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The length of a rectangle is increasing at a rate of 8 cm/s and its width is increasing at a rate of 3 cm/s . When the length is 20 cm and the width is 10 cm, how fast is the area of the rectangle increasing?

To set up the equation, I have $A=lw$. Differentiate both sides of the equation, I have $\dfrac{dA}{dt}=\dfrac{dl}{dt}l+\dfrac{dw}{dt}w=\dfrac{dA}{dt}=(8.20)+(10.3)=510$.

The solution manual says $140$ and the equation is set$\dfrac{dA}{dt}=\dfrac{dw}{dt}l+\dfrac{dl}{dt}w$. Why does the equation is set this way? I thought that the rate of increase of the length must be paired with length and the rate of change of width should be paired with width? So what's going on here?

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  • $\begingroup$ Why do you think that "the rate of increase of the length must be paired with length and the rate of change of width should be paired with width"? $\endgroup$ – Michael Rybkin Mar 1 at 23:50
  • $\begingroup$ It is a good question that I don't have an answer. At least to me that seemed reasonable. But now you have helped me to clear things up, so thank you! $\endgroup$ – James Warthington Mar 1 at 23:58
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Hint. Because you have to use the chain rule (then the product rule). Both $w$ and $l$ are functions of time. So when you differentiate with respect to $t$, you have to apply the chain rule as it is a function of time. Do the same with $l$. Just like taking the derivate of $h(x) g(x)$.

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  • $\begingroup$ No, I mean why it isn't $\dfrac{dl}{t}l+\dfrac{dw}{dt}w$, this way the rate of change of length is paired with length, and the rate of change of width is paired with width. $\endgroup$ – James Warthington Mar 1 at 23:23
  • $\begingroup$ Because when you take the derivate with respect to l, you have to treat w as a constant. And l is a function with respect to t so you can use the chain rule ( I included a picture in my original answer) $\endgroup$ – Sina Babaei Zadeh Mar 1 at 23:34
  • $\begingroup$ Thank you! The Lagrangian notation is a bit clearer for me in this case. $\endgroup$ – James Warthington Mar 1 at 23:40
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    $\begingroup$ The picture shows something completely wrong, and neither the problem nor what you did have to do with the chain rule. We must use the product rule instead: $$A(t) = w(t)l(t) \Rightarrow A'(t) = w'(t)l(t) + w(t)l'(t)$$ $\endgroup$ – Wolfgang Kais Mar 1 at 23:49
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    $\begingroup$ @BorKari … then, the result would be $\frac{dl}{dt} + \frac{dw}{dt} = 3+8$? No, that's wrong. Also, generalizations never help to make things more clear. :-) $\endgroup$ – Wolfgang Kais Mar 2 at 0:01
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First of all, the expression that you got is wrong. It should be this:

$$ \dfrac{dA}{dt}=\dfrac{dl}{dt}w+\dfrac{dw}{dt}l=8\cdot 10+3\cdot20=140\ cm^2/s $$

Secondly, here's how you arrive at that expression: $$ A(t)=l(t)\cdot w(t)\\ A'(t)=\left[l(t)\cdot w(t)\right]'=l'(t)\cdot w(t)+l(t)\cdot w'(t) $$

You simply use the product rule because that's how you differentiate a product of two functions. Each quantity there is a function of time and you differentiate the whole thing with respect to time because you're trying to find the rate of change of area with respect to time.

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$A(t)=l(t) \cdot w(t)$, so the product rule gives $$\frac{dA}{dt} = \frac{dw}{dt}l + \frac{dl}{dt}w = 3 \cdot 20 + 8 \cdot 10 = 60 + 80 = 140$$

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I think you want to know why the terms $l$ and $\frac{dl}{dt}$ are not related and the terms $l$ and $\frac{dw}{dt}$ are related in the formula for area of a rectangle. Let us consider a simple case of a rectangle with length $l$ and width $w$. enter image description here Now let us increase the length by an amount $dl$ and the width by an amount $dw$. So what will be the change in the area? You can see that as soon as you increase both length as well as the width the rectangle changes from $OPRS$ to rectangle $OLMN$. The change in the area is the additional area of the rectangle $RBMC$.

The area of the rectangle initially was $A=lw$.

The area of the rectangle after increasing the length and width becomes $A+dA$.

$$A+dA=(l+dl)(w+dw)$$ $$A+dA=lw+dl(w)+l(dw)+(dl)(dw)$$ $$lw+dA=lw+dl(w)+l(dw)+(dl)(dw)$$ $$dA=dl(w)+l(dw)+(dl)(dw)$$ Now, we assume two things, first this change in length as well as width is very small and tends to $0$. So we can ignore the term, $(dl)(dw)$ in the formula to get: $$dA=dl(w)+l(dw)$$ Now maybe you are getting the pattern of the formula and why the rate $dw$ is related with $l$ instead of $w$. The second assumption is we change the length and width in a very short period of time $dt$ which is very small and tends to $0$. So we can calculate the rate of change of area as : $$\frac{dA}{dt}=w\frac{dl}{dt}+l\frac{dw}{dt}$$ Actually, the changes are denoted by $\triangle$, and then we let this change tend to $0$, so the change $\triangle l \rightarrow dl$. The rest of the calculations are easy and have been told wonderfully by other members.

Hope this helps...

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