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Homework Help: Related Rates / Calculating Rate of Change

  1. May 1, 2012 #1
    1. The problem statement, all variables and given/known data

    A lake is polluted by water from a plant located on its shore. Ecologist determine that when the level of pollutant is x parts per million (ppm), there will be F fish in the lake. When there are 4,000 fish in the lake, the pollution is increasing at a rate of 1.4ppm/year. At what rate is the fish population changing at that time?

    2. Relevant equations

    F = [itex]\frac{32,000}{3+\sqrt{x}}[/itex]

    3. The attempt at a solution

    Ok so if I understand correctly, Im trying to find the change in F over the change in time (one year) or:[itex]\frac{dF}{dt}[/itex]. They give me [itex]\frac{dx}{dt}[/itex] = 1.4. But I dont know how to use that to find [itex]\frac{dF}{dt}[/itex]. Any ideas?
     
  2. jcsd
  3. May 1, 2012 #2

    Mark44

    Staff: Mentor

    Use the chain rule: dF/dt = dF/dx * dx/dt.
     
  4. May 1, 2012 #3
    I tried that, but isnt the derivative of F just equal to 1? I really cant understand this problem. Its on the practice test for our upcoming final and I have never even seen anything like it before. But I really would like to understand it in case something like it shows up on our final though.

    Cause Ive used that form before for like price elasticity when [itex]\frac{dx}{dp}[/itex] * [itex]\frac{p}{x}[/itex]. But i dont understand how it would work here.

    And by the way, this is a related rates problem right? And not an exponential growth model?
     
  5. May 1, 2012 #4

    Mark44

    Staff: Mentor

    No. Show me how you calculated dF/dx.
     
  6. May 1, 2012 #5

    Mark44

    Staff: Mentor

    Yes, this is a related rates problem.

    You are given the relationship between pollution level x and the number of fish F, and need to find the rate at which the fish population is changing, dF/dt.
     
  7. May 1, 2012 #6
    Oh I see, [itex]\frac{df}{dx}[/itex] = [itex]\frac{-3x^{-3/2}}{2}[/itex]? So [itex]\frac{df}{dx}[/itex] * [itex]\frac{dx}{dt}[/itex] = [itex]\frac{-3x^{-3/2}}{2}[/itex] * 1.4?
     
  8. May 1, 2012 #7

    Mark44

    Staff: Mentor

    Just how did you go from
    F = 32000/(3 + √x)

    to dF/dx = (-3/2)x-3/2?

    Also, you are treating dx/dt as if it were a constant (1.4). It isn't. This is the rate of change of pollutant at a particular time, the time when there are 4,000 fish. You will also need to figure out what the value of x is at this same time.
     
  9. May 1, 2012 #8
    I derived it. 32000 & 3 are constants so they cancel out and you are left with 1/[itex]\sqrt{x}[/itex] with is 1/x[itex]^{\frac{1}{2}}[/itex] which is just x[itex]^{\frac{-1}{2}}[/itex] Then you add one to the exponent and multiply it in front of x which gives you (-3/2)x[itex]^{-3/2}[/itex].

    Oh ok, so I rewrite the original equation in terms of x in order to find out how many pollutants are in the lake when there are 4000 fish.

    That gives me (3+[itex]\sqrt{x}[/itex])*(4000) = 32000. Which simplifies to x = 25 ([itex]\sqrt{x}[/itex] = 5) Is that right?
     
  10. May 1, 2012 #9

    Mark44

    Staff: Mentor

    You're kidding, right?

    What you said is so incorrect, it's hard to know what to tell you.

    What differentiation rules have you heard of?

    BTW, you don't "derive" a function - you differentiate it to get the derivative.
    At least you're on the right track, but x = 25 and √x = 5 are both incorrect.

    x does not represent the number of pollutants in the lake - presumably there is one pollutant. x represents the parts per million of that pollutant.
     
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