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Calculus Related Rates Problems

  1. Oct 31, 2011 #1
    Two carts, A and B, are connected by a rope 39 ft long that passes over a pulley P (see the figure). The point Q is on the floor h = 12 ft directly beneath P and between the carts. Cart A is being pulled away from Q at a speed of 2.5 ft/s. How fast is cart B moving toward Q at the instant when cart A is 5 ft from Q?
     
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  3. Oct 31, 2011 #2
    I'm not too sure where to go with this but I think I got the start:

    (x^2+12^2)^1/2+(y^2+12^2)^1/2=39

    Can somewhere tell me if this is right and where to go from here?
     
  4. Nov 1, 2011 #3

    I like Serena

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    Welcome to PF, gina4930! :smile:

    You mentioned something about a figure?
    I'm afraid it's too hard for me to deduce what you're doing without a figure. :redface:
     
  5. Nov 1, 2011 #4

    HallsofIvy

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    I assume your picture shows two carts, on a level surface, connected by a rope passing over a pulley at some point above and between them. The height of the pulley is 12 ft. Let x be the distance from Q to one cart and y the distance from Q to the other cart. Then the hypotenuse of the right triangle formed by the first cart, P, and Q, is given by [itex]\sqrt{x^2+ 144}[/itex]. The hypotenuse of the right triangle formed by the second cart, P, and Q, is given by [itex]\sqrt{y^2+ 144}[/itex]. The two hypotenuses form the entire length of the rope: [itex]\sqrt{x^2+ 144}+ \sqrt{y^2+ 144}= 39[/itex]. That's the formula you have. Well, done!

    That is the "static" formula. To get a "dynamic" formula, relating the rates of motion, differentiate both sides with respect to t, using the chain rule. Your final formula will involve both dx/dt and dy/dt. You are given one and asked to find the other.
     
  6. Nov 1, 2011 #5

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    Hey HoI! :wink:

    That looks like a very plausible interpretation of the problem!
     
  7. Nov 1, 2011 #6
    When I differentiated the equation, I got:
    1/2(x^2+144)^-1/2*(2x*dx/dt) + 1/2(y^2+144)^-1/2*(2y*dy/dt)=0
    then I substituted in x=-5 , dx/dt= -2.5 , and y= (533)^1/2
    When I solved for dy/dt I got two totally different answers. The first time I got -1.08 and the second time I got -.04. According to my online assignment, neither of these are correct. Can anyone point out my mistake? Thanks!
     
  8. Nov 1, 2011 #7

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    Looking good...

    But how did you get y=(533)^1/2?
     
  9. Nov 1, 2011 #8
    I solved for y using the original equation. I thought using (533)^1/2 is more accurate then putting 23.1.
     
  10. Nov 1, 2011 #9

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    Ah well, after correcting a mistake of my own, I get y=√532, which is still different from yours.

    And after that I get dy/dt=-1.084 m/s.

    Since the problem asks how fast cart B is moving toward Q, the answer should be 1.084 m/s.
    Can it be that it should be without a minus sign?
     
  11. Nov 1, 2011 #10
    I did not even think to put it without a minus sign. Thank you. Thank was the correct answer. I greatly appreciate it.
     
  12. Nov 1, 2011 #11

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