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Math Help - Difficult Related Rates

  1. #1
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    Difficult Related Rates

    1)An oil storage tank has the shape of a paraboloid with a radius of 5ft and a height of 9ft. Oil flows into the tank at the rate of 8 cubic feet/min. The volume of the tank is V=(50*pi*h^(3/2))/3.

    a)Find the volume of a full tank
    b)How long would it take to fill the tank if it was initially empty?
    c)How fast is the height of oil increasing when h=4?

    I need help on where to start. I am usually pretty good at this stuff but I am lost on this one. Like finding the volume? Is that dV/dT?

    2)A container has the shape of an open right circular cone. The height of the container is 10cm and the diameter or the opening is 10cm. Water in the container is evaporating so that its depth, h, is changing at a constant rate of -3/10 cm/hr.

    a)Find the volume V of water in the container when h=5 cm.
    b)Find the rate of change of the volume of water in the container, with respect to time, when h=5 cm.

    I need help on where to start.
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  2. #2
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    [quote]1)An oil storage tank has the shape of a paraboloid with a radius of 5ft and a height of 9ft. Oil flows into the tank at the rate of 8 cubic feet/min. The volume of the tank is V=(50*pi*h^(3/2))/3.

    a)Find the volume of a full tank
    b)How long would it take to fill the tank if it was initially empty?
    c)How fast is the height of oil increasing when h=4?

    V=\frac{50{\pi}h^{\frac{3}{2}}}{3}

    \frac{dV}{dt}=25{\pi}h^{\frac{1}{2}}\cdot \frac{dh}{dt}

    For part a, just use h=9 in V to find the full volume.

    part b, Use the given 8 cubic feet/min and the result from part a.

    part c, Use h=4 and dV/dt=8 and solve for dh/dt

    2)A container has the shape of an open right circular cone. The height of the container is 10cm and the diameter or the opening is 10cm. Water in the container is evaporating so that its depth, h, is changing at a constant rate of -3/10 cm/hr.

    a)Find the volume V of water in the container when h=5 cm.
    b)Find the rate of change of the volume of water in the container, with respect to time, when h=5 cm.
    Use similar triangles to eliminate a variable. This problem is an old cliche related rates. If you google you will probably find something similar.

    \frac{r}{h}=\frac{5}{10}=\frac{1}{2}\Rightarrow r=\frac{h}{2}

    Now, sub into the volume of cone formula, differentiate to get dV/dt. Then, enter in your h=5 to find dV/dt at that height. You are given dh/dt=-3/10
    Last edited by galactus; November 23rd 2008 at 02:16 PM.
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  3. #3
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    thanks so much...i still am lost on b tho...like what should the equation look like? i got all the other answers except for b
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  4. #4
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    b for which problem?.
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  5. #5
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    Quote Originally Posted by galactus View Post
    b for which problem?.
    1b please
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  6. #6
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    Well, let's think about it. We know the total volume of the tank from part a.

    If water is going in at 8 ft^3/min wouldn't you just divide 8 into the result from part a to get the total minutes to fill?.

    Think about your units. The tank volume is in cubic feet. The rate is in cubic feet per minute.

    So, \frac{ft^{3}}{\frac{ft^{3}}{min}}=min

    Divide by 60 if you want hours.
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  7. #7
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    man thanks so much you dont know how much i appreciate this
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