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Homework Help: A spherical iron ball covered with ice

  1. Oct 27, 2007 #1
    hi everyone this is a related rates question no one in the class could solve it. I could not find how to find the equation that relate all the things in the question.the question is:

    A spherical iron ball 8 in. in diameter is coated with a layer of ice of uniform thickness. If the ice melts at the rate of 10 in^3/min, answer the following questions.

    (a) How fast is the thickness of the ice decreasing when the ice is 2 in. thick?

    (b) How fast is the outer surface area of the ice decreasing?



    anyone can help?
    thank you
     
  2. jcsd
  3. Oct 27, 2007 #2
    a) The volume of a sphere with respect to the radius is [tex]V=(\frac{4\pi}{3})r^3[/tex]. Solve for [tex]r[/tex] then differentiate with respect to time.
    b) The surface area of a sphere with respect to the radius is [tex]\alpha=4\pi r^2[/tex]. Write the surface area in terms of the volume and again, differentiate with respect to time.

    EDIT: I just re-read the problem and the fact that the ice is melting, not the iron-ball, changes the way you should consider the 2 in. thickness of the ball. The radius of the iron-ball (4 in.) should be added to 2 in. ice to find the rate.
     
    Last edited: Oct 27, 2007
  4. Oct 9, 2008 #3
    is the answer for this one 1/14.4 cm/min? ^^
     
  5. Oct 9, 2008 #4
    a) You must realize that the rate of melting given is not for the iron ball + ice... It is only for the ice. Write a formula for [tex]V_i_c_e[/tex], given the total radius [tex]r_b_a_l_l_+_i_c_e[/tex] is [tex]4 + w[/tex]; when [tex]w[/tex] is the width (thickness) of the ice. Then differentiate with respect to time.
    b) This is a bit more complicated. Again, start with the formula for surface area, [tex]\alpha=4\pi (4 + w)^2[/tex]. Solve for [tex]w[/tex] with respect to [tex]V[/tex], then differentiate with respect to time.

    PS. Your answer is [tex]\pi[/tex] times greater than the answer.
     
    Last edited: Oct 9, 2008
  6. Oct 9, 2008 #5
    hmmm. i don't seem to understand ... could you please elaborate a) ? i get the same answer
     
  7. Oct 9, 2008 #6
    a)
    [tex]V_i_c_e = V_e_n_t_i_r_e _b_a_l_l - V_i_r_o_n = \frac{4\pi}{3}(4+w)^3 - \frac{4\pi}{3}(4)^3 = \frac{4\pi}{3}((4+w)^3 - (4)^3)[/tex]

    Differentiate with respect to time:
    [tex]\frac{dV_i_c_e}{dt} = \frac{4\pi}{3}(3(4+w)^2\frac{dw}{dt})[/tex]

    [tex]-10 = \frac{4\pi}{3}(3(4+2)^2\frac{dw}{dt})[/tex]

    [tex]\frac{dw}{dt} = -\frac{5}{72\pi} = -\frac{1}{45.2389342} = -0.0221048532\frac{m}{s}[/tex]
     
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