Expert: Scotto - 7/30/2009

Question
QUESTION: I'm helping someone at yahoo! answers with this question:

Water is flowing into a hemispherical bowl of radius 10 cm at constant rate. When the height of the water in the bowl is h cm. The volume of water is pi/3(30h^2-h^3)cm^3. Given that the bowl is completely filled with water in 20 seconds, find the rate at which
i)The water is flowing into the bowl
*ii) the water level is rising when halfway to the top

Any help appreciated, thank ^^

I came up with :

i) volume= rate * time
let r=rate
the height is the radius, so
pi/3(30h^2-h^3)cm^3 = r* 20 sec
pi/3[30(10)^2-(10)^3]cm^3 = r * 20 sec
pi/3[30(100)-(1000)]cm^3 = r* 20 sec
pi/3[3000-1000]cm^3 = r * 20 sec
2000 pi/3 cm^3 = r * 20 sec
r = [2000 pi/3 cm^3]/20 sec = 100 pi/3 cm^3/sec
ii) when the height is 5,
the volume is pi/3(30h^2-h^3)cm^3 = pi/3[30(5)^2-(5)^3]cm^3 = pi/3[30(5)^2-(5)^3]cm^3 = pi/3[30(25)-125]cm^3 = pi/3[750-125]cm^3 =
625 pi/3 cm^3

let T=time passed at height of 5.

final volume/(partial volume) = 20 sec/T sec

2000 pi/3 cm^3/625 pi/3 cm^3 = 20/T
2000/625 = 20/T
3.2 = 20/T
T= 20/3.2 = 5.25 sec

then rate of rising= 625 pi/3 cm^3/5.25 sec

Is this correct? Please let me know. Thank you.

Sincerely,
Jason G.



ANSWER: The bowl is the bottom half of a sphere.
This means the total volume V is half that of a sphere, or ½ (4πr³)/3, which is 2πr³/3,
Since we are told that it fills in 20 sec, that would indicate that there πr³/30 cm/s.

There is a right triangle between the vertical line in the center of the sphere, the horizontal line from the center to the outside d, and they hypotenuse connecting the two, which is the radius r of the sphere.  We are told that it is halfway full.  This means that the vertical distance is 5.

Thus, 5² + d² = 10².

Since the volume V = π(30h² - h²)/3, dV/dh = π(20h – h²).

Note this is the amount of liquid.  It starts w flow 0 and would end with flow 0 if we had a complete sphere.  However, we only have half a sphere.

i) Since V = (2πr²/3) cm³, and it takes 20 sec to fill, the rate = volume/time = (2πr³/3)/20, and r = 10.
Thus, rate = (2π(10³)/3)/20 = (2000π/3).

ii) The level at which it is rising = (incoming flow)/(surface area).

To get the suface area, we need the radius at this height, which will do right now.

There is a right triangle with 5 as the height, d as it’s horizontal length from the center to the edge, and the radius of the bowl as the diagonal.  The radius is known to be 10.  This means t² + d² = 10² => 25 + d² = 10² => d² = 75 => d = √75.  Now to take the increase/area = (200π/3)/(75) = 8/9 cm/s.

I would have gone slower, but it is far to hot near this computer (approaching 100° F), so I will just send this.  I'm not even proof-reading the document, but I am sweating as I sit here.  Please right back with any corrections.


---------- FOLLOW-UP ----------

QUESTION: Hi. I just corrected your errors:

2.The bowl is the bottom half of a sphere.
This means the total volume V is half that of a sphere, or ½ (4πr³)/3, which is 2πr³/3,
Since we are told that it fills in 20 sec, that would indicate that there πr³/30 cm/s.

There is a right triangle between the vertical line in the center of the sphere, the horizontal line from the center to the outside d, and they hypotenuse connecting the two, which is the radius r of the sphere. We are told that it is halfway full. This means that the vertical distance is 5.

Thus, 5² + d² = 10².

Since the volume V = π(30h² - h³)/3, dV/dh = π(20h – h²).

Note this is the amount of liquid. It starts w flow 0 and would end with flow 0 if we had a complete sphere. However, we only have half a sphere.

i) Since V = (2πr³/3) cm³, and it takes 20 sec to fill, the rate = volume/time = (2πr³/3)/20, and r = 10.
Thus, rate = (2π(10³)/3)/20 = (2000π/3)/20 = 100π/3

Or since h=10, the volume is π(30h² - h³)/3 = π[30(10)² - 10³]/3 = π[30(100) - 1000]/3 = π[3000 - 1000]/3 = (2000π/3)
and the rate = (2000π/3)/20 = 100π/3

ii) The level at which it is rising = (incoming flow)/(surface area).

To get the surface area, we need the radius at this height, which will do right now.

There is a right triangle with 5 as the height, d as it’s horizontal length from the center to the edge, and the radius of the bowl as the diagonal. The radius is known to be 10. This means t² + d² = 10² => 25 + d² = 10² => d² = 75 => d = √75. The surface area at d is πd². Now to take the increase/area = (100π/3)/(75π) = 4/9 cm/s.

Thank you for your help.

Answer
So I see that you have it.
Your welcome.

I'll gladly answer any more.  It is getting better today - it's only going to get to the 90's.
Yesterday was the all time record for any time of the year.  I believe it got up to 102° at the airport.  Now that's exactly 56° centigrade (or celsius, as they seem to be calling it nowadays).

Since I took chemistry in college, its centigrade, since cent, in latin, is for 100.
Think about it - 1 cent is 1/100th of a dollar, a centimeter is 1/100th of a meter, a centiliter is 1/100th of a liter, a century is 100 years.  In centigrade, there are 100 degrees between freezing and boiling.

Doesn't this make some sense?


While I'm on the subject, do you know we're all worth something?
I mean, everyone I know has some kind of common cents ...

I've one last question - what's the same about a buck and something that smells good?

Answer: A buck has a hundered and the smelly stuff has good in.
That is a hundreds cents and good insense - makes sense, right?