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Math Help - Speed of skydiver

  1. #1
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    Speed of skydiver

    Hi. I have trouble integrating the following problem.

    The equation is basically m\frac{dv}{dt} = mg - kv^2 with the condition when t = 0, v = 50m/s

    I know this is a separable DE, so rearranaging it becomes

    \int \frac{mdv}{mg-kv^2} = \int dt

    Now I have trouble integrating the left side. Trig substitution doesn't work, so it seems like it'll end up being logs. Any help? Thanks.
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  2. #2
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    Factorise the denominator as a difference of two squares, then apply partial fractions.
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    Forum Admin topsquark's Avatar
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    Or note that
    \displaystyle \int \frac{dx}{1 - x^2} = tanh^{-1}(x) + C

    -Dan
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    I think I'm doing this wrong, because the values I can when I substitute time is really wrong.

    \int \frac{mdv}{mg-bv^2} = \int dt

    \frac {\sqrt m \coth^{-1} (\frac {\sqrt b v}{\sqrt {gm}}}{\sqrt {bg}}= t + c

    When t = 0, v = 52m/s

    c = \frac {\sqrt m \coth^{-1} (\frac {52\sqrt b}{\sqrt {gm}}}{\sqrt {bg}}

    Substituting this back

    \frac {\sqrt m \coth^{-1} (\frac {\sqrt b v}{\sqrt {gm}}-\sqrt m \coth^{-1} (\frac {52\sqrt b}{\sqrt {gm}}}{\sqrt {bg}} = t

    Solving for v

    v = \frac{\sqrt {gm} \coth (\coth^{-1}(\frac {52\sqrt b}{\sqrt {gm}}) + \frac {t\sqrt {bg}}{\sqrt m}) }{\sqrt b}

    Let
    g = 9.8
    m = 140
    b = 110

    When I anything other than t = 0 as the value for t, the value of v gets really close to 3.53, which is the value of \frac {\sqrt {gm}}{\sqrt b}

    What did I do wrong?
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by chengbin View Post
    I think I'm doing this wrong, because the values I can when I substitute time is really wrong.

    \displaystyle \int \frac{mdv}{mg-bv^2} = \int dt

    \displaystyle \frac {\sqrt m \coth^{-1} (\frac {\sqrt b v}{\sqrt {gm}}}{\sqrt {bg}}= t + c
    You can use coth^(-1) instead of tanh^(-1) but note that
    \displaystyle coth^{-1}(x) = tanh^{-1} \left ( \frac{1}{x} \right )

    (By the way, if you put \displaystyle just after the [tex] tag it'll make the LaTeX statements easier to read.)

    The solution to the integral above is
    \displaystyle \int \frac{m~dv}{mg - bv^2} = \sqrt{ \frac{m}{bg}}~tanh^{-1} \left ( v \cdot \sqrt{\frac{b}{mg}} \right )

    In terms of coth^(-1)
    \displaystyle \int \frac{m~dv}{mg - bv^2} = \sqrt{ \frac{m}{bg}}~tanh^{-1} \left ( v \cdot \sqrt{\frac{b}{mg}} \right ) = \sqrt{ \frac{m}{bg}}~coth^{-1} \left ( \frac{1}{v} \cdot \sqrt{\frac{mg}{b}} \right )

    which is why your numbers are off.

    -Dan
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  6. #6
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    Quote Originally Posted by topsquark View Post
    You can use coth^(-1) instead of tanh^(-1) but note that
    \displaystyle coth^{-1}(x) = tanh^{-1} \left ( \frac{1}{x} \right )

    (By the way, if you put \displaystyle just after the [tex] tag it'll make the LaTeX statements easier to read.)

    The solution to the integral above is
    \displaystyle \int \frac{m~dv}{mg - bv^2} = \sqrt{ \frac{m}{bg}}~tanh^{-1} \left ( v \cdot \sqrt{\frac{b}{mg}} \right )

    In terms of coth^(-1)
    \displaystyle \int \frac{m~dv}{mg - bv^2} = \sqrt{ \frac{m}{bg}}~tanh^{-1} \left ( v \cdot \sqrt{\frac{b}{mg}} \right ) = \sqrt{ \frac{m}{bg}}~coth^{-1} \left ( \frac{1}{v} \cdot \sqrt{\frac{mg}{b}} \right )

    which is why your numbers are off.

    -Dan
    Hi. Thanks for the correction, but my numbers are still off.

    This is what I get.

    \displaystyle \sqrt \frac {m}{bg} \coth^{-1} \bigg (\frac {1}{v} \cdot \sqrt \frac {mg}{b}\bigg ) = t + c

    \displaystyle c = \sqrt \frac {m}{bg} \coth^{-1} \bigg (\frac {1}{52} \cdot \sqrt \frac {mg}{b}\bigg )

    \displaystyle t = \sqrt \frac {m}{bg} \bigg (coth^{-1} \frac {1}{v}\cdot \sqrt \frac {mg} {b} \bigg } - \coth^{-1} \bigg (\frac {1}{52} \cdot \sqrt \frac {mg}{b}\bigg ) \bigg )

    \displaystyle v = \frac {\sqrt \frac {mg}{b}}{\coth \bigg ( \coth^{-1}\bigg ( \frac {1} {52} \cdot \sqrt \frac {mg} {b} \bigg ) + \frac {t} {\sqrt \frac {m}{bg}}\bigg )}

    But when I substitute a value for t, the value quickly drops from 52 to < 1 from t = 0 to t = 1.
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  7. #7
    Forum Admin topsquark's Avatar
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    I don't know how to help any further. When I try for a value of the constant I keep betting imaginary numbers. (See here.) As all other numbers in the problem should be arbitrary I have to conclude your value of b = 110 is far too large. I have been trying to find typical values of b to compare, but have so far been unsuccessful. The integral is correct, but I can't get any further than that.

    -Dan

    I'm thinking this over again. Your object is already traveling faster than the terminal speed at t = 0. All problems of this type I have seen start with the object falling from rest. For some reason this is throwing the equation off. I'll look into it and get back to you when (if!) I solve it.
    Last edited by topsquark; March 20th 2011 at 06:09 PM.
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  8. #8
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    Using what I suggested...

    \displaystyle \frac{1}{mg - kv^2} = \frac{1}{(\sqrt{mg})^2 - (\sqrt{k}\,v)^2} = \frac{1}{(\sqrt{mg} - \sqrt{k}\,v)(\sqrt{mg}+\sqrt{k}\,v)}.

    Now applying Partial Fractions

    \displaystyle \frac{A}{\sqrt{mg} - \sqrt{k}\,v} + \frac{B}{\sqrt{mg} + \sqrt{k}\,v} = \frac{1}{(\sqrt{mg} - \sqrt{k}\,v)(\sqrt{mg} + \sqrt{k}\,v)}

    \displaystyle \frac{A(\sqrt{mg} + \sqrt{k}\,v) + B(\sqrt{mg} - \sqrt{k}\,v)}{(\sqrt{mg} - \sqrt{k}\,v)(\sqrt{mg} + \sqrt{k}\,v)} = \frac{1}{(\sqrt{mg} - \sqrt{k}\,v)(\sqrt{mg} + \sqrt{k}\,v)}

    \displaystyle A(\sqrt{mg} + \sqrt{k}\,v) + B(\sqrt{mg} - \sqrt{k}\,v) = 1

    \displaystyle A\sqrt{mg} + A\sqrt{k}\,v + B\sqrt{mg} - B\sqrt{k}\,v = 1

    \displaystyle (A - B)\sqrt{k}\,v + (A + B)\sqrt{mg} = 0v + 1

    \displaystyle (A - B)\sqrt{k} = 0 \implies A - B = 0 \implies A = B

    and \displaystyle (A + B)\sqrt{mg} = 1 \implies A + B = \frac{1}{\sqrt{mg}} \implies 2B = \frac{1}{\sqrt{mg}} \implies B = \frac{1}{2\sqrt{mg}} \implies A = \frac{1}{2\sqrt{mg}}.


    So \displaystyle \frac{1}{mg - kv^2} = \frac{1}{2\sqrt{mg}(\sqrt{mg} - \sqrt{k}\,v)} + \frac{1}{2\sqrt{mg}(\sqrt{mg} + \sqrt{k}\,v)}

    Therefore

    \displaystyle \int{\frac{m}{mg - kv^2}\,dv} = \frac{m}{2\sqrt{mg}}\int{\frac{1}{\sqrt{mg} - \sqrt{k}\,v} + \frac{1}{\sqrt{mg} + \sqrt{k}\,v}\,dv}

    \displaystyle = \frac{m}{2\sqrt{kmg}}}\int{\frac{\sqrt{k}}{\sqrt{m  g} - \sqrt{k}\,v} + \frac{\sqrt{k}}{\sqrt{mg} + \sqrt{k}\,v}\,dv}

    \displaystyle = \frac{m}{2\sqrt{kmg}}\left(-\ln{|\sqrt{mg} - \sqrt{k}\,v|} + \ln{|\sqrt{mg} + \sqrt{k}\,v|}\right) + C

    \displaystyle = \frac{m}{2\sqrt{kmg}}\ln{\left|\frac{\sqrt{mg} + \sqrt{k}\,v}{\sqrt{mg} - \sqrt{k}\,v}\right|} + C.


    So the solution of your DE is

    \displaystyle t = \frac{m}{2\sqrt{kmg}}\ln{\left|\frac{\sqrt{mg} + \sqrt{k}\,v}{\sqrt{mg} - \sqrt{k}\,v}\right|} + C, and since you know that when \displaystyle t = 0, v = 50...

    \displaystyle 0 = \frac{m}{2\sqrt{kmg}}\ln{\left|\frac{\sqrt{mg} + 50\sqrt{k}}{\sqrt{mg} - 50\sqrt{k}}\right|} + C

    \displaystyle C = -\frac{m}{2\sqrt{kmg}}\ln{\left|\frac{\sqrt{mg} + 50\sqrt{k}}{\sqrt{mg} - 50\sqrt{k}}\right|}.


    Therefore \displaystyle t = \frac{m}{2\sqrt{kmg}}\ln{\left|\frac{\sqrt{mg} + \sqrt{k}\,v}{\sqrt{mg} - \sqrt{k}\,v}\right|} - \frac{m}{2\sqrt{kmg}}\ln{\left|\frac{\sqrt{mg} + 50\sqrt{k}}{\sqrt{mg} - 50\sqrt{k}}\right|}

    \displaystyle t = \frac{m}{2\sqrt{kmg}}\ln{\left|\frac{(\sqrt{mg} - 50\sqrt{k})(\sqrt{mg} + \sqrt{k}\,v)}{(\sqrt{mg} + 50\sqrt{k})(\sqrt{mg} - \sqrt{k}\,v)}\right|}
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  9. #9
    Forum Admin topsquark's Avatar
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    I guess the different forms can be useful! Thank you for saving me hours of work on this.

    (sighs) Now I'm going to have to figure out why one solution worked and the other didn't.

    I'm going to bed.

    -Dan
    Last edited by topsquark; March 20th 2011 at 07:08 PM.
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  10. #10
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    Thanks for all your hard work.

    @topsquark

    This is not a problem from a textbook. I'm trying to model someone falling, but has reached terminal velocity already, hence when t = 0, v = 50. I wanna know his speed 2 seconds after he releases his parachute at t = 0.

    The value for k is from the drag equation, \frac {1}{2}\rho v^2C_dA

    p is 1.22 (density of air)
    Cd = 2 (large drag for a parachute)
    A = 100m^2 (area of the parachute)

    The fact you experts are having trouble with this is making me doubt if I'm doing this question correctly.
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  11. #11
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by chengbin View Post
    Thanks for all your hard work.

    @topsquark

    This is not a problem from a textbook. I'm trying to model someone falling, but has reached terminal velocity already, hence when t = 0, v = 50. I wanna know his speed 2 seconds after he releases his parachute at t = 0.

    The value for k is from the drag equation, \frac {1}{2}\rho v^2C_dA

    p is 1.22 (density of air)
    Cd = 2 (large drag for a parachute)
    A = 100m^2 (area of the parachute)

    The fact you experts are having trouble with this is making me doubt if I'm doing this question correctly.
    I see now. Yes a parachute would have a rather large value of b. Solve Prove It's solution for v and plug in 2 s for t. You will find that the speed is drastically reduced.

    -Dan
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  13. #13
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    I don't see what's so strange about it...
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  14. #14
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    Quote Originally Posted by Prove It View Post
    I don't see what's so strange about it...
    It drops from 50 to 3.53 in an instant. I don't think a parachute slows you down from 50m/s to 3.53m/s in an instant. That's why there is a relatively high minimum altitude you need to deploy your parachute while skydiving.
    Last edited by chengbin; March 21st 2011 at 05:19 PM.
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  15. #15
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by chengbin View Post
    It drops from 50 to 3.53 in an instant. I don't think a parachute slows you down from 50m/s to 3.53m/s in an instant.
    Ever used a parachute? Take a look at that b value. At a first glance the numbers look about right to me as well.

    -Dan
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