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Homework Help: Area of a square

  1. Jan 11, 2009 #1

    I hope I got 8 and 9 right. I am however stuck on 10. I can't seem to figure out how to convert x or a side to perimeter and represent the change in area with respect to time.
  2. jcsd
  3. Jan 11, 2009 #2
    Now that I look at it If I solve for da/dt I should get 2x and if I plug the constant rate increase of one side "x" in I get .8 And since the Perimiter is just all 4 sides added up I should get a forth of the x or .2 (b).
  4. Jan 11, 2009 #3


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    Gold Member

    Are you familiar with other related rates problems? You are given dx/dt and need to find dA/dt. Use the fact that dA/dt = (dA/dP)(dP/dt) to solve the problem. Remember P = 4x and A = x^2.

    Edit: I think your answer is right.
  5. Jan 11, 2009 #4
    ok thanks
  6. Jan 11, 2009 #5
    How could I go about setting problem 11 up? I messed up when I wrote change in volume when it was just volume because I haven't taken the derivative yet?
  7. Jan 12, 2009 #6


    Staff: Mentor

    You have the rote mechanics of taking derivatives of simple functions, but it's evident that you don't understand the chain rule. For problem 11, you have dV/dt = 4/3 pi r^3, which is incorrect: for a sphere, V = 4/3 pi r^3. You also show some numbers but no indication of how you got them.

    As already mentioned, for a sphere V = 4/3 pi r^3.

    The differential of V, dV, is
    dV = dV/dr * dr.

    Time doesn't play a role in this problem do you don't want or need dV/dt.
    Can you calculate dV/dr?
    The differential of r, dr, is approximately equal to [itex]\Delta[/itex]r, which you should be able to figure out from the information given in this problem.
  8. Jan 12, 2009 #7


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    Science Advisor

    You have on your paper
    "[tex]\frac{dA}{dt}= 2x[/tex]"

    That is incorrect. It should be
    "[tex]\frac{dA}{dt}= 2x\frac{dx}{dt}[/tex]"

    And you are given that dx/dt= 0.4 cm/s so
    [tex]\frac{dA}{dt}= 0.8 x[/tex]

    Now combine that with the fact that P= 4x.
  9. Jan 12, 2009 #8
    Ok so should I plug in the change in the radius (.02cm) in order to get the change in volume in number 11.

    Also in number 10, I now realize that I took the derivative with respect to time and not with respect to x so I would get 2x(dx/dt).
  10. Jan 12, 2009 #9
    Actually now that I look at it again, wouldn't I plug in 10.02cm for r instead because it is the new volume and get (b)?
  11. Jan 12, 2009 #10


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    You could get the exact by calculating the volume at r= 10.2 and r= 10 and subtracting but that is not what is asked.

    You could do it either by setting r= 10, dr= 0.2 or by setting r= 10.2, dt= -0.2. You will get different answers but this only an "approximation". It is far easier to use r= 10, dr= 0.2.
  12. Jan 12, 2009 #11


    Staff: Mentor

    I took the liberty of cleaning up some typos, to aid the OP's understanding. My changes are shown in bold.
  13. Jan 13, 2009 #12
    Alright thanks, I think I have a further understanding of this now. I did actually when I first started the problem, plug 10 and 10.02 in to get the change in volume. Of course that wasn't an available answer and it wasn't the "approximate" answer. But after taking the derivative of the sphere volume formula with respect to radius I yield

    dV\dr = 4pi(r)^2
    I then plug in the radius 10 and the change in the radius .02.
    dV/.02 = 4pi(10)^2
    dV = 25.133

    Also Is there a reason I use 10 instead of 10.02? Is it because 10 was the original radius and we add .02 to the radius to find the change?
  14. Jan 13, 2009 #13


    Staff: Mentor

    Yes, that's the right answer (to 3 decimal places). Yes, 10 is the original radius, and dr = .02, both of which you need to find the change in volume.

    One slight change I would make is to work with differentials rather than the derivative dV/dr.
    IOW, instead of this equation:
    dV/dr = 4 pi r^2,
    I would write:
    dV = 4 pi r^2 dr
    and then substitute the known information. You get the same result.
  15. Jan 13, 2009 #14
    Thanks for the confirmation. I also like the way you wrote that, have to use that format from now on.
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