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Homework Help: Differential of a 1st ODE help understanding notation

  1. Apr 12, 2012 #1
    1. The problem statement, all variables and given/known data

    The problem regards a ball thrown vertically, there is a model of the motion that we worked out, from the original equation

    a(t) = -(g/b^2)(v^2+b^2)

    With some help from another forum member I integrated with regard to t (dv/dt?) this to


    Where v is velocity, v_0 is initial velocity and b and g are constants

    The question then states - By writing a = v dv/dx, solve the resulting differential equation

    2. Relevant equations

    I think this is somehow related to the chain rule as v dv/dx is equal to (dv/dt)(dt/dx)

    3. The attempt at a solution

    The first integration is velocity with respect to time. I know that v dv/dx is a standard result found with the chain rule, in this case (dv/dt)(dt/dx).

    The problem I am having is where x comes into this, there is no mention of distance in any of the formulae I am using. How do I apply this to the problem?
  2. jcsd
  3. Apr 12, 2012 #2


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    Assuming that x is the distance, from some starting point, that an object is after time t. The v= dx/dt and, of course, a= dv/dt.

    From the chain rule, we have a= dv/dt= (dv/dx)(dx/dt). Now, that would be true for x any function of t but since v= dx/dt, that becomes dv/dt= v(dv/dx). Motion in which the force (and so acceleration) is independent of time are very common in physics (essentially due to the idea that "if I do an experiment this morning rather than this afternoon, I should get the same result") so that technique (called "quadrature" since the integral of v dv is [itex](1/2)v^2[/itex]) allows us to remove time from the problem completely.
  4. Apr 13, 2012 #3
    Thanks for the explanation, but I have to say I dont really follow your point. If I throw a ball and am interested in its position, then isnt knowing the time essential? or at least relatively speaking. The ball would be the same height above the ground x seconds after you threw it, I agree that I would expect gravity to be the same in the morning and afternoon :)

    I also dont follow how the chain rule applies in this case, because even before we started there was no mention of the time variable, and now it appears and disappears again in one line...
  5. Apr 14, 2012 #4
    I was chewing this over, am I correct in thinking this - because we have an expression for velocity and acceleration, and with the expressions we know exactly what the ball is going to do, and because of this then the time is not needed. I am having trouble with your response, no doubt the info is all there, I just have to figure it out! :) Thanks for your help anyway!
  6. Apr 14, 2012 #5
    Here is my attempt at a solution, can someone please check it

    From earlier in the question


    Substitute this in

    v(dv/dx) = -(g/b^2)(b^2+v^2)

    Separate terms

    dv/(v.g) = -dx

    Integrate and get

    x = -1/g ln v
  7. Apr 14, 2012 #6


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    Hi Mark! :smile:

    (try using the X2 button just above the Reply box :wink:)

    If you have a = f(v),

    you can either write dv/dt = f(v) and integrate to get v as a function of t

    or write vdv/dx = f(v) and integrate to get v as a function of x

    it just depends which result you want :smile:
    noooo … how did you get that? :confused:
  8. Apr 14, 2012 #7
    Hi Tim!

    Thanks for the reply, its been a while! :)

    The question asks for the highest point. We have an expression for acceleration

    a = -(g/b2)(v2+b2)

    where g = acc. due to gravity, m = mass of ball. D=diameter of ball b = mg/0.2D2

    I assume my approach up to the point you went nooooooo is ok :)

    I got that by trying to isolate the terms

    v(dv/dx) = -(g/b2)(v2+b2)

    v(dv/dx)/(v2+b2) = -(g/b2)

    (dv/dx)/(v+b2) = -(g/b2)

    -dv/(vg) = dx

    Then I integrated to get x in terms of v

    I assume that we want an expression that when you enter v=0, you get the value of the highest point. For the rest I am pretty much scratching my head, as my answer contains ln(0), so the ball would go pretty high :)
  9. Apr 14, 2012 #8


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    (did you divide that v2 on the bottom by the v on the top? you can't do that!! :redface:)

    no, v(dv)/(v2+b2) = -(g/b2)dx …

    ∫ v/(v2+b2) dv = -∫ (g/b2) dx :smile:
  10. Apr 14, 2012 #9
    Oops, yea, that v/v2 should have been 1/v :redface:

    So the equation as it stands so far is

    v(dv)/(v2+b2) = -(g/b2)dx (which i had in one of my first attempts :biggrin: )

    Integrating this

    1/2 ln (v2 + b2) = -gx/(b2

    A few steps later I get the expression for x to be

    x = -b2((ln(v2)+ln(b2))/2g)

    But then I have again the same problem, that my expression contains ln(0). Is my assumption correct, that the maximum height of the ball is found where velocity = 0? Because if I have a 1/v anywhere in my expression, when I integrate it, I will be left with ln(0) again.
  11. Apr 14, 2012 #10


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    yes (and you must also add a constant) :smile:
    your arithmetic is seriously bad :redface:

    you can't go from ln(v2 + b2) to ln(v2)+ln(b2)
  12. Apr 14, 2012 #11
    I thought this was an ok move to be honest. Is this just for logs I cant do this?

    3(5 + 7) could be 3 x 12 = 36 or 15 + 21 which also equals 36

    So this far is correct although I cant remember how I did it and now I wonder where the v on the LHS goes to -

    integrate v(dv/dx)

    1/2 ln (v2 + b2) = -gx/(b2)

    I want x on its own so multiply both sides by b2 and divide by -g?
  13. Apr 14, 2012 #12


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    no, for everything (except straight multiplication) you can't do it!!
    do exp to both sides (using eln(A) = A) :smile:

    (but add on the constant first)
  14. Apr 15, 2012 #13
    I redid the integration and got an expression for x without taking the exp. It is still incorrect, can you please explain why I would take the exp of both sides. I am wondering if I should solve this using the particular integral method instead.

    (1/2)ln(b2+v2)+C1 = -(g/b2)+C2

    Rearranging for x

    x = -((b2/2)ln(b2+v2)+C3)/g

    The ball is launched upwards at 20 ms-1, the other variables are given, and with this expression, the balls height is found to be 380 metres, which is clearly wrong, although an improvement over the infinity + 18 I got yesterday.
  15. Apr 15, 2012 #14


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    Hi Kawakaze! :smile:
    But wouldn't you prefer to write v as a function of x, rather than x as a function of v ? :wink:
    i] you only need one constant, so you could have gone straight to C3 (on either side of the equation)

    ii] write A = (1/2)ln(C3), then that becomes …
    (1/2){ln(b2+v2)+ln(A)] = -(gx/b2)

    ie (1/2)ln(A(b2+v2)) = -(gx/b2) :smile:

    (and then do exp for both sides if you want v as a function of x)
  16. Apr 15, 2012 #15
    Good Evening, Tim!

    Before I begin, thanks! Secondly, what is the significance of the constant? If I do not know it, then I have a big problem, I thought maybe this would be the initial height but if it isnt, then how can I calculate this?
  17. Apr 15, 2012 #16


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    you need to do the "initial" of both sides of the equation …

    (1/2)ln{(b2+v2)/(b2+vo2)} = -(g(x-xo)/b2)
  18. Apr 16, 2012 #17
    Hi Tim,

    I got the equation to the following.

    x = -((b2/2)ln(b2+v2)+C)/g

    The initial values, are b2 = 1090 (b=root(mg/0.2D2)), v = 20, g = 9.81

    When I re-arrange for C I get

    C = xg/((b2/2)ln(b2+v2)) = 0.0036m

    The two things that stick out here, 1) the initial value when taking x as 0, gives no useable answer, the ball is thrown from 1.5m off the ground. 2) When I substitute this value for C in, I still get a crazy answer. This is weird as my tutor also told me these integrals are correct. Maybe its the fact i did 10hrs at work today and its now 00.15am
  19. Apr 16, 2012 #18


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    Hi Mark! :smile:
    I don't understand what you've done here. :confused:

    The initial condition is that v = 20 when x = 0.

    So when x = 0, the ln must be 0, so the thing inside the ln must be 1

    ie it must be (b2+v2)/(b2+vo2).
  20. Apr 16, 2012 #19
    I dont understand what you mean. If x=0 then the top half of the equation is zero. So the only solution that could work is if the bottom half equals zero. But this is not correct, because if I put C=0 into the next part of the question where we find its max height, I get a crazy answer.

    This is frustrating, can you please at least confirm the algebra is not where the fault is? Thanks!
  21. Apr 16, 2012 #20


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    What top half? :confused:

    I only see a left and a right …
    the left is 0, so the right must be 0, the right is a factor times a ln, so the inside of the ln must be 1
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