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Question
hi scotto,
 
"suppose that water is being emptied from a spherical tank of radius 10.  if the depth of the water is 5 ft and decreasing at a rate of 3 ft/sec, at what rate is the radius of the top of the water decreasing?"
my ap teacher gave us this question on the last test but no one got it right (i skipped it outright). he's going to put a similar problem in another test in 2 days (!) and i have no idea where to even begin. an answer in its entirety would be much appreciated!  thank you!

Answer
Let the bottom of the tank be at h=0, so the depth from the bottom is h.  This makes the distance to the center 10-h.

The radius at the depth is 5', and is the bottom leg of a right triangle.  The radius of the spehere is 10, and is the hypoteneuse of the same triangle.  This means the distance down is
√(100-25) = 5√3.

Now the surface area of a circle is π*r².  The variable r is found in the equation of a right triangle as r² + (10-h)² = 10².  This makes r² = 10² - (10-h)² = 100 - 100 + 20h - h².  Cancelling the 100's, we get r² = 20h - h².

Note that r and h are both functions of t, so it should be r²(t) = 20h(t) - h²(t).

Taking the derivative with respect to t gives 2r(t)(dr/dt) = 20 dh/dt - 2h(t)(dh/dt).
The value of t is not know, but r(t) = 5, h(t) = 10 - 5√3, and dh/dt = -3.
Putting these in, dr/dt can be found.  

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Scotto

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Any kind of calculus question you want. I also have answered some questions in Physics (mass, momentum, falling bodies), Chemistry (charge, reactions, symbols, molecules), and Biology.

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Experience in the area: I have tutored students in all areas of mathematics for over 25 years. Education/Credentials: BSand MS in Mathematics from Oregon State University, where I completed sophomore course in Physics and Chemistry. I received both degrees with high honors. Awards and Honors: I have passed Actuarial tests 100, 110, and 135.

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