# Mega Help on Related Rates (Calculus)

• Oct 7th 2007, 07:40 PM
Super Mallow
Mega Help on Related Rates (Calculus)
I have 4 problems. I have literally spent 5-6 hours on this, asked a Calculus III student for help, etc. I cannot get these answers or even get close to them.

Obviously, I'm in a need of a lot of help, which any is appreciated. I legitimately tried this, I'm not looking for an easy way out, but I do need to get solutions to them in order to understand them by tomorrow morning

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?

a)-1/24Pi
b)Square Root of 26y - y^2
c) 5/288Pi

2) A balloon is rising vertically above a level, straight road at a constant rate of 1 ft/sec. Just when the balloon is 65 feet about the ground, a bicycle moving at a constant rate of 17 ft/sec passes under it. How fast is the distance s(t) between the bicycle and the balloon increasing 3 sec later?

11

3) Coffee is draining from a conical filter nto a cylindrical coffee pot at the rate of 10 in^3/min.
a) How fast is the level in the pot rising when the coffee in the cone is 5 inches deep?
b) How hast is the level in the cone falling then?

[A diagram on the page shows the height and diameter of the cone is 6 inches; diameter of pot is also 6 inches]

4) You are videotaping a race from a stand 123 feet away from the track, following a car that is moviing at 180 mi/h (264 ft/sec). How fast will your camera angle theta be changing when the car is right in fron of you? A half second later?

Like I said, I really did try and try to get help, but my Calculus III buddy and I could not figure this out. I really am in some need of help.

Huge thanks to anyone in advance for any help
• Oct 8th 2007, 02:06 AM
ticbol
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?

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