Results 1 to 3 of 3

Math Help - Water flowing into hemisphere rates

  1. #1
    Member jacs's Avatar
    Joined
    Jan 2006
    From
    Sydney
    Posts
    107

    Water flowing into hemisphere rates

    Water is flowing into a hemispherical container of radius 10cm at a constant rate of 6cm per second. It is known that the forumla for the volume of a solid segment cut off a sphere is
    <br />
V = \frac{\pi}{3}h^2(3r - h)<br />
    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

    <br />
r= \sqrt{2hr - h^2} <br />

    and

    <br />
\frac{8}{3}cm^2/s<br />

    I tried doing it using similar triangles and got nothing at all like that
    thanks
    jacs
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Apr 2005
    Posts
    1,631
    Quote Originally Posted by jacs
    Water is flowing into a hemispherical container of radius 10cm at a constant rate of 6cm per second. It is known that the forumla for the volume of a solid segment cut off a sphere is
    <br />
V = \frac{\pi}{3}h^2(3r - h)<br />
    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

    <br />
r= \sqrt{2hr - h^2} <br />

    and

    <br />
\frac{8}{3}cm^2/s<br />

    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.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member jacs's Avatar
    Joined
    Jan 2006
    From
    Sydney
    Posts
    107
    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
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. rate of flowing water
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 30th 2010, 03:05 PM
  2. Replies: 3
    Last Post: March 20th 2010, 04:18 PM
  3. related rates, water running into trough
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 11th 2009, 04:30 PM
  4. Replies: 6
    Last Post: September 20th 2009, 03:53 PM
  5. Replies: 7
    Last Post: February 6th 2009, 02:45 PM

Search Tags


/mathhelpforum @mathhelpforum