Thread: Water flowing into hemisphere rates

Water is flowing into a hemispherical container of radius 10cm at a constant rate of 6cm³ per second. It is known that the forumla for the volume of a solid segment cut off a sphere is

where r is the radius of the sphere and h is the height of the segment.

a. Find the rate at which the height of the water is rising when the water height is 2cm.

b. Find the radius of the cirucular water surface when the height is h, and hence find the rate of increase of the surface area when the water height is 2cm.

I have managed to get through part a with no problem, but am totally stuck on part b. The book claims that



and



I tried doing it using similar triangles and got nothing at all like that
thanks
jacs

Originally Posted by jacs

Water is flowing into a hemispherical container of radius 10cm at a constant rate of 6cm³ per second. It is known that the forumla for the volume of a solid segment cut off a sphere is

where r is the radius of the sphere and h is the height of the segment.

a. Find the rate at which the height of the water is rising when the water height is 2cm.

b. Find the radius of the cirucular water surface when the height is h, and hence find the rate of increase of the surface area when the water height is 2cm.

I have managed to get through part a with no problem, but am totally stuck on part b. The book claims that



and



I tried doing it using similar triangles and got nothing at all like that
thanks
jacs

a) the dh/dt when h=2cm.

V = (pi/3)(h^2)[3r -h] ---the r is not changing, is a constant.
V = (pi/3)[3*10*h^2 -h^3]
V = (pi/3)[30h^2 -h^3]
Differentiate both sides with respect to time t,
dV/dt = (pi/3)[60h*dh/dt -(3h^2)*dh/dt]
Since dV/dt is 6cc/sec, and h=2cm at the time in question, then,
6 = (pi/3)[60*2*dh/dt -3*2^2 *dh/dt]
6 = (pi/3)[120 -12]dh/dt
6 = (108pi/3)dh/dt
dh/dt = 6 /(108pi/3) = 18/(108pi) = 1/(6pi) cm/sec ---------answer.

---------------
b.1) The radius of the water surface when depth of water is h.

Imagine, or draw the figure on paper.
It is a semicircle, with the water level being below the diameter. Call the radius of the water level as R, to differentiate or make it different from the radius r of the semicircle. Draw a radius r to one end of the water level. A right triangle is formed, with these:
---hypotenuse = r
---vertical leg = r-h
---horizontal leg = R
By Pythagorean Theorem,
r^2 = (r-h)^2 +R^2
R^2 = r^2 -(r-h)^2
The righthand side is in the form a^2 -b^2 which is (a+b)(a-b),so,
R^2 = (r +(r-h))(r -(r-h))
R^2 = (2r -h)(h)
R^2 = 2rh -h^2 ---------------***
R = sqrt(2rh -h^2) --------------------------answer.

------------
b.2) The dA/dt of the water surface when h=2cm.

A = pi(R^2)
A = pi(2rh -h^2) ----where r is 10cm constantly.
A = pi(20h -h^2)
Differentiate both sides with respect to time t,
dA/dt = pi(20*dh/dt -2h*dh/dt)
Since h=2cm, and dh/dt = 1/(6pi) cm/sec, at the time in question, then,
dA/dt = pi[20*1/(6pi) -2*2*1/(6pi)]
dA/dt = pi[20/(6pi) -4/(6pi)]
dA/dt = pi[16/(6pi)]
dA/dt = 8/3 sq.cm/sec ------------answer.

Thank you so much. I would have never seen that (despite the subtraction in a square root being a big neon sign screaming Pythagoras at me....lol)


It all falls together beautifully now and it makes perfect sense...you made it look so easy.


once again, my thanks and appreciation

jacs