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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?

So on internet I found a solution but I didn't do that way and I am still thinking that I am not wrong but the answer is not the same. I am gonna write both the solutions which I found on int and by myself and I will be waiting your help.

Which I found on the int: $A=lw$ then take derivative $\frac{dA}{dt}= \frac{dl}{dt}.w + l.\frac{dw}{dt}$

using given number $\frac{dA}{dt}= (8)(10) + (20)(3)$

My answer: given numbers--> $\frac{dl}{dt}= 8$, $\frac{dw}{dt}=3$, $l=20$, $w =10$

so $A=wl$ when I wanna write $w$ in terms of $l$ ----> $l=2w$

so $A=2w*w$ when I take derivative of it ---> $\frac{dA}{dw}= 4w $

according to chain rule $\frac{dA}{dt}= \frac{dA}{dw}\frac{dw}{dt}$

when I put the numbers ----> $4w*3$ and we know that $w=10$

It should be 120. I think I found my mistake but still couldn't understand why. I write $w$ in terms of l but if I do the other way then the result is 160. What am I doing wrong?

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  • $\begingroup$ Why isn't it $8*3 = 24$. If the area is $lw$ then the area after a second will be $(8l)(3w) = 24wl$ and so on. am I missing something? $\endgroup$ – Yanko Jan 17 at 20:59
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    $\begingroup$ $l$ is twice $w$ only at that particular time and it's not a general functional relationship between these two values. $\endgroup$ – Matteo Jan 17 at 21:01
  • $\begingroup$ Your expression for $\frac{dA}{dt}$ is correct. It should be $140$. $\endgroup$ – John Douma Jan 17 at 21:03
  • $\begingroup$ @Yanko, $8$ and $3$ are rates of increase of each size of the rectangle. So your expression is not correct. The rate of increase of the area is correctly $$\frac{dA}{dt} = w\frac{dl}{dt} + l\frac{dw}{dt}$$. $\endgroup$ – Matteo Jan 17 at 21:10
  • $\begingroup$ @Yanko after one second the area is $(l+8)(w+3)$. Not $(8l)(3w)$. $\endgroup$ – fleablood Jan 17 at 21:22
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One millisecond later, the sides are $20.008$ and $10.003$ and the relation $l=2w$ is no more true.

The rate of increase of the area must be close to

$$\frac{20.008\cdot10.003-20\cdot10}{10^{-3}}=140.024.$$

With one microsecond, we get

$$\frac{20.000008\cdot10.000003-20\cdot10}{10^{-6}}=140.000024.$$

This confirms the answer $140$.

The reason why your method doesn't work is because

$$\frac{20}{10}\ne\frac{8}{3}.$$

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While it is true that at this moment in time $A = 2w^2 = 200$
The length and width are not changing uniformly.

If they were then it would be correct to say $A = 4w \frac {dw}{dt}$

But as they are changing at different rates, you need to use the chain rule.

Perhaps a visualization will help.

enter image description here

We have the rectangle at time $t$ and at time $t+1$ and and the red and green rectangles are the approximate change at some intermediate time.

the green areas sum to $l (dw)$ and red areas $w (dl)$

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  • $\begingroup$ thank you! I think I understand now but I can't open the link, I think it's kind of bug. Do you mind uploading the picture another website. $\endgroup$ – Disintegrators Jan 17 at 21:55
  • $\begingroup$ More precisely, the rates shouldn't be equal but proportional to the respective sides. $\endgroup$ – Yves Daoust Jan 17 at 22:02
  • $\begingroup$ @Disintegrators I thought I had. Looks fine on my screen. $\endgroup$ – Doug M Jan 17 at 22:35
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You need to think of $l$ and $w$ as functions of time.

$l(t) = L + 8t$ where $L$ is the initial length. And $w(t) = W+3t$ and, yes, $\frac {dl}{dt}=8$ and $\frac {dw}{dt} = 3$ but $l \ne 20$ and $w\ne 10$. (That'd mean they are constant functions. They aren't.) $l(t_0) = 20$ and $w(t_0) = 10$ at time $t_0$.

To make things simple we can let $t_0=0$ and $l(t) = 20 + 8t$ and $w(t) = 10 + 3t$ and $l(0) =20$ and $w(0) = 10$.

Now we just can't say $l(t) = 2w(t)$ because that just isn't true. You could say $ {l(t)} = \frac {20+8t}{10+3t}{w(t)}$ and $A(t) = \frac {20+8t}{10+3t}{w(t)}^2$ but... that just makes things complicated. (If I had more time and energy I'd be curious to try to figure that and see if it is the same but.... I don't.)

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