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Related Rates triangle area

  1. Jan 16, 2007 #1
    1. A rectangle has a constant area of 200 square meters and its length L is increasing at the rate of 4 meters per second. Find the width W at the instant the width is decreasing at the rate of 0.5 meters per second.

    2. Know: dL/dt= 4 m/s, dW/dt=.5 m/s Want: W

    3. LW=A. Then DL/dt * DW/dt = 200 m^2. How do you get W from this?

    I have some other questions too. Related rates are hard.
  2. jcsd
  3. Jan 16, 2007 #2


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    dL/dt*dw/dt= 200 can't possibly be correct, can it? 4*0.5 is not 200! Recheck your differentiation- particularly the "product rule"! Also, it is the area that is the constant 200, not the rate of change. What is the derivative of a constant?
  4. Jan 16, 2007 #3
    Never mind, I totally forgot about the product rule.
  5. Jan 16, 2007 #4
    Okay, I got stuck again.
    DL/dt*W+ L*DW/dt = 0.
    4*W + L* -.5 = 0. How do you find L? Once you find L, isn't it obvious what W will be? Then you wouldn't have to go through this whole rate problem.
  6. Jan 16, 2007 #5
    use the two equations

    solve the original area equation and differential equation simultaneously

    That is,

    A = wl
    0 = 4w - .5l

    solve the above for w.
  7. Jan 22, 2007 #6
    Kay, thanks.
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