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Optimization homework problem help

  1. Aug 7, 2006 #1
    two vertical poles PQ and ST are secured by a rope PRS going from the top of the first pole to a Point R on the ground to the top of the second polle. Show that the shortest length of rope occurs when [tex]\theta_1=\theta_2[/tex]

    I have found eqn and found the derivative, what do I do to show that it is a min?
     
  2. jcsd
  3. Aug 7, 2006 #2

    berkeman

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    Staff: Mentor

    Post a copy of your equation and the derivative. Your equation should show the length of the rope as a function of theta_1 (or theta_2 if you prefer, or whatever). For there to be a minimum length, the length equation will show that the length L starts at some value, then decreases some up to some point, and then increases from there to the end. If you think about drawing the function L as a funtion of theta (put L on the vertical and theta on the horizontal), the function L(theta) will look like a "U", right? When you differentiate L(theta) with respect to theta, what is special about the low point in the L(theta) curve? What can you say about the sign of the derivative for the first part of the U versus the end of the U? What must the derivative do at the bottom of the U as the sign of the derivative is changing from negative to positive?
     
  4. Aug 7, 2006 #3
    Have you already found the critical points? The points where that make the first derivative zero or undefined. Then to show that there is a minimum at a critical point you can either evaluate the second derivative at that point and if it is positive then there is a minimum, or look at the value of the derivative on both sides of the critical point and if it is negative for values smaller than the the value of the critical point and positive for values larger then you have a minimum.
     
  5. Aug 7, 2006 #4
    [tex]L_{tot}=L_1+L_2[/tex]
    [tex]L=QRT, L_1=PR, L_2=RS, x=QR, \theta_1=PRQ, \theta_2=SRT[/tex]
    [tex]x=L_1cos(\theta_1)[/tex]
    [tex]L-x=L_2cos(\theta_2)[/tex]



    so...[tex]\sqrt{(PQ)^2+x^2}=L_1[/tex]

    [tex]\sqrt{(ST)^2+(RT)^2}=L_2[/tex]

    then find [tex]L_{tot}[/tex] and find the derivative..i find this even more confusing now
     
    Last edited: Aug 7, 2006
  6. Aug 7, 2006 #5

    berkeman

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    Sorry, I got lost there. I'd be inclined to keep L in terms of theta, as the problem is asking you to prove something about theta_1 and theta_2, so you'll have to end up there anyway. Can you maybe re-write the equations starting out something like:

    [tex] Let L = QRT, l_1 = PR, l_2 = RS, x = QR, \theta_1 = PRQ, \theta_2 = SRT [/tex]

    [tex]x = l_1 cos(\theta_1) [/tex]

    [tex]L - x = l_2 cos(\theta_2) [/tex]

    [tex]l_total = l_1 + l_2 = ..... [/tex]
     
  7. Aug 7, 2006 #6
    i edited my above, i find this more confusing now though
     
  8. Aug 7, 2006 #7

    berkeman

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    Well, maybe it -- sorry if it is. But go ahead and solve it anyway. Change RT --> L-x, do the addition and differentiation, set to zero and solve for x. Hopefully L and PQ and ST will fall out of the equations....
     
  9. Aug 7, 2006 #8
    so my derivative is:

    [tex]l_{tot}= \frac{x}{\sqrt{P^2Q^2+x^2}}+\frac{-1+x}{\sqrt{S^2T^2+L^2-2Lx+x^2}}[/tex]..now i just solve for x?

    yyyyyyeah..theres a ton of letters, I dont think they are all gonna cancel out.
     
    Last edited: Aug 7, 2006
  10. Aug 7, 2006 #9

    berkeman

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    That's not Ltot, it's the derivative of Ltot with respect to x (well except for a couple factor of 2 issues). So it is L'tot, and you would set that equal to zero to find out what the x value is for the minimum, and then figure out the angles from the x. I agree it looks messy at the moment -- maybe there's an easier formulation....
     
  11. Aug 7, 2006 #10

    Gokul43201

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    P, Q, S and T are not variables. You can't separate P & Q and write PQ^2 = P^2Q^2 - that is meaningless.

    So what you should have is:
    [tex]dl_{tot}/dt = \frac{x}{\sqrt{PQ^2+x^2}}+\frac{-L+x}{\sqrt{ST^2+(L-x)^2}} = 0 [/tex]

    [tex]\implies \frac{x}{\sqrt{PQ^2+x^2}} = \frac{L-x}{\sqrt{ST^2+(L-x)^2}} [/tex]

    Square both sides (don't bother expanding the square of (L-x)) and cross multiply...and answer pops out in hardly a couple of steps.
     
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