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Math Help - Related Rates Question

  1. #1
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    Related Rates Question

    A tight rope is attached 30 ft above the ground between buildings A and B, which are 50 ft apart. A guy is walking on the rope at a constant rate of 2 ft/s, and is illuminated by a spotlight (S) 70 ft above on building A

    ....S_______50_________
    ....|.............................|
    ....|.............................|
    ....|.............................|
    .70|.............................|
    ....|.............................|
    ....|.............................|
    ..A|__________________|B <tightrope
    ....|.............................|
    ....|.............................|
    .30|.............................|
    ....|.............................|

    1. How fast is the shadow of the tightrope walker's feet moving along the ground when he is midway between the buildings?

    I got 20/7 ft/s. I used similar triangles and didnt even have an X in the deriv (just x') so the rate would always be the same no matter where the guy is on the rope....is that right?

    2. How far from point A is he when the shadow of his feet reaches the base of B building?

    I got 35 feet, using similar triangles

    3. How fast is the shadow of the tightrope walker's feet moving up the wall of the B building when he is 10 feet from point B?

    I got 4.375 ft/s

    am i right?



    and another: Car B is 30 miles directly east of Car A and begins moving west at 90 mph. At the same time, car A begins movie north at 60 mph.
    a. after 1/10 of an hour has elapsed, at what rate are the cars getting closer together.
    used distance formula and got 277.4033 mph. (that doesnt seem right)

    b. what will be the minimum distance between the cars, and at what time t does the minimum distance occur
    Last edited by stones44; February 11th 2009 at 04:53 PM.
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  2. #2
    Senior Member mollymcf2009's Avatar
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    1. How fast is the shadow of the tightrope walker's feet moving along the ground when he is midway between the buildings?

    I got 20/7 ft/s. I used similar triangles and didnt even have an X in the deriv (just x') so the rate would always be the same no matter where the guy is on the rope....is that right?


    2. How far from point A is he when the shadow of his feet reaches the base of B building?

    You did your calculation correctly,but you calculated how far he is from building B and it asks for how far he is from building A. So 50ft - 35 ft = 15 ft

    3. How fast is the shadow of the tightrope walker's feet moving up the wall of the B building when he is 10 feet from point B?

    I got 4.375 ft/s

    Let x = distance between walker and point B
    Let y = distance from shadow to top of bulding B
    dx = rate of walker
    dy = rate of shadow moving up building B

    I have:
    y = 17.5 (from similar triangles)
    x = 10 (given)
    dx = 2 ft/s
    dy = ?

    Assuming you obtained your function using the derivative of the pathagorean theorum:

    y^2 = 70^2 + x^2

    2y dy = 2x dx

    So,

    2(17.5)dy = 2(10)(2)

    dy = \frac{40}{35} = 1.14 ft/s

    Double check me, (as I am so tired I'm about to drop) but I'm pretty sure this is right.



    and another: Car B is 30 miles directly east of Car A and begins moving west at 90 mph. At the same time, car A begins movie north at 60 mph.
    a. after 1/10 of an hour has elapsed, at what rate are the cars getting closer together.
    used distance formula and got 277.4033 mph. (that doesnt seem right)

    For this problem, you will need to find dz. You are given two rates in the problem, dx = 90 mph & dy = 60mph

    2ydy +2xdx = 2zdz

    Ok, so first,

    Let's get our x and y values. We know how long they travelled and how fast they were going:

    distance = rate x time

    For each car,
    d_A = (60mph)(.1hr) = 6mi
    d_B = (90mph)(.1hr) = 9mi

    So,
    y = 6 miles
    x = 30-9 = 21 miles (car B backtracked on our 30 mile distance by 9 miles)

    To find z, use pathagorean theorum. You can do that part.

    Then plug and chug for dz, which should be the rate after 1/10 hour.

    2ydy +2xdx = 2zdz

    Hope this helps. I have to go to bed, sorry I can't help with the last part!
    Last edited by mollymcf2009; February 10th 2009 at 09:30 PM.
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  3. #3
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    for the car one, i followed you and got 103.0204..that doesnt make sense tho does it? car B is getting closer to car A faster than if car A was stationary...hmmm

    and for the shadow one, the final part, x does not equal 10. it equals 40. and i used the derivative of the similar triangles equation. and the other part it is 35 not 15. X is the distance from A to the walker.
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  4. #4
    Senior Member mollymcf2009's Avatar
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    That'll teach me to be on here when I'm dog tired! Sorry about that, glad you figured it out!
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  5. #5
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    so can someone help with the second (car) problem?

    and maybe confirm my first problem's answers also
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  6. #6
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    anyone? i need this soon\

    for the second problem, i made dx/dt -90, and got an answer of ~ -70, which negative means getting closer so the answer would be 70 mph right?
    and then for the next part of that problem i got 3/13 hours and 16.4 something miles, but i dont think thats right
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