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Homework Help: Related Rates Problem: Cube

  1. Apr 19, 2007 #1
    1. The problem statement, all variables and given/known data

    The side of a cube increases at 1 cm / s. How fast is the diagonal of the cube changing when the side is 1 cm?

    2. Relevant equations


    [itex] a^2+b^2=c^2 [/itex]
    Implicit Differentiation

    3. The attempt at a solution

    I'm attempting to find the diagonal of the cube through Pythagorean Theory where I find the diagonal of one side of the cube to give me an equation for side length 'A', after finding side length 'A' I know that the height of the cube is 'x' centimeters so I can then again use Pythagorean Theory to find the equation of the diagonal going through the cube.

    So I set all sides of the cube equal to 'x' and solve for a diagonal of one of the sides of the cube in order to get a side length ( 'A' ) of the diagonal within the cube. Using Pythagorean I come to the equation of:


    [itex] \sqrt2x^2 = C [/itex]

    After that I know that the height of the cube is 'x' so once again using Pythagorean I can derive an equation for the diagonal through the cube.

    [itex](\sqrt2x^2)^2 + x^2 = D^2[/itex]

    [itex]3x^2 = D^2[/itex]
    [itex]\sqrt3x^2 = d[/itex]

    Now this is where I become confused in order to find the rate of change at 1 cm. Do I then find the derivative of [itex]\sqrt3x^2 = D[/itex] through implicit differentiation?


    [itex]D = \sqrt3x^2[/itex]
    [itex]D' = (3x)/\sqrt3x^2 (dLength/dt)[/itex]

    ... I'm rather lost on what to do from here. Any help would be greatly appreciated.
    Last edited: Apr 19, 2007
  2. jcsd
  3. Apr 19, 2007 #2
    Should this be in this thread or Calculus?? Just for next time.
  4. Apr 19, 2007 #3
    I think you are there dD/dt=6x*dx/dt
    lets try it for 1.05, using your eqn for D at x=1.05, D=3.3075 deltaD=.3075
    using formula where x=1 and dx=.05 6*.05=.30 in good agreement in spite of a 5 percent change.
  5. Apr 19, 2007 #4
    Sorry, I think that my final equation was wrong ... was I right or am I right now??
  6. Apr 19, 2007 #5
    I thought you were right before.
  7. Apr 19, 2007 #6
    Well I didn't have dD/dt=6x*(dx/dt) because I had differentiated incorrectly I believe.

    My final equation should be:

    [itex](dD/dt) = (3x)/(\sqrt(3x^2))*(dx/dt) [/itex]
  8. Apr 19, 2007 #7
    oh , i see my blunder,,
    D^2=3x^2. hence D=x(sqrt(3))
    so dD/dt=1.73(dx/dt) not sure how you came up with what you posted.
  9. Apr 19, 2007 #8
    I seem to possess that skill with these problems .. early today I proved that 0 = -37 hahaha ... oh i hate when I forget that sqrt can be negative. No it's alright, I have solved the problem on my own .. thank you so much though!
  10. Apr 19, 2007 #9
    hey I'm struggling with the same issue on another thread. Serendipity must have put on here tonite on this thread. I like that square roots of negative numbers exist, it provides symmetry to the universe and an endless demand for textbooks as well...:rolleyes: and just last nite, on another thread over in intro physics that forgetting this set up a logical contradiction that neither student nor teacher could reconcile.
    Last edited: Apr 19, 2007
  11. Apr 20, 2007 #10


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    To this point you are correct. However, when you take the square root of both sides you do NOT get
    but rather
    [tex]\sqrt{2}x= C[/tex]
    Or did you mean for that to be [itex]\sqrt{2x^2}[/itex]. If so, put everything you want inside the square root in "{ }" braces. But for positive x, [itex]\sqrt{x^2}= x[/itex] anyway.

    Again, that should be
    [tex]\sqrt{3}x= D[/tex]
    (Also, don't switch from "D" to "d" in the middle!)

    Yes, you are getting quite lost. For one thing, where did that "Length" come from? Don't forget what you variables stand for. You have already decided to use "x" as the length of a side. Assuming you mean [itex]\sqrt{3x^2}[/itex] then write it as [itex](3x^2)^{1/2}[/itex] and its derivative is [itex](1/2)(3x)^{-1/2}= 3x/\sqrt{3x^2}[/itex] as you have. However, [itex]3/\sqrt{3}= \sqrt{3}[/tex] and [itex]x/\sqrt{x^2}= x/x= 1[/itex] as long as x is positive, as it is here. That is, you have [itex]D'= \sqrt{3}[/itex]. (Had you written [itex]\sqrt{3x^2}[/itex] as [itex]\sqrt{3}x[/itex], that would have been obvious!) By the way, since D is written as an explicit function of x (D= f(x)), that is not "implicit" differentiation.

    You want to find how fast the diagonal is changing: that is dD/dt and you know how fast the length of a side is changing: dx/dt. The "obvious" way to do that is to start with an equation involving D and x (and D= \sqrt{3}x will work fine!) and differentiate both sides with respect to t. That would involve the chain rule rather than implicit differentiation: if D= f(x), then
    dD/dt= dD/dx (dx/dt). In other words, the equation you really need is
    dD/dt= \sqrt{3} (dx/dt). Okay, if dx/dt= 1 cm/s what is dD/dt?
    Last edited by a moderator: Apr 20, 2007
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