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Math Help - Related rate

  1. #1
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    [SOLVED] Related rate

    Hi guys !

    There's a related rate problem I really can't figure out. I think I have an idea, but I'm not too sure about it. Heres the situation :
    A ball is dropped at t=0 at 15 m high, 9 m from a lamp. This lamp has a height of 15 m too. Knowing that the height of the ball can be known with the formula y(t)=15-4.9t, at what speed goes the shadow of the ball at t=0.5 (ie. 0.5s after being dropped)

    Here is my drawing of the situation.
    Related rate-relatedrate.png


    I called x the distance between under the ball and the shadow of it. I first calculated what height the ball would be after 0.5s, which gave me :
    y(0.5) = 15-4.9(0.5) = 13.775 m

    Then, I decided to calculate its speed after at this moment. This would logically be the rate dy/dt, I thought. It gave me :
    dy/dt = (13.775-15)/(0.5) = -2.45 m/s

    I tried to figure out how I could relate y (height of the ball) and x, and the only thing I thought of was : Let's use similar triangle rules ! The biggest triangle was 15 m*(9+x) m and the smallest, y*x. I used the following formula :
    x/(x+9) = y/15

    I isolated x :
    x = (9y)/(15-y)

    And then, I derivated it in function of the time.
    dx/dt = [135(dy/dt)]/[(15-y)]

    I knew that dy/dt = -2.45 m/s and that y = 13.775. So the final answer was :
    dx/dt = -220.4 m/s

    So, at t=0.5, the shadow of the ball moves toward the lamp at the speed of 220.4 m/s.

    Did I do anything wrong ? Is the answer correct or logical/plausible ?

    Thanks a lot !

    (Sorry, I don't know anything about LaTEX)
    Last edited by Zaide; December 2nd 2011 at 12:23 PM.
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  2. #2
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    Re: Related rate

    All good except for

    Quote Originally Posted by Zaide View Post
    Then, I decided to calculate its speed after at this moment. This would logically be the rate dy/dt, I thought. It gave me :
    dy/dt = (13.775-15)/(0.5) = -2.45 m/s
    I guess you're reasoning geometrically on the assumption of constant speed - your straight line sketch? ... but that's against x not t!

    Why not differentiate y(t), or use v = u + at.
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  3. #3
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    Re: Related rate

    Quote Originally Posted by tom@ballooncalculus View Post
    All good except for



    I guess you're reasoning geometrically on the assumption of constant speed - your straight line sketch? ... but that's against x not t!

    Why not differentiate y(t), or use v = u + at.
    Well, I was considering that at t=0.5, the speed of the ball was -2.45 m/s. I know it's not constant. If it would have been at t=1, i would've use -4.9 m/s.
    Si you would just differentiate y(t) = -4.9t+15 ? Don't really get it. How do I put the distance of the shadow (x) in this ?!
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  4. #4
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    Re: Related rate

    Quote Originally Posted by Zaide View Post
    Well, I was considering that at t=0.5, the speed of the ball was -2.45 m/s.
    Why?

    I know it's not constant.
    But it would have to have been, to justify using distance/time to calculate it.

    So you would just differentiate y(t) = -4.9t+15 ? Don't really get it.
    I said do it! So do it, do it, do it!

    Quote Originally Posted by Zaide View Post
    How do I put the distance of the shadow (x) in this ?!
    Like I said, you did everything right apart from the velocity at half a second.
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  5. #5
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    Re: Related rate

    I don't really get what you're trying to say. How would you do it ? I just started doing these related rates, and there are a few things I don't understand.
    Do I derivate y to get dy/dt = -9.8t, and then replace the dy/dt in the equation I had ? Like that :
    dx/dt = [135(dy/dt)]/[(15-y)]
    dx/dt = [135(-9.8t)]/[(15-y)]
    dx/dt = -440.8 m/s ?
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  6. #6
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    Re: Related rate

    Looks good to me.
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  7. #7
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    Re: Related rate

    Well, thanks a lot !
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