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Posers and Puzzles

Posers and Puzzles

  1. Subscriber joe shmo On Vacation
    Strange Egg
    13 Oct '08 01:19 / 3 edits
    I haven't done these types of problems yet, so the math I have done on it so far may be incorrect, but i'll let you be the judge of that.

    This is just for fun

    Ok, here it is broken into two parts

    start with a cube(A) of side length x, nested inside a cube(B) of side lenght (x+1), at what side lengths will their rates of change in volume be equal

    Given: dx(A)/dt(A) = 3 units per second
    and dx(B)/dt(B) = 2 units per second

    start by differentiating

    dV(A)/dt(A) = (3x^2)( dx(A)/dt(A))
    &
    dV(B)/dt(B) = (3x^2 + 6x + 3 )(dx(B)/dt(B))

    then set dV(A)/dt(A) = dV(B)/dt(B) and solve for x

    9x^2 = 6x^2 + 12x + 6

    3x^2 - 12x -6 = 0

    x^2 -4x - 2 = 0

    using the quadratic formula I come up with

    2 + sqrt(6).

    ruling out the negative solution.

    so the side lenght of (A) = 2 + sqrt(6)

    and Side lenght of (B) = 3 + sqrt(6)

    Now, my second question. If i were to ask, what time will the rates of volume be the same, assuming start times were the same, would i need to add another piece of information, like one of the original starting volumes to deduce this?
  2. Subscriber joe shmo On Vacation
    Strange Egg
    13 Oct '08 01:31
    Originally posted by joe shmo
    I haven't done these types of problems yet, so the math I have done on it so far may be incorrect, but i'll let you be the judge of that.

    This is just for fun

    Ok, here it is broken into two parts

    start with a cube(A) of side length x, nested inside a cube(B) of side lenght (x+1), at what side lengths will their rates of change in volume be equal
    ...[text shortened]... to add another piece of information, like one of the original starting volumes to deduce this?
    if my notation doesnt make sence, I'll be happy to clarify what I thought it meant..
  3. Standard member wolfgang59
    Infidel
    13 Oct '08 14:21
    I cant follow this. If the rates of change of the sides are constant then the rates of change of the volumes will be constant. Namely 8 and 27. They are never going to be equal!

    The volume (V) is equal to x^3 so

    DV/dx = 3x^2

    DV/dt = DV/dx * dx/dt
    = (3x^2) * 3
    = 9x^2

    I'm unclear as to the realtionship between the two cubes - Is the bigger cube constantly 1 unit larger than the smaller?

    ie does y = x+1 (y is side length of larger cube)
  4. Subscriber joe shmo On Vacation
    Strange Egg
    13 Oct '08 23:03
    Originally posted by wolfgang59
    I cant follow this. If the rates of change of the sides are constant then the rates of change of the volumes will be constant. Namely 8 and 27. They are never going to be equal!

    The volume (V) is equal to x^3 so

    DV/dx = 3x^2

    DV/dt = DV/dx * dx/dt
    = (3x^2) * 3
    = 9x^2

    I'm unclear as to the realtionship between the two cubes - ...[text shortened]... constantly 1 unit larger than the smaller?

    ie does y = x+1 (y is side length of larger cube)
    sorry, I think I meant to ask, when are their rates of change in volume with respect to time equal

    And yes, the side of the larger cube is (x+1)

    so the two volumes are V= x^3 & V = (x+1)^3

    the cubes are growing at different rates relative to each other, so the larger cube isnt always 1 unit larger

    given enough time V= x^3 should become greater than V = (x+1)^3 right?
  5. Standard member wolfgang59
    Infidel
    14 Oct '08 07:45
    Originally posted by joe shmo
    sorry, I think I meant to ask, when are their rates of change in volume with respect to time equal

    And yes, the side of the larger cube is (x+1)

    so the two volumes are V= x^3 & V = (x+1)^3

    the cubes are growing at different rates relative to each other, so the larger cube isnt always 1 unit larger

    given enough time V= x^3 should become greater than V = (x+1)^3 right?
    How can x^3 be greater than (x+1)^3 ????

    Your problem I think is that you are using x as a variable as well as using it as a discrete value.
  6. Subscriber joe shmo On Vacation
    Strange Egg
    14 Oct '08 08:26
    Originally posted by wolfgang59
    How can x^3 be greater than (x+1)^3 ????

    Your problem I think is that you are using x as a variable as well as using it as a discrete value.
    yeah, your right about that..all logic escapes me somtimes. haha

    This may seem like a strange question to ask but, I have a hunch that you are right about me using a discrete value. Can you explain to me, if possible, what you think the connection I'm making is, and its flaw?

    Its probably best to use examples to answer this, if you are willing to try it.
  7. Standard member wolfgang59
    Infidel
    14 Oct '08 12:08 / 1 edit
    You have 2 variables which vary with time; the length of small cube (lets call it 'a' and the length of the larger (lets call that 'b'

    Lets call their respective volumes A and B.

    We are given that

    da/dt = 3; a=3t + C (where C is a constant)
    db/dt = 2; b=2t + K (where K is a constant)
    initial value t=0, a=x, b=x+1

    At time t, a=C = x and b=K=x+1

    Therefore a=3t+x and b=2t+x+1


    A=a^3; DA/da = 3a^2

    B=b^3; DB/db = 3b^2

    We want to find what t equals when dA/dt = dB/dt

    dA/dt = dA/da * da/dt = (3a^2) * 3
    dB/dt = dB/db * db/dt = (3b^2) * 2

    So what is t when 9a^2 = 6b^2 ?
    Substitute for a and b.

    3 (3t + x)^2 = 2 (2t+x+1)^2

    Then solve for t in terms of x.
    REMEMBER that x IS NOT A VARIABLE!!!
  8. Subscriber joe shmo On Vacation
    Strange Egg
    14 Oct '08 18:30
    Originally posted by wolfgang59
    You have 2 variables which vary with time; the length of small cube (lets call it 'a' and the length of the larger (lets call that 'b'

    Lets call their respective volumes A and B.

    We are given that

    da/dt = 3; a=3t + C (where C is a constant)
    db/dt = 2; b=2t + K (where K is a constant)
    initial value t=0, a=x, b=x+1

    At time t, a=C = x and b= ...[text shortened]... x)^2 = 2 (2t+x+1)^2

    Then solve for t in terms of x.
    REMEMBER that x IS NOT A VARIABLE!!!
    thanks wolfgang, That is ( as far as i can tell ) the mathematics I was thinking of, or at least, trying to think of!!!LOL

    I love, and at the same time hate, how this site totally distracts me from the problems I should be doing. haha

    thanks again
    Eric