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Math Help - Related rates problem

  1. #1
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    Related rates problem

    One of the practice problems in my book is giving me trouble.

    It asks

    "A hemispherical bowl of radius 10 cm contains water to a depth of h cm. Find the radius of the surface of the water as a function of h."

    I can see that the radius gets smaller as h gets smaller but I'm not how to go about finding an equation. The back of the book says that the answer is r=(20h -h^2)^(1/2) and I think they used the circle equation but I'm not sure. Does anyone have any hints?
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  2. #2
    Super Member Quacky's Avatar
    Joined
    Nov 2009
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    Windsor, South-East England
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    Re: Related rates problem

    The bowl has a radius of 10cm. The distance from the height of the water to the top centre of the hemisphere (from where the radius is measured) is (10-h)cm

    I drew a diagram in front of me for this, and you can do it using Pythagoras.

    Draw the hemisphere. Draw the level of the water at any distance (10-h)cm below the top and label the (10-h) distance. Then draw a radius from the centre top of the hemisphere to the point on the hemisphere that touches the surface of the water. If you have understood me, you should now have a right triangle. The base is r, the height is (10-h) and the hypotenuse is 10.

    I hope you'll forgive me for my dreadful artistic skills, but I have done a diagram here:

    Spoiler:


    I'll check this in a second, it's sort of a leap of faith on my part, but hopefully you should get the expected result.

    Edit: Yeah, I've managed to obtain their solution this way.
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