Expert: Paul Klarreich - 7/30/2009

Question
HI. I was helping this person with a question on Yahoo! Answers:

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

I came up with:

) 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 right? Please let me know. Thank you.

Jason G.


Answer
Questioner: Jason
Country: United States
Category: Calculus
Private: No
Subject: Help with this problem
Question: HI. I was helping this person with a question on Yahoo! Answers:

Water is flowing into a hemispherical bowl of radius 10 cm at A constant rate. When the height of the water in the bowl is h cm COMMA the volume of water is pi/3(30h^2-h^3)cm^3.

>> I'll take your word for it.

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

I came up with:

) 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 right? Please let me know. Thank you.

Jason G.

........................................
V = pi/3(30h^2-h^3)

Now when  h = r = 10,

V = pi/3(30(100) - 1000)

V = pi/3(3000 - 1000)

V = pi/3(2000)

If it takes 20 seconds to fill the bowl, V = 20r

pi/3(2000) = 20r
................
r = 100 pi/3 = dV/dt
................
The rest is a routine related rates example:

V = pi/3(30h^2-h^3)

dV/dt = pi/3(60h - 3h^2) dh/dt

dV/dt = pi(20h - h^2) dh/dt

Now substitute  dV/dt = 100pi/3,  and  h = 5:

100pi/3 = pi(20(5) - (5)^2) dh/dt

100pi/3 = pi(100 - 25) dh/dt

100pi/3 = 75 pi dh/dt

4/3 = 3 dh/dt

dh/dt = 4/9