt was that to use the chain rule? I really don't get this at all and it is incredibly frustrating, I have about 14 hours of homework to do today and I am too frustrated to continue already. I need to take a break and come back in a bit. – user138246 Oct 2 '11 at 16:38
@Henning, indeed. – Did Oct 2 '11 at 16:47

pedja's answer does seem to be expressed in a somewhat complicated way.

Let $A$ be the area in square centimeters. Let $s$ be the length of the side in centimeters. Let $t$ be time in seconds.

Then we are given $\dfrac{ds}{dt} = 6$.

We recall that $A = s^2$.

We want $\dfrac{dA}{dt}$ when $A=16$.

$$\frac{dA}{dt} = \frac{d}{dt} s^2 = 2s \frac{ds}{dt}.$$

When $A=16$ then $s=4$ and $ds/dt = 6$. So $$2s\frac{ds}{dt} = 2\cdot4\cdot 6.$$

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This is incredibly frustrating but I just don't follow what is happening. Isn't the derivative of s 2s? Where does 6 come in? – user138246 Oct 2 '11 at 17:38
You would be less frustrated if you tried to answer the step-by-step questions I am asking you in the comments. Just my two cents. – Did Oct 2 '11 at 17:52
@jordan, the derivative of $s^2$ with respect to s is $2s$. However, we want the derivative of $s^2$ with respect to t, so we have to multiply it by $\frac{ds}{dt}$. I know it seems like magic, but it follows from the chain rule for derivatives. – Drew Christianson Oct 2 '11 at 18:27
Your step by step questions were just making me look like an idiot, like I can't read english, so I stopped reading them. – user138246 Oct 2 '11 at 18:31
@Jordan, I see. You could have said so earlier (you know, simple politeness, sparing others' time, and everything). – Did Oct 2 '11 at 19:14