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Ladder problem, no equation to use for dy/dt

  1. Oct 12, 2009 #1
    1. The problem statement, all variables and given/known data
    http://img132.imageshack.us/img132/8048/prob2r.jpg [Broken]


    2. Relevant equations
    dy/dt = dx/dt * dy/dx


    3. The attempt at a solution
    would dy/dx = 0? because the ladder is 11 m, the derivative of 11 is 0..
    and then would dy/dt be the rate of the top of the ladder sliding down?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Oct 12, 2009 #2

    lanedance

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    you need to find y in terms of x first, thinking about triangles

    then use your chain rule
     
  4. Oct 12, 2009 #3
    You know that [tex]\frac{dy}{dt}[/tex] is equal to [tex]\frac{dy}{dx}*\frac{dx}{dt}[/tex]

    You know [tex]\frac{dx}{dt}[/tex] is 3. So all that's left to do is solve for [tex]\frac{dy}{dx}[/tex]

    By the pythagorean theorem, [tex]a^2+b^2=c^2[/tex]. In this case, on the y-axis you have your a, on the x-axis you have your b and the hypotenuse you have c.

    Therefore, [tex]y^2+x^2=c^2[/tex], so [tex]y=\sqrt{c^2-x^2}[/tex]. Differentiate with respect to x.

    [tex]\frac{dy}{dx}=-x/\sqrt(121-x^2)[/tex]

    Substitute 8.316 for x, and solve for [tex]\frac{dy}{dx}[/tex]
     
  5. Oct 12, 2009 #4
    for [tex]\frac{dy}{dx}=-x/\sqrt(121-x^2)[/tex] of [tex]y=\sqrt{c^2-x^2}[/tex]
    wouldnt it be first (c-x)/(sqrt(c^2-x^2)) then plug in numbers?
    here is what i got so far:
    http://img94.imageshack.us/img94/8545/prob2o.jpg [Broken]
     
    Last edited by a moderator: May 4, 2017
  6. Oct 12, 2009 #5

    lanedance

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    might be even easier to use implict differntiation wrt x on the following equation
    [tex]x^2+y^2=c^2[/tex]

    note c is constant and y = y(x)
     
  7. Oct 13, 2009 #6
    Theres a problem in this problem: What is the starting position of the ladder? Is it initially fully vertical? If it is, what makes it slide down?
     
  8. Oct 13, 2009 #7
    so you are saying that in the picture that i uploaded, the only thing that is wrong is where i wrote [tex]x^2+y^2=11^2[/tex] instead of [tex]x^2+y^2=c^2[/tex]?
    everything else is good?
     
  9. Oct 13, 2009 #8

    lanedance

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    no c = 11 as you said but i was suggesting to try implicit differntiation on the initial equation, though teh way you have looked at it is fine
     
  10. Oct 14, 2009 #9

    lanedance

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    this is a related rates problem, the speed & ladder position are given, so those questions are irrelevant
     
  11. Oct 14, 2009 #10
    I maybe wrong with my assumption lanedance but I think there is a difference in speed as the top of the ladder approaches 7.2' if it starts from 10' or if it starts from 7.3', visualize the scenario.
     
  12. Oct 14, 2009 #11

    lanedance

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    the way I read it, the problem states the bottom of the ladder slides at a constant rate, so the initial position has no impact
     
  13. Oct 14, 2009 #12
    well, i did everything you mentioned already..
    and i get dy/dt = 3.464
    so what am i doing wrong?
     
  14. Oct 14, 2009 #13

    lanedance

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    [tex] x^2 + y^2 = 11^2 [/tex]
    y = 7
    [tex] x = \sqrt{11^2-7^2} = \sqrt{72} = \sqrt{2.2^2 3^2} = 6.\sqrt{2}[/tex]
    x' = 3
    differentiate wrt t
    [tex] 2xx' + 2yy' = 0 [/tex]
    rearrange for y'
    [tex] y' = -xx'/y = -(6\sqrt{2})(3)/7[/tex]

    think you just missed the negative
     
  15. Oct 14, 2009 #14
    well it was negative at first for both the rate change and for dy/dt but then the system say it should be positive...
    yeah that works.. i thought they both supposed to be the same answer.
    thanks for help! that makes sense now!
     
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