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Math Help - Help with a work problem and a one-to-one explanation:

  1. #1
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    Help with a work problem and a one-to-one explanation:

    Hi guys. Thanks in advance for the help last time, and I'm wondering if you guys could help me out here.

    1. (Work problem)

    The Drain-o-Max above ground pool pumping system is designed to do up to 50000(pi) ft-lbs of work on a single charge. It attaches directly to the side of your above ground pool and pumps the water right out over the side of the pool.

    What is the radius of the largest 4 foot tall circular above ground pool that the Drain-o-Max can empty on a single charge? (Assume pool is initially full to the brim and the weight density of water is 62.5 ft-lbs.)

    2. a) show that f(x) = 4x^3 + 5x + 1 is one-to-one.
    b) Find g'(1), where g is the inverse function to f(x) = 4x^3 + 5x + 1.

    For #1, I'm really lost and have no idea where to begin my answer.

    For #2, for a), I found the derivative and then saw that it was a parabola. Is there something to do with the Mean Value Theorem to prove that the original eqn. is 1-to-1?

    I have no idea how to do b either.

    Thanks in advance guys!
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  2. #2
    MHF Contributor Calculus26's Avatar
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    #2a f ' (x) >0 for all x hence increasing and one to one




    #2b

    Use the theorem on the derivative of the inverse:

    if f(a) = c

    Then if g = f^(-1)

    g ' (c) = 1/ (f ' (a))

    For

    f(x) = 4x^3 + 5x + 1

    f(0) = 1

    g ' (1) = 1/ f'(0) = 1/ 5
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  3. #3
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by XFactah416 View Post
    Hi guys. Thanks in advance for the help last time, and I'm wondering if you guys could help me out here.

    1. (Work problem)

    The Drain-o-Max above ground pool pumping system is designed to do up to 50000(pi) ft-lbs of work on a single charge. It attaches directly to the side of your above ground pool and pumps the water right out over the side of the pool.

    What is the radius of the largest 4 foot tall circular above ground pool that the Drain-o-Max can empty on a single charge? (Assume pool is initially full to the brim and the weight density of water is 62.5 ft-lbs.)

    2. a) show that f(x) = 4x^3 + 5x + 1 is one-to-one.
    b) Find g'(1), where g is the inverse function to f(x) = 4x^3 + 5x + 1.

    For #1, I'm really lost and have no idea where to begin my answer.

    For #2, for a), I found the derivative and then saw that it was a parabola. Is there something to do with the Mean Value Theorem to prove that the original eqn. is 1-to-1?

    I have no idea how to do b either.

    Thanks in advance guys!
    Note that the work done on the water by the pump can be given by

    W=62.5\int(volume)(distance)

    So, from context we infer that

    1. V=\pi{x}^2\Delta{y} if you were to impose your coordinate axes with the origin at the center of the bottom of the pool.

    2. d=4-y.

    So, get everything in terms of y and you should be all set.
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  4. #4
    MHF Contributor Calculus26's Avatar
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    See attachment
    Attached Thumbnails Attached Thumbnails Help with a work problem and a one-to-one explanation:-work.jpg  
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  5. #5
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    Quote Originally Posted by Calculus26 View Post
    See attachment
    Hrrm.. calculus, the radius is not constant though. In fact, it is asking for the radius in this problem. BTW, thank you very much for #2. It makes so much more sense now that I figured out what I was doing wrong.

    VonNemo, I understand what you're doing.. but how would you get (pi)x^2 in terms of y? I clearly understand the 4-y part as the distance.

    Perhaps my picture is just drawn completely wrong and I can't visually understand what is happening.
    Last edited by XFactah416; January 18th 2010 at 04:07 PM.
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  6. #6
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    50000\pi = \int_0^4 62.5(\pi r^2)(4-y) \, dy<br />

    50000\pi = 62.5 \pi r^2 \int_0^4 (4-y) \, dy<br />

    800 = r^2 \int_0^4 (4-y) \, dy

    800 = r^2 \left[4y - \frac{y^2}{2}\right]_0^4

    800 = r^2 (16 - 8)

    100 = r^2


    using a physics approach ...

    center of mass of the pool is being raised 2 ft

    mg (2) = 50000 \pi

    mg = 25000 \pi

    62.5 \cdot \pi r^2 (4) = 25000 \pi

    250 r^2 = 25000

    r^2 = 100
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  7. #7
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    Quote Originally Posted by skeeter View Post
    50000\pi = \int_0^4 62.5(\pi r^2)(4-y) \, dy<br />

    50000\pi = 62.5 \pi r^2 \int_0^4 (4-y) \, dy<br />

    800 = r^2 \int_0^4 (4-y) \, dy

    800 = r^2 \left[4y - \frac{y^2}{2}\right]_0^4

    800 = r^2 (16 - 8)

    100 = r^2

    using a physics approach ...

    center of mass of the pool is being raised 2 ft

    mg (2) = 50000 \pi

    mg = 25000 \pi

    62.5 \cdot \pi r^2 (4) = 25000 \pi

    250 r^2 = 25000

    r^2 = 100
    OK, wow, I'm just out of it.

    Thanks in advance guys. I think I just completely took the wrong approach to this problem and didn't really understand what it was asking for. Stupid me.
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