It's good that you posted them (4 new Problems) in this new thread.

But why nobody touched any of them yet?

Because they are too many.

One, if the Problem needs long solution, or two per posting only.

Some of us will not solve a Problem if we cannot solve the others too.

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1) Water is flowing at a rate of 6 m^3/min from a reservoir shaped like a hemispherical bowl of radius 13 m, shown here in profile. Answer the following questions, given that the volume of water in a hemispherical bowl of radius are is V=(Pi/3)y^2(3R-y) when the water is Y meters deep

a) At what rate is the water level changing when the water is 8 meters deep?

b) What is the radius r of the water's surface when the water is y meters deep?

c) At what rate is the radius r changing when the water is 8 meters deep?

**Answers from Book**

a)-1/24Pi

b)Square Root of 26y - y^2

c) 5/288Pi

a) At what rate is the water level changing when the water is 8 meters deep?

We are to find dy/dt.

Given:

R = 13m

dV/dt = 6 cu.m/min.

V = (pi/3)(y^2)(3R -y)

V = (pi/3)(y^2)(3*13 -y)

V = (pi/3)(39y^2 -y^3)

Differentiate both sides with respect to time t,

dV/dt = (pi/3)[78y dy/dt -3y^2 dy/dt]

6 = (pi/3)[78*8 dy/dt -3(8^2)dy/dt]

6 = (pi/3)[432 dy/dt]

6 = 144pi dy/dt

dy/dt = 6/(144pi) = 1/(24pi) m/min --------------answer.

b) What is the radius r of the water's surface when the water is y meters deep?

If r is horizontal, s is vertical, and (0,0) is the center of the vertical circle of the hemispherical bowl, then the equation of that circle is

r^2 +s^2 = 13^2

If y is the depth of the water, then the water surface is (13 -y) from (0,0).

Or, s = (13 -y)

So,

r^2 +(13 -y)^2 = 13^2

r^2 +13^2 -26y +y^2 = 13^3

r^2 -26y +y^2 = 0

r^2 = 26y -y^2

r = sqrt(26y -y^2) m ------the radius of the water surface.

c) At what rate is the radius r changing when the water is 8 meters deep?

r = sqrt(26y -y^2)

r = [26y -y^2]^(1/2)

dr/dt = (1/2)[26y -y^2]^(-1/2) *[26dy/dt -2y(dy/dt)]

dr/dt = [13 dy/dt -y dy/dt] / sqrt[26y -y^2]

dr/dt = [13(1/(24pi)) -8(1/(24pi))] / sqrt[26*8 -8^2]

dr/dt = 5/(24pi) / sqrt[144]

dr/dt = 5/(24pi) /12

dr/dt = 5/(288pi) m/min ----------------answer.

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For the other 3 Problems, try posting them separately in new threads. I will answer them there if I have time, and if nobody answered them yet when I have the chance.