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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?2011-10-02
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    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)2011-10-02
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    The area of the initial square is 16 cm${}^2$ but you do not know the length of the side?2011-10-02
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    Oh yeah I knew that it is 4.2011-10-02
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    Yes it is. Next step: what is the length of the side of the square at time $t$?2011-10-02
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    I think it would be $length= \sqrt{t36}$2011-10-02
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    Argh... Let us come back to what we know: the length at time 0 is 4 cm; the length increases at 6cm/s; hence the length at time t is...2011-10-02
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    I don't get it, if the time is zero there is no length?2011-10-02
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    You said yourself that **AT** time 0 the length was 4 cm. The question is to know what is the length **AT** time t. For example the length at time 1 s is...2011-10-02
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    Wouldn't that be time 1?2011-10-02
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    You are asked a length. What. Is. The. Length. At. Time t? Let me try once more to help you: you are given only two things, the length at time 0 (that is, 4 cm) and the rate of increase of the length (that is, 6 cm/s), right? Hence you have to use them and them only...2011-10-02

2 Answers 2

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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.2011-10-02
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    @Jordan, keep telling yourself that, and you will certainly never learn it.2011-10-02
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    @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.2011-10-02
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    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.2011-10-02
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    @Henning, indeed.2011-10-02
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    @jordan what do you mean "to use the chain rule". The equation $\frac{dP}{dt}=2a\frac{da}{dt}$ follows from this line of reasoning: The problem is phrased entirely in rates of change (both givens and the desired answer). So we need to move the equation for the area of a static sphere ($P = a^2$) into one with rates of change. Specifically, we're concerned with change relative to time (t). So, what's the rate of change of P relative to t? Merely $\frac{dP}{dt}$ by definition of the derivative. Whats the rate of change of $a^2$ relative to t? Well, the rate of change of $a^2$ w.r.t. to a2011-10-02
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    @jordan is 2a, and the rate of change of a w.r.t. t is $\frac{da}{dt}$ so, the rate of change of the right hand side w.r.t. t is $2a\frac{da}{dt}$. This does follow from the chain rule, but I thought an English explanation might prove more useful to do.2011-10-02
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    I have no idea what you are talking about. Moving the equation for the area of a static sphere.2011-10-02
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    @jordan Sorry, mistyped square as sphere. My bad, didn't mean to confuse you further.2011-10-02
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    Oh ok. So I think I might get it.2011-10-02
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    Good! If you have more questions, ask away. Don't forget to use the @_____ if you want to ask someone something particular about their explanation.2011-10-02
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    Henning's quite right. Saying "this is complicated" in situations like this is almost always self-fulfilling...2011-10-02
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    I know I get frustrated with math, but I have a ton of homework to do and this is one of the first problems I attempted and I wasn't even able to do any of them. That is beyond frustrating for me.2011-10-02
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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?2011-10-02
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    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.2011-10-02
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    @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.2011-10-02
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    Your step by step questions were just making me look like an idiot, like I can't read english, so I stopped reading them.2011-10-02
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    So it is implicit differentiation and it would be 2s sprime?2011-10-02
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    @jordan Yes, kind of. In this case, $s\prime$ is somewhat ambiguous because the problem deals with rates of change . $s\prime(t)$ might be better notation. However, the best is what everyone answering your questions has been using: $\frac{ds}{dt}$. That way you know what you differentiating with respect to what. But yes, it is implicit differentiation. // Two notes of etiquette. One, If you're responding to someone in comments, use @ (their name) so they know you responded. Two, giving up is not they way to go. Everyone wants to help you _learn_ and sometimes that means basic step-by-step.2011-10-02
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    @Jordan, I see. You could have said so earlier (you know, simple politeness, sparing others' time, and everything).2011-10-02
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    It's not implicit differentiation, but it's the same kind of use of the chain rule that you see when you do implicit differentiation. $\dfrac{d}{ds} s^2 = 2s$, but $\dfrac{d}{dt} s^2 = 2s\cdot\dfrac{ds}{dt}$. That's an instance of the chain rule.2011-10-03
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    @Jordan, very interesting comment. If you communicate with people who try to help you by *hitting* them, nothing more needs be said.2011-10-03
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    If you help people by putting them down then it isn't really helping.2011-10-04
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    Just for the record: I DID NOT put you down. The elementary questions I asked (if tackled one by one and in the right order) are the easiest way to reach the solution that I could think of. These are elementary BY DESIGN. You left deliberately (and without notification) this (carefully selected) path to the solution. The effect (apart from alienating people) is that you are back to square one of your frustration. But the path is still there, if and when you choose to forget all this nonsense of looking-like-an-idiot which worries you so much and just hinders your learning process...2011-10-07