project

i went back and look over it here is my work so far im confused by what is negative a
pic math.jpg
 
OK - the height is measured from the bottom of the sphere and you got the proper radius r
r2 = 22 - (2-h)2 = 4h - h2,
and the proper cross sectional area. You are now at
\(\displaystyle \pi\) (4h - h2) h' = -a \(\displaystyle \sqrt{20\, h}\)

a is just the cross sectional area of the hole but you have to be careful about the units. Since h is in meters, we should have the radius of the hole in meters also. Note that 1 meter is 100 cm.
 
OK - the height is measured from the bottom of the sphere and you got the proper radius r
r2 = 22 - (2-h)2 = 4h - h2,
and the proper cross sectional area. You are now at
\(\displaystyle \pi\) (4h - h2) h' = -a \(\displaystyle \sqrt{20\, h}\)

a is just the cross sectional area of the hole but you have to be careful about the units. Since h is in meters, we should have the radius of the hole in meters also. Note that 1 meter is 100 cm.

so would have to converted radius of the circular hole is 1 cm
into meters which is 0.01 and what would happen to pi would you divide out by both sides?
 
so would the final equation be (4h-h^2)h'=0.01/pi square 20h
No. But you are correct in that the radius of the hole is 0.01 m. However, a is the area of the hole, not the radius. What is the area of a circle with 0.01 m radius
 
so after we find a which is (4h-h2)h'=-0.0001sqaured20h would you have to take the derivative in order to solve for b which is "how long will it take for the water to completely drains? the reason im asking it doesnt give you the time so would it be 0 because im confused on how to start it and dont know where to begin
 
i still dont understand does it have something to do with in the begin of the problem
 
i still dont understand does it have something to do with in the begin of the problem
Yes, it does. You now have the formulas
\(\displaystyle A(h)\, \frac{dh}{dt}\, =\, -0.0001\, \sqrt{20h}\)
\(\displaystyle A(h)\, =\, 4h\, -\, h^2\)
\(\displaystyle V(h)\, =\, \underset{0}{\overset{h}{\int}}\, A(u)\, du\)

And you are given that the spherical tank is half full of water at time zero [its volume is half that of the corresponding sphere]. You should be able to solve for h as a function of t [actually t as a function of h] and V as a function of h. Thus obtain when it will be empty
 
i having trouble with problem 3 and i have notes i think i mess up somewhere
 
disregard the question about 3 sorry i have another question about 4 that is im different and my equation look like these

(4h-h2)(20h)-1/2=-0.0001t+C

we Distributive the 20h square correct and i already know what the t is 0 and h=2.
 
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disregard the question about 3 sorry i have another question about 4 that is im different and my equation look like these

(4h-h2)(20h)-1/2=-0.0001t+C

we Distributive the 20h square correct and i already know what the t is 0 and h=2.

So what is your question?
 
my question is distributive the 20h^-1/2 to (4h-h^2) is right because im pretty sure you do it just my friend did his a little differently then me im trying to compare
 
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