Water is slowly leaking out of a bowl at a rate of pi in3/min, as shown in the f

Question

Water is slowly leaking out of a bowl at a rate of pi in3/min, as shown in the figure below. The bowl is a perfect hemisphere with radius 10 inches. If the water level is y inches, the volume of water in a hemispherical bowl of radius R is V = pi / 3 y2(3R- y). At what rate is the water level changing when the water is 8 inches deep? What is the radius r of the water's surface when the water is y inches deep? When the water level is 4 inches deep, the water level is changing at a rate of 1/64 inches per minute. At what rate is the radius of the water changing at that rime?

Explanation / Answer

a)dV/dt= pi/3 2 y dy/dt ( 3R -y) + pi/3 y^2 (-dy/dt)

-pi = pi/3 *2*8*dy/dt*(3*10-8) - pi/3*8^2 dy/dt

dy/dt = -1/96 = -0.0104

b)

using pythagoreon theorem

(R -y)^2 + r^2 = R^2
r = sqrt(R^2 - (R-y)^2)

c)
use (R-y)^2 + r^2 = R^2

take derivative

2(R-y)(-dy/dt) + 2 r dr/dt = 0
2*(10-4)*(1/64) + 2*sqrt(10^2 - (10-4)^2)*dr/dt = 0
dr/dt=-3/256

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