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53% CM, Thu 12 30 PM Q. E Firefox File Edit View History Bookmarks Tools Window Help e O O WeBWork Abramson MA Op https webwork asl edu MAT 265 Fall 2014/Section 2.7/8/?user bdalmuha&key; oBD5 sl1F96Qgxft. v C CY Yahoo Section 2.7: Problem 8 Problem 8 Password/Email Prev Up Next Grades (1 pt) Water is leaking out of an inverted conical tank at a rate of 11300.0 cubic centimeters per min at the same time that water is Problems being pumped into the tank at a constant rate. The tank has height 11.0 meters and the diameter at the top is 3.0 meters. If the water level is rising at a rate of 20.0 centimeters per minute when the height of the water is 1.0 meters, find the rate at which water is being Problem 1 pumped into the tank in cubic centimeters per minute. Problem 2 dV Then you know that if Vis volume of water, 11300.0 Problem 3 Note: Let "R" be the unknown rate at which water is being pumped in Problem 4 Use geometry (similar triangles?) to find the relationship between the height of the water and the volume of the water at any given time. Recall that the volume of a cone with base radius r and height his given by nr2h Problem 5 Problem 6 Problem 7 Preview Answers Submit Answers Problem 8 You have attempted this problem 0 times. You have unlimited attempts remaining Display Options View equations as Email instructor ages Math Jax Show saved answers? Yes No Use Equation Editor? Yes No Apply options

Explanation / Answer

First, we need everything in common terms (centimeters)

H = 11 m = 1100 cm
D = 3 m = 300 cm

h = 1m = 100 cm


Now, we need to set up a formula for volume:

V = (1/3) * pi * r^2 * h

Now, we need to find r when h = 100

r = 0 when h = 0
r = 150 when h = 1100

(1100 - 0) / (150 - 0) = 110 / 15 = 22 / 3

h = (22/3) * r
r = (3/22) * h

V = (1/3) * pi * r^2 * h
V = (1/3) * pi * (3/22)^2 * h^2 * h
V = (1/3) * pi * (9/484) * h^3
dV/dt = 3 * (1/3) * pi * (9/484) * h^2 * dh/dt
dV/dt = (9 * pi / 484) * h^2 * dh/dt

Now, we know that 11300 cm^3 of water is draining out every minute, so we need to modify our volume derivative:

dV/dt = (9 * pi / 484) * h^2 * dh/dt - 11400

h = 100
dh/dt = 20
dV/dt = ?

dV/dt = (9 * pi / 484) * 100^2 * 20 - 11300
dV/dt = (10000 * 20 * 9 * pi - 484 * 11300) / 484
dV/dt = 383.609

So, 383.609 cm^3 is being pumped in every minute.

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