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Math Help - Find height given volume

  1. #1
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    Find height given volume

    So I heard an interesting problem on the radio, can't crack it. A trucker's gas gauge is broken, he knows that the tank diameter is 20 inches, if he dips the dipstick and gas is at half the length obviously it's half-full. What height does he need to mark off on the dipstick in order to measure when the tank is 1/4 full?

    I figure let's forget about the unnecessary third dimension and solve this for a unit circle. I tried thinking of this as a circle where I'm solving for the top 1/4 area and I don't know the length from the "fluid level" to the top of the tank; nor the length from the x = 0 line to the point where the fluid level meets the edge of the tank. So I try setting up my function as x^{2} + (y - b)^{2} = 1 where b is the necessary number to set the fluid-level at the x-axis. I then need to integrate from 0 to a where a is the length of the fluid-level line from the x = 0 line to the edge of the tank. Integrating \displaystyle \int_{0}^{a} \Big( b + \sqrt{1 - x^{2}} \Big) dx and setting it equal to \pi / 8 is a bit laborious and gives an answer that doesn't really seem to help.

    I've heard that there is a purely trigonometric solution, but I don't know any trigonometry that deals with area except the basic formula A = \pi r^{2}.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by ragnar View Post
    So I heard an interesting problem on the radio, can't crack it. A trucker's gas gauge is broken, he knows that the tank diameter is 20 inches, if he dips the dipstick and gas is at half the length obviously it's half-full. What height does he need to mark off on the dipstick in order to measure when the tank is 1/4 full?

    I figure let's forget about the unnecessary third dimension and solve this for a unit circle. I tried thinking of this as a circle where I'm solving for the top 1/4 area and I don't know the length from the "fluid level" to the top of the tank; nor the length from the x = 0 line to the point where the fluid level meets the edge of the tank. So I try setting up my function as x^{2} + (y - b)^{2} = 1 where b is the necessary number to set the fluid-level at the x-axis. I then need to integrate from 0 to a where a is the length of the fluid-level line from the x = 0 line to the edge of the tank. Integrating \displaystyle \int_{0}^{a} \Big( b + \sqrt{1 - x^{2}} \Big) dx and setting it equal to \pi / 8 is a bit laborious and gives an answer that doesn't really seem to help.

    I've heard that there is a purely trigonometric solution, but I don't know any trigonometry that deals with area except the basic formula A = \pi r^{2}.
    From the Scham Mathematical handbook:
    Attached Thumbnails Attached Thumbnails Find height given volume-area-segment.jpg  
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  3. #3
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    However, just saying that the gas tank "has a diameter of 20 inches" doesn't tell us the shape or orientation. Captain Black assumed a cylindrical tank "on its side", which is probably most likely, but a spherical tank or a cylindrical tank with vertical axis are also possible.
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