Results 1 to 3 of 3

Math Help - How to set this volume problem?

  1. #1
    Member
    Joined
    Dec 2008
    Posts
    228

    How to set this volume problem?

    Let V be the volume of the region bounded between two expanding concentric spheres. at time t = 0 the spheres have radii of 6 cm and 40 cm. The radius of the smaller sphere is increasing at a rate of 8 cm/sec, while the radius of the larger sphere is increasing at a rate of 2 cm/sec. At what value of t (t>0) will the volume V be a maximum?

    I had no idea to set this problem up because I don't know how to install the variable "t". The book set up the equation as v = (4/3)pi(40+2t)^3 - (4/3)pi(6+8t)^3.

    I'm just hoping that someone can make sense out of the equation, how the radii were replaced by expressions with "t".
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member sinewave85's Avatar
    Joined
    Oct 2006
    From
    Lost in a series of tubes.
    Posts
    218
    Quote Originally Posted by Kaitosan View Post
    Let V be the volume of the region bounded between two expanding concentric spheres. at time t = 0 the spheres have radii of 6 cm and 40 cm. The radius of the smaller sphere is increasing at a rate of 8 cm/sec, while the radius of the larger sphere is increasing at a rate of 2 cm/sec. At what value of t (t>0) will the volume V be a maximum?

    I had no idea to set this problem up because I don't know how to install the variable "t". The book set up the equation as v = (4/3)pi(40+2t)^3 - (4/3)pi(6+8t)^3.

    I'm just hoping that someone can make sense out of the equation, how the radii were replaced by expressions with "t".
    Hi, Kaitosan.

    First, lets look at how they introduced t into the equation. We are told, "At time t = 0 the spheres have radii of 6 cm and 40 cm. The radius of the smaller sphere is increasing at a rate of 8 cm/sec, while the radius of the larger sphere is increasing at a rate of 2 cm/sec." That means that you can describe the radii of the circles in terms of t since the radius is a function of time. Let R be the radius of the eternal circle and r be the radius of the internal circle.

    R = 40 cm(the starting radius) + 2 cm/sec (the rate of expansion) *t sec(the time) = 40 + 2t

    r = 6 cm(the starting radius) + 8cm/sec (the rate of expansion) *t sec(the time) = 6 + 8t

    Those values for R and r are substituted into the standard equations for the volume of a circle.

    V = \frac{4}{3}\pi R^3 - \frac{4}{3}\pi r^3

    V = \frac{4}{3}\pi(40 + 2t)^3 - \frac{4}{3}\pi(6 + 8t) ^3

    To find the maximum value for V, you need to differentiate that equation in order to find and test the critical numbers.

    V^{\prime} = \frac{4}{3}\pi(3)(40 + 2t)^{(3 -1)}(2) - \frac{4}{3}\pi(3)(6 + 8t)^{(3 - 1)}(8)

    V^{\prime} = 8\pi(40 +2t)^{2} - 32\pi(6 + 8t)^{2}

    0 = 8\pi(40 +2t)^{2} - 32\pi(6 + 8t)^{2}

    0 = -63t^2 -56t + 364

    t = -2.8889, 2 \mbox{ by the quadratic formula}

    The only critical value of t > 0 is 2, and testing will confirm that V is maximized at t = 2 seconds. Hope that helps.
    Last edited by sinewave85; April 8th 2009 at 12:12 PM. Reason: Noticed a few typos
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Dec 2008
    Posts
    228
    Thanks, that helps. I'm just not used to that kind of problem.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: January 10th 2009, 06:49 AM
  2. problem regarding volume
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: November 4th 2008, 08:07 PM
  3. Another Volume Problem
    Posted in the Calculus Forum
    Replies: 4
    Last Post: October 4th 2008, 03:53 PM
  4. volume problem
    Posted in the Calculus Forum
    Replies: 3
    Last Post: June 6th 2008, 11:05 PM
  5. volume problem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: December 9th 2007, 04:43 PM

Search Tags


/mathhelpforum @mathhelpforum