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Math Help - Check this problem out!

  1. #1
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    Angry Check this problem out!

    Im reviewing for my calculus final exam and I was just going through some problems that I didnt understand and never got back to them. Im trying to solve this one out and I cant get it right.

    A swimming pool is 22 ft wide, 60 ft long, 2 ft deepat the shallow end, and 10 ft deep at its deepest point. If the pool is being filled with water at a rate of 0.8ft^3/min, how fast is the water level rising when the depth at the deepest point is 7 ft??

    I keep getting .00115ft/min and i know thats not the right answer.
    Can some one give me a hand?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ((ECHO)) View Post
    Im reviewing for my calculus final exam and I was just going through some problems that I didnt understand and never got back to them. Im trying to solve this one out and I cant get it right.

    A swimming pool is 22 ft wide, 60 ft long, 2 ft deepat the shallow end, and 10 ft deep at its deepest point. If the pool is being filled with water at a rate of 0.8ft^3/min, how fast is the water level rising when the depth at the deepest point is 7 ft??

    I keep getting .00115ft/min and i know thats not the right answer.
    Can some one give me a hand?
    i got something close to that, 0.00069264, is that the correct answer?
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  3. #3
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    Quote Originally Posted by Jhevon View Post
    i got something close to that, 0.00069264, is that the correct answer?
    I also got something similar: .00071294893. Not sure if the difference between my answer and Jhevon's is due to rounding errors during working though.
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  4. #4
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    My apologies I repeated my calculation and got 0.00069264069 as well. Rounding error the first time:

    Let the height = h. Let S be the length of the slope of the pool as the depth changes from 2 feet to 10 feet. Then Cos(theta) = 8/60
    theta = Cos(^-1) (8/60)
    = 82.33774434 * (let * be degrees) here on referred to as A
    Thus Cos (A) = h/S

    Let the length of the triangular prism (where the whole triangular prism has dimensions: Length 60 feet, height 8 feet, width 22 feet, slope 60.5309838 feet)
    (SQUARE ROOT OF: ((h^2)/cos^2 (A) - h^2)) gives LENGTH L. (Pythagorus' theorem)

    Thus volume of triangular prism given by:
    V = (1/2)(h)((SQUARE ROOT OF: ((h^2)/cos^2 (A) - h^2)))22
    = 11(h^2)((SQUARE ROOT OF: (1- cos^2 (A))/cos^2 (A)))
    dV/dt = .8
    dh/dt = (dh/dV)(dV/dt)
    = (dh/dV) .8
    dV/dh = 22h((SQUARE ROOT OF: (1- cos^2 (A))/cos^2 (A)))
    dh/dV = 1/22h((SQUARE ROOT OF: (1- cos^2 (A))/cos^2 (A)))
    dh/dt = (.8)1/22h((SQUARE ROOT OF: (1- cos^2 (A))/cos^2 (A)))
    where A = 82.33774434 * (let * be degrees)
    Let h = 7
    dh/dt = 0.00069264069
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  5. #5
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    Hello, ((ECHO))!

    I get the same answer is behemoth and Jhevon . . .


    A swimming pool is 22 ft wide and 60 ft long.
    It is 2 ft deep at the shallow end, and 10 ft deep at the other end.
    If the pool is being filled with water at a rate of 0.8 ft³/min,
    how fast is the water level rising when the depth at the deepest point is 7 ft?
    Code:
                            60
      -   * - - - - - - - - - - - - - - - - - - - - *
      :   |                                         |
      2   |                                         | 2
      :   |                 60                      |
      - A + - - - - - - - - - - - - - - - - - - - - * B
      :   |                                   *
      :   |            x                *
      : C +-----------------------* D
      8   |:::::::::::::::::*
      :   |h :::::::::*
      :   |:::::*
      - E *

    Since \Delta BAE \sim \Delta DCE, we have: . \frac{x}{h} = \frac{60}{8}\quad\Rightarrow\quad x = \frac{15}{2}h .[1]
    . . Hence: . \frac{dx}{dt} = \frac{15}{2}\!\cdot\!\frac{dh}{dt} . [2]

    The volume of water is: . V \:=\:\frac{1}{2}xh(22) \:=\:11xh

    . . Then: . \frac{dV}{dt}\;=\;11x\!\cdot\frac{dh}{dt} + 11h\!\cdot\!\frac{dx}{dt} . [3]


    When h = 7, [1] gives us: . x \:=\:\frac{15}{2}(7) \:=\:\frac{105}{2}

    Substitute into [3]: . 0.8 \;=\;11\left(\frac{105}{2}\right)\left(\frac{dh}{d  t}\right) + 11(7)\left(\frac{15}{2}\!\cdot\!\frac{dh}{dt}\righ  t)

    Hence: . 1155\left(\frac{dh}{dt}\right) \:=\:0.8\quad\Rightarrow\quad \frac{dh}{dt}\:=\:\frac{0.8}{1155}

    Therefore: . \frac{dh}{dt} \:=\:0.000692641<br />
ft/min.

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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, ((ECHO))!

    I get the same answer is behemoth and Jhevon . . .


    Code:
                            60
      -   * - - - - - - - - - - - - - - - - - - - - *
      :   |                                         |
      2   |                                         | 2
      :   |                 60                      |
      - A + - - - - - - - - - - - - - - - - - - - - * B
      :   |                                   *
      :   |            x                *
      : C +-----------------------* D
      8   |:::::::::::::::::*
      :   |h :::::::::*
      :   |:::::*
      - E *

    Since \Delta BAE \sim \Delta DCE, we have: . \frac{x}{h} = \frac{60}{8}\quad\Rightarrow\quad x = \frac{15}{2}h .[1]
    . . Hence: . \frac{dx}{dt} = \frac{15}{2}\!\cdot\!\frac{dh}{dt} . [2]

    The volume of water is: . V \:=\:\frac{1}{2}xh(22) \:=\:11xh

    . . Then: . \frac{dV}{dt}\;=\;11x\!\cdot\frac{dh}{dt} + 11h\!\cdot\!\frac{dx}{dt} . [3]


    When h = 7, [1] gives us: . x \:=\:\frac{15}{2}(7) \:=\:\frac{105}{2}

    Substitute into [3]: . 0.8 \;=\;11\left(\frac{105}{2}\right)\left(\frac{dh}{d  t}\right) + 11(7)\left(\frac{15}{2}\!\cdot\!\frac{dh}{dt}\righ  t)

    Hence: . 1155\left(\frac{dh}{dt}\right) \:=\:0.8\quad\Rightarrow\quad \frac{dh}{dt}\:=\:\frac{0.8}{1155}

    Therefore: . \frac{dh}{dt} \:=\:0.000692641<br />
ft/min.

    This is the approach i used, behemoth100 did something different using trig functions. it seemed to work out though
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