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Increasing area

  1. Apr 9, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the rate of change of the area of a square whose side is 8 cm long, if the side length is increasing at 2cm/min


    2. Relevant equations
    A(t) = xy
    x(0)=8
    y(0)=8
    dx/dt=2
    dy/dt=2


    3. The attempt at a solution
    dA/dt = dx/dt * x * dy/dt * y
    Using the product rule...
    dA/dt = dx/dt * x * y + dy/dt * y * x
    What happens now? Does the dx/dt * x = 1?

    or is it..
    dA/dt = dx/dt * d/dx * x * dy/dt * d/dy * y
    Using the product rule..
    dA/dt = dx/dt * d/dx * x * y + dy/dt * d/dy * y * x
    Therefore the d/dx = 1, therefore..
    dA/dt = dx/dt * y + dy/dt * x
    dA/dt = 2*8 + 2*8
    dA/dt = 32 cm^2/min ?? If it is correct is what I'm thinking correct (ie my working?)


    Help, thanks!
     
    Last edited: Apr 9, 2012
  2. jcsd
  3. Apr 9, 2012 #2
    I got 32t + 4t^2 - I did the d(xy)/dt = dx/dy*y + dy/dt*x thing like you did, which gave me 32t.

    I wrote out the values for the next few minutes and tacked on the 4t^2. I wouldn't know how to do it the proper way, been awhile.

    So for the first minute the increase is 32 + 4 = 36, which is for a square of 10 by 10 (36 + initial 64)

    Then for the next minute it would be 64 + 16 = 80, (80 + initial 64 = 144; 12 by 12)

    Edit: Also, it's area so it's cm^2/min - so yea it would be 32cm^2/min
     
    Last edited: Apr 9, 2012
  4. Apr 9, 2012 #3
    Huh? Wouldn't it work a simpler way:

    [itex]A(t)=\left(8+2t\right)^2[/itex]
    [itex]A'(t)=2\left(8+2t\right)\cdot2[/itex]
    [itex]A'(t)=32+8t[/itex]?

    here t is in minutes.
     
    Last edited: Apr 9, 2012
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