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  1. #1
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    related rates

    I need help with this peoblem:


    A rope that is 30 feet long is tied to a weight and strung through a pulley that is 8 feet above the weight and is tied to the bumber of a car that is 2 feet from the ground. The car moves away from the weight at a rate of 5 feet/sec. How fast is the weight rising when it is 4 feet from the ground?

    Forgive me if this makes no sense, but this is exactly how my professor worded the question and he thinks is makes perfect sense.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by zachb View Post
    I need help with this peoblem:


    A rope that is 30 feet long is tied to a weight and strung through a pulley that is 8 feet above the weight and is tied to the bumber of a car that is 2 feet from the ground. The car moves away from the weight at a rate of 5 feet/sec. How fast is the weight rising when it is 4 feet from the ground?

    Forgive me if this makes no sense, but this is exactly how my professor worded the question and he thinks is makes perfect sense.
    ok, i think this is the right interpretation for your problem (see the diagram below--ALWAYS DRAW A DIAGRAM WHEN DOING RELATED RATES!)

    so obviously there is a right angled triangle here, so the formula we will be using is Pythagoras' formula.

    notice that x is increasing at the same rate y is decreasing, so dx/dt = -dy/dt

    also notice that when the weight is four feet from the ground, y=4, x = 26 (since its a 30 foot rope, they have to add up to 30) and by pythagoras' formula, z = 25.69

    we are told the the car moves 5 feet/s so dz/dt is increasing at 5 ft/s

    we are looking for dy/dt when y = 4, here goes:

    x^2 = y^2 + z^2
    => 2x dx/dt = 2y dy/dt + 2z dz/dt
    => 2x (-dy/dt) = 2y dy/dt + 2z dz/dt
    => - 2x dy/dt - 2y dy/dt = 2z dz/dt
    => dy/dt (-2x - 2y) = 2z dz/dt
    => dy/dt = [2z dz/dt]/(-2x - 2y)
    now plug in all the values

    => dy/dt = [2(25.69)(5)]/(-2(26) - 2(4)) = 256.9/(-52 - 8) = 256.9/(-60)
    => dy/dt = - 4.28 ft/s

    note that dy/dt is negative since the length of y according to our diagram is decreasing.

    so the rate is just 4.28 ft/s
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  3. #3
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    No, this is the type of car I want.
    (I will without jokes, drive that thing. I do not know why people today say that car is ugly. I really think this has a nice look to it).
    Attached Thumbnails Attached Thumbnails related rates-picture11.gif  
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    i seem to have made an error in my thought process above. note that as the length y = 4, the length x will not be hypotenuse of the triangle, rather, only a part of x will be. i leave it to you to correct this error, you have the frame work of the solution

    or maybe it will work as it is, i dont know

    i'll think about it later if you dont figure it out
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    i think replacing y with y + 2 will do the trick, since this will be the height that maintains the right angled triangle. so when y = 4, the height that maintains the triangle is 6ft, which is y + 2.

    what do you guys think (speaking to the geniuses of the forum)?
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  6. #6
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    Quote Originally Posted by Jhevon View Post
    i think replacing y with y + 2 will do the trick, since this will be the height that maintains the right angled triangle. so when y = 4, the height that maintains the triangle is 6ft, which is y + 2.

    what do you guys think (speaking to the geniuses of the forum)?
    You're correct on this one. The weight is initially 2ft above the ground while the pully is initially 8 ft above the weight. Since you've said that y is the distance from the weight to the pully, if the weight is 4ft above the ground then y is 2ft higher than its initial 2ft above the ground and so the distance from the weight and the pully is 2ft shorter than its initial length: y = 8ft - 2ft = 6ft.

    Since y = 6, a lot of the other numbers in the equation, such as x and z, also change.

    The total distance of x + y is 30ft, so x = 24. Using the pythagorean theorem, z = sqrt(24^2 - 6^2) = sqrt(540) = 23.24ft

    Quote Originally Posted by Jhevon View Post
    x^2 = y^2 + z^2
    => 2x dx/dt = 2y dy/dt + 2z dz/dt
    => 2x (-dy/dt) = 2y dy/dt + 2z dz/dt
    => - 2x dy/dt - 2y dy/dt = 2z dz/dt
    => dy/dt (-2x - 2y) = 2z dz/dt
    => dy/dt = [2z dz/dt]/(-2x - 2y)
    now plug in all the values
    dy/dt = [2(23.24)(5)]/(-2*24-2*6) = -3.87 ft/s

    But the rate that the weight is rising from the ground is 3.87 ft/s

    P.S. As a side note, the diagram is a little misleading. The pully is 8ft above the weight, not 8ft above the ground. Besides that, very nice diagram.
    Last edited by ecMathGeek; March 22nd 2007 at 11:20 PM.
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  7. #7
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    Thanks everyone for your help. This was actually a question I had on an exam yesterday and I posted here after the exam to see just how WRONG I was.

    ALWAYS DRAW A DIAGRAM WHEN DOING RELATED RATES!
    Yes. I wasted a lot of time during the exam trying to draw the picture, but it just didn't make any sense to me. I eventually got the picture right, but I only had five minutes left.

    By looking at the work you did (everything before you plugged in the values) I see that I was actually on the right track. Of course, I second guessed myself and started over and got the whole thing wrong.
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ecMathGeek View Post
    P.S. As a side note, the diagram is a little misleading. The pully is 8ft above the weight, not 8ft above the ground. Besides that, very nice diagram.
    Yeah, i expected the diagram to be misleading since it was based on my initial logic which was faulty--but not in the way you mentioned. i figured initially the weight was on the ground, so being 8ft above the weight was basically being 8 ft above the ground. then the car drives and raises it to 4ft, which is the instant we are worried about. but maybe that logic was faulty as well, i don't know
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  9. #9
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    Re:

    Quote Originally Posted by zachb View Post
    I need help with this peoblem:


    A rope that is 30 feet long is tied to a weight and strung through a pulley that is 8 feet above the weight and is tied to the bumber of a car that is 2 feet from the ground. The car moves away from the weight at a rate of 5 feet/sec. How fast is the weight rising when it is 4 feet from the ground?

    Forgive me if this makes no sense, but this is exactly how my professor worded the question and he thinks is makes perfect sense.
    If your professor didn't draw the diagram for you on the test then he is a real moron. This problem takes more left-handed artistic ability, than sheer math skill. What a numskull...
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  10. #10
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by qbkr21 View Post
    If your professor didn't draw the diagram for you on the test then he is a real moron. This problem takes more left-handed artistic ability, than sheer math skill. What a numskull...
    I agree with you in that a question of this nature should not be used on a test (unless of course it was a test on related rates and it was the ONLY question--which would be cruel), however, it does not require as much artisitic skill as i made it seem. I was feeling jovial at the time and decided to place a car in the diagram and draw the pulley and all that stuff, but it really wasn't necessary. drawing a simple right-angled triangle with the proper dimensions is sufficient to think about the solution to this problem
    Last edited by Jhevon; April 10th 2007 at 06:55 AM.
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