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Partial Drivative- Inclined Plane Problem

  1. Oct 8, 2008 #1
    1. The problem statement, all variables and given/known data
    A ping pong ball is projected (with velocity v)from a foot of an inclined plane with an inclination of [tex]\beta[/tex] . The initial velocity of projection makes an angle [tex]\alpha[/tex] with the surface of th incline. Find the relation beteen [tex]\alpha[/tex] and [tex]\beta[/tex] such that the range is maximum.

    3. The attempt at a solution
    Time of flight=[tex]\frac{2v sin \alpha}{g cos \beta}[/tex]

    Range(R)=[tex]\frac{2 v^{2} sin \alpha cos \alpha}{g cos \beta}-\frac{4 tan \beta v^{2} sin ^{2} \alpha}{g cos \beta}[/tex]

    Now the range is a function of both [tex]\alpha[/tex] and [tex]\beta[/tex]. I have been searching how to do the derivative. I learned from mathworld.wolfram that I need to consider one quantity constant and differentiate with respect to the other. and then do the same by keeping the other constant. Then by adding these partial derivatives I can get the complete derivative of the function concerned (which in my case is the range of the projectile).

    [tex]\frac{\delta R}{d\alpha}=\frac{2 v^{2} cos 2\alpha}{g cos \beta}-\frac{4 v^{2} tan \beta sin 2\alpha}{g cos \beta}[/tex]


    [tex]\frac{\delta R}{d\beta}=\frac{v^{2} sin 2\alpha tan \beta}{g cos \beta}-\frac{4 v^{2} sin^{2} \alpha(2tan^{2}\beta+1)}{g cos \beta}[/tex]

    Adding these two and putting them equal to zero does not yield the desired result. I am stuck. Please help me with this.
     
  2. jcsd
  3. Oct 9, 2008 #2
    One method is to resolve the velocity in the plane parallel to the incline slope and perpendicular to the inclined slope.

    wait while i try it out
     
  4. Oct 9, 2008 #3
    I done just that. I have resolved it along the surface of incline and perpendicular to it.

    Yup! take your time!! :approve:
     
  5. Oct 10, 2008 #4
    x = ucos[tex]\beta[/tex]t - 0.5gsin[tex]\alpha[/tex]t[tex]^{2}[/tex] ----1
    y = usin[tex]\beta[/tex]t - 0.5gcos[tex]\alpha[/tex]t[tex]^{2}[/tex] ---- 2
    At the point of impact, y = 0
    Hence, t = (2usin[tex]\beta[/tex]) / (gcos[tex]\alpha[/tex]) --- 3

    put 3 into 1
    and differentiate with respect to [tex]\beta[/tex]

    you shld end up with

    tan 2[tex]\beta[/tex] = cot[tex]\alpha[/tex]

    thus, [tex]\beta[/tex] = pi/4 - [tex]\alpha[/tex]/2

    EDIT: switch all the alpha with beta and beta with alpha
     
  6. Oct 10, 2008 #5

    gabbagabbahey

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    You have an error in your Range; you should have

    [tex]R=\frac{2 v^{2} sin \alpha cos \alpha}{g cos \beta}+\frac{2 tan \beta v^{2} sin ^{2} \alpha}{g cos \beta} [/tex]

    It looks like you forgot to multiply your at^2 term by (1/2), and it should be +(1/2)at^2, not minus. (Assuming your g is -9.8m/s^2)

    The total derivative of R is given by:

    [tex]dR=\frac{\partial R}{\partial \alpha}d \alpha +\frac{\partial R}{\partial \beta}d \beta [/tex]

    At a local max/min the total derivative will be zero, so each partial derivative must also vanish. Compute the partial derivatives and set them equal to zero.
     
  7. Oct 30, 2008 #6
    Can you please explain how you got that relation?
     
  8. Nov 4, 2008 #7

    gabbagabbahey

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    Hmmm...did you understand my last post? Is there anything in there that you don't understand?
     
  9. Nov 4, 2008 #8
    To an extent...
    You were right that i forgot to multiply that by 1/2 but there would be a minus sign there.

    I didn't understand how you got the total derivative?
    And then is it necessary that if z=x+y and z=0, then how come can we say that x=0, and y=0 are the only solutions???

    And I am facing problems in deriving the required relation between alpha and beta!
     
  10. Nov 4, 2008 #9

    gabbagabbahey

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    The total derivative of a function of two variables, say [itex]f(x,y)[/itex] is by definition:

    [tex]df=\left| \frac{\partial f}{\partial x} \right| d x +\left| \frac{\partial f}{\partial y} \right| d y[/tex]

    Since [itex]x[/itex] and [itex]y[/itex] are independent variables, so are there total derivatives [itex]dx[/itex] and [itex]dy[/itex]; and so the only way that [itex]df[/itex] can be zero is if each of the partial derivatives [tex]\frac{ \partial f}{\partial x}[/tex] and [tex]\frac{ \partial f}{\partial y}[/tex] are zero.

    Do you follow this?
     
  11. Nov 4, 2008 #10
    Yup! I do it now. As x is independent of y at all times, therefore we cannot force it to become a function of y to get the minimum value. right?
    just like saying if
    x i+ y j=0 i+ 0 j
    in vector notation, then x=0 and y=0. right >
     
  12. Nov 4, 2008 #11

    gabbagabbahey

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    More or less, yes.

    So in this case your function is [itex]R(\alpha,\beta)[/itex] and so [tex]\frac{\partial R}{\partial \alpha}[/tex] and [tex]\frac{\partial R}{\partial \beta}[/tex] must both be zero at the local max/min...what do you get when you compute these partial derivatives....where are they zero?
     
  13. Nov 4, 2008 #12
    SOLVED

    Thanks a lot!! I am very grateful to you. I have got my result.
     
  14. Nov 4, 2008 #13

    gabbagabbahey

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    Congrats! :smile:
     
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