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Math Help - particle motion with calculus

  1. #1
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    particle motion with calculus

    A particle moves on the x-axis in such a way that its position at time t, t>0, is given by x(t) = (ln x)^2. At what value of t does the velocity of the particle attain maximum?

    Suppose a population of bears grows according to the logistic differential equation

    dP/dt = 2P - 0.01P^2

    where P is the number of bears at time t in years. Which of the following statements are true?

    I. The growth rate of the bear population is greatest at 100
    II. If P > 200, the population of bears is decreasing.
    III. lim t -> inf P(t) = 200

    A conical tank is being filled with water at the rate of 16 ft^3/min. The rate of change of the depth of the water is 4 times the rate of change of the radius of the water surface. At the moment when the depth is 8ft and the radius of the surface is 2ft, the area of the surface is changing at what rate?


    Out of 28 different questions, I couldn't figure out how to do these.

    Thanks for your help
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by LeoBloom. View Post
    A particle moves on the x-axis in such a way that its position at time t, t>0, is given by x(t) = (ln x)^2. At what value of t does the velocity of the particle attain maximum?
    Is this supposed to be x(t) = \left ( ln(t) \right ) ^2? This doesn't make sense otherwise...

    x = \left ( ln(t) \right ) ^2

    v = \frac{dx}{dt} = 2 ln(t) \cdot \frac{1}{t}

    For v to be a maximum we require that the first derivative of v is 0:
    \frac{dv}{dt} = 0 = \frac{2}{t^2} - \frac{2~ln(t)}{t^2}

    I get t = e as the only critical point. I'll leave it to you to prove that this is an absolute maximum.

    -Dan
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    Quote Originally Posted by topsquark View Post
    For v to be a maximum we require that the first derivative of v is 0

    Why is this the case? That would be when acceleration is 0 ... hmm ... I think I just answered my own question. Thanks alot Dan.
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  4. #4
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    For the conical tank:

    A handy thing to know is that the rate of change of the volume equals the cross-sectional area at that instant times the rate of change of the height.

    In other words, \frac{dV}{dt}=A(t)\frac{dh}{dt}...[1]

    A(t)=\frac{\frac{dV}{dt}}{\frac{dh}{dt}}...[2]

    \frac{dh}{dt}=\frac{\frac{dV}{dt}}{A(t)}...[3]

    The area of the surface when r=2 would be 4{\pi}

    From [3], 4{\pi}=\frac{16}{\frac{dh}{dt}}

    \frac{dh}{dt}=\frac{4}{\pi}

    But remember, \frac{dh}{dt}=4\frac{dr}{dt}.

    So, \frac{dr}{dt}=\frac{1}{\pi}

    Now, can you finish?.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by LeoBloom. View Post
    Suppose a population of bears grows according to the logistic differential equation

    dP/dt = 2P - 0.01P^2

    where P is the number of bears at time t in years. Which of the following statements are true?

    I. The growth rate of the bear population is greatest at 100
    II. If P > 200, the population of bears is decreasing.
    III. lim t -> inf P(t) = 200
    Well, let's take this case by case.

    I. The growth rate is at its greatest when
    \frac{d^2P}{dt^2} = 0

    So
    \frac{d^2P}{dt^2} = \left ( 2 - 0.02P \right ) \frac{dP}{dt} = \left ( 2 - 0.02P \right ) (2P - 0.01P^2) = 0

    I get values of P = 0, 100, 200. The question is, which gives the greatest growth rate? Plugging the numbers in I get P = 100. So this one is true.

    II. When P > 200
    \frac{dP}{dt} < 0
    so this one is true.

    III. If there is a limit for P at very large times then we know that
    \frac{dP}{dt} = 0

    This corresponds to a P equal to P = 0 or P = 200. If P = 0 at t = 0 then P = 0 always. That is the only way I can get the P = 0 solution to work at all. Thus the answer has to be P = 200. So III is correct also.

    -Dan
    Last edited by topsquark; May 21st 2008 at 04:10 AM.
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    top, thanks so much for your help, I understand it now. But I. is true as well since the d2p/dt2 = 2 - 0.02p, 0 = 2 - 0.02p making p = 100 and it increases pre 100 and decrease after 100, so it is greatest at P = 100. I don't understand why you said I. isnt true, mistake maybe?
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by LeoBloom. View Post
    top, thanks so much for your help, I understand it now. But I. is true as well since the d2p/dt2 = 2 - 0.02p, 0 = 2 - 0.02p making p = 100 and it increases pre 100 and decrease after 100, so it is greatest at P = 100. I don't understand why you said I. isnt true, mistake maybe?
    My apologies, I was thinking the first one said at P = 200 like the second one. So yes, I. is true also.

    -Dan
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