How fast is the area of rectangle increasing?












3












$begingroup$


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?










share|cite|improve this question











$endgroup$












  • $begingroup$
    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








  • 3




    $begingroup$
    $l$ 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 $$frac{dA}{dt} = wfrac{dl}{dt} + lfrac{dw}{dt}$$.
    $endgroup$
    – 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
















3












$begingroup$


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?










share|cite|improve this question











$endgroup$












  • $begingroup$
    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








  • 3




    $begingroup$
    $l$ 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 $$frac{dA}{dt} = wfrac{dl}{dt} + lfrac{dw}{dt}$$.
    $endgroup$
    – 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














3












3








3


0



$begingroup$


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?










share|cite|improve this question











$endgroup$




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?







calculus






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 17 at 20:57









user144410

1,0412719




1,0412719










asked Jan 17 at 20:40









DisintegratorsDisintegrators

161




161












  • $begingroup$
    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








  • 3




    $begingroup$
    $l$ 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 $$frac{dA}{dt} = wfrac{dl}{dt} + lfrac{dw}{dt}$$.
    $endgroup$
    – 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


















  • $begingroup$
    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








  • 3




    $begingroup$
    $l$ 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 $$frac{dA}{dt} = wfrac{dl}{dt} + lfrac{dw}{dt}$$.
    $endgroup$
    – 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
















$begingroup$
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






$begingroup$
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






3




3




$begingroup$
$l$ 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






$begingroup$
$l$ 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




$begingroup$
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 $$frac{dA}{dt} = wfrac{dl}{dt} + lfrac{dw}{dt}$$.
$endgroup$
– Matteo
Jan 17 at 21:10






$begingroup$
@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 $$frac{dA}{dt} = wfrac{dl}{dt} + lfrac{dw}{dt}$$.
$endgroup$
– 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




$begingroup$
@Yanko after one second the area is $(l+8)(w+3)$. Not $(8l)(3w)$.
$endgroup$
– fleablood
Jan 17 at 21:22










3 Answers
3






active

oldest

votes


















1












$begingroup$

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



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



With one microsecond, we get



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



This confirms the answer $140$.



The reason why your method doesn't work is because



$$frac{20}{10}nefrac{8}{3}.$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    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 $w (dl)$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      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



















    0












    $begingroup$

    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 $wne 10$. (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.)






    share|cite|improve this answer









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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      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



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



      With one microsecond, we get



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



      This confirms the answer $140$.



      The reason why your method doesn't work is because



      $$frac{20}{10}nefrac{8}{3}.$$






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        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



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



        With one microsecond, we get



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



        This confirms the answer $140$.



        The reason why your method doesn't work is because



        $$frac{20}{10}nefrac{8}{3}.$$






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          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



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



          With one microsecond, we get



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



          This confirms the answer $140$.



          The reason why your method doesn't work is because



          $$frac{20}{10}nefrac{8}{3}.$$






          share|cite|improve this answer









          $endgroup$



          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



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



          With one microsecond, we get



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



          This confirms the answer $140$.



          The reason why your method doesn't work is because



          $$frac{20}{10}nefrac{8}{3}.$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 17 at 22:09









          Yves DaoustYves Daoust

          133k676232




          133k676232























              0












              $begingroup$

              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 $w (dl)$






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                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
















              0












              $begingroup$

              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 $w (dl)$






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                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














              0












              0








              0





              $begingroup$

              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 $w (dl)$






              share|cite|improve this answer











              $endgroup$



              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 $w (dl)$







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Jan 17 at 21:21

























              answered Jan 17 at 21:16









              Doug MDoug M

              45.4k31954




              45.4k31954












              • $begingroup$
                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


















              • $begingroup$
                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
















              $begingroup$
              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




              $begingroup$
              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




              $begingroup$
              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




              $begingroup$
              @Disintegrators I thought I had. Looks fine on my screen.
              $endgroup$
              – Doug M
              Jan 17 at 22:35











              0












              $begingroup$

              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 $wne 10$. (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.)






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                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 $wne 10$. (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.)






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  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 $wne 10$. (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.)






                  share|cite|improve this answer









                  $endgroup$



                  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 $wne 10$. (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.)







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 17 at 21:57









                  fleabloodfleablood

                  1




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