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Homework Help: Differentiation of surface area of a right cylinder

  1. Nov 1, 2011 #1
    1. The problem statement, all variables and given/known data
    The total surface area of a right circular cylinder is given by the formula A = 2pir(r + h) where r is the radius and h is the height.
    a) Find the rate of change of A with respect to h if r remains constant.
    b) Find the rate of change of A with respect to r if h remains constant.
    c) Find the rate of change of h with respect to r if A remains constant.

    2. Relevant equations
    A = 2pir(r + h)

    3. The attempt at a solution
    I thought i had this all right but then realized that what i had done for part a was wrong, i think that part b and c are right but i will just put my answers up to check.
    a) A = 2pir(r + h)
    r = c (constant)
    A = 2pic(c + h)
    A = 2pi(c2 + ch)
    A/2pi = c2 + ch
    however, i do not know at this point, how i would isolate c!
    What i did was bring the A/2pi over and get c2 + ch - A/2pi = 0 and use the quadratic formula. From that i got that c = (h +- h)/4pi and reason that c cannot be zero or else the surface area would be zero so c must be h/2pi. I think that this process of my work is wrong because i dont think you can use the quadratic formula and solve for c because c is a constant.
    We know that
    A = 2pic(c + h)
    A = 2pic2 + 2pich
    dA/dh = 2pic = 2pi(h/2pi) = h
    As you can see i dont think that what i did was correct, please help me with this one, a) is what i need most help with

    b) h = c (constant)
    A = 2pir(r + c)
    A/2pir = r + c
    c = (A/2pir) - r
    dA/dr = 2pi(r + c) + 2pir = 4pir + 2pic = 4pir + 2pi((A/2pir) - r) = 4pir + A/r - r

    c) A = c (constant)
    c = 2pir(r + h)
    c/2pir - r = h
    dh/dr = (-c/2pir2) - 1
    = (-(2pir(r + h))/2pir2) - 1
    = ((- r - h) / r) - 1
    = (-2r - h) / r
  2. jcsd
  3. Nov 2, 2011 #2


    Staff: Mentor

    Nope, not right, and it's incorrect from almost the very beginning. What they're looking for is dA/dh, or if you know about partial derivatives, [itex]\frac{\partial A}{\partial h}[/itex].

    b and c are incorrect, as well.

    For a, suppose you had A = 2π*6(h + 6). What would dA/dh be? If it was A = 2πk(h + k), with k a constant, what would dA/dh be?

    For b, suppose you had A = 2πr(r + 3). What would dA/dr be? If is was A = 2πr(r + k), what would dA/dr be?
  4. Nov 2, 2011 #3
    A = (2pi * 6)(h+6)
    = (12pi)(h + 6)
    = 12pih + 72pi
    dA/dh = 12pi

    A = (2pi * k) (h + k)
    = 2pikh + 2pik2
    dA/dh = 2pi * k

    A = 2pir(r + 3)
    = 2pir2 + 6pir
    da/dr = 4pir + 6pi

    A = 2pir(r+k)
    = 2pir2 + 2pirk
    dA/dr = 4pir + 2pik

  5. Nov 2, 2011 #4


    User Avatar
    Science Advisor

    Where in the world did the "6" come from?

    and now where did "k" come from?

    where did this "3" come from?

    where did "k" come from?

    Why not just use the original variables "r" and "h" rather than introducing numbers or other symbols?
  6. Nov 2, 2011 #5
    but isn't that what i did in my originial answer, the only difference was that i was trying to isolate r or h so that when i got the derivative i could plug their original values back in to the derivative to simplify the derivative.
  7. Nov 2, 2011 #6


    Staff: Mentor

    They came from the revised problems that I suggested in post 2.

  8. Nov 2, 2011 #7


    Staff: Mentor

    Right. So in the original problem, with r considered to be constant, you have A = 2πr(h + r), so dA/dh = 2πr.


    Should be dA/dr, not da/dr, but otherwise OK.
    Both fine. Now if A = 2πr(r + h), and h is considered to be a constant, what is dA/dr?
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