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forum.virtualchallengemeets.com • View topic - 12F
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12F

PostPosted: Thu Mar 14, 2013 9:50 am
by calcgal
Can someone help with these problems?

12-F 57 A conical vat is 2 m tall and 1 m in diameter. It sits on the ground on its base. The initially empty vat is filled with water such that the water height changes at a constant rate of 0.1 m/s. What is the rate of change of the water volume when the water height is 1.3 m? (The units for the solution is m^3/s)

I got .0332, but the answer key says .00962

I'm also getting the wrong answer for the geometry problem 12F-60. I got 1150, but the key says 993.

Thanks!

Re: 12F

PostPosted: Thu Mar 14, 2013 6:53 pm
by 101dalmatians

Re: 12F

PostPosted: Thu Mar 14, 2013 8:22 pm
by Fredfredburger
Concerning 57, You differentiated the volume of a cone, but since its base is on the ground it's actually a frustum. The radius of your first base is 1/2 m and the top radius you can find using the scaling principle you used for the cone. The answer is approximately .0962.

Re: 12F

PostPosted: Thu Mar 14, 2013 8:48 pm
by calcguy
12F-57 is a related rates problem. Volume = πh/3[0.5² + (0.5 - h/2)² + 0.5(0.5 - h/2)]
dV/dt = π/4(h² - 2h + 1) dh/dt
Plug in h = 1.3 and dh/dt = 0.1 and you get dV/dt = 0.00707. Either the answer key is wrong or I made a mistake somewhere.

12F-60. The 753 is a tangent segment so it is a leg of the right triangle with x, the side of the square, as the hypotenuse, and R, the radius of the sector, as the other leg. Set the square area equal to the sector area: x² = 0.75πR² which gives R = 0.651x.
Now use the Pythagorean theorem: x² - (0.651x)² = 753² which gives x = 993.

Re: 12F

PostPosted: Thu Mar 14, 2013 9:06 pm
by darksaber21