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B I need help with a related rates question

  1. Jul 24, 2016 #1
    A spherical snowball melts at a rate proportional to its surface area. Show that the rate of change of the radius is constant.

    Two ratios are proportional if they change equally and are related by a constant of proportionality? Not sure about this definition, but please correct it if you can.

    My initial thoughts were to set up an equation and differentiate it with respect to time.

    S=surface area, r=radius, k=constant of proportionality

    S=4pir^2=s

    (dS/dt)=2*4*pi*r*(dr/dt)

    (dr/dt)=[(dS/dt)]/[(8pir)]

    At this point, I do not know what to do.

    The words I am using to describe the situation: The rate at which the snowball melts (dS/dt) is proportional to the radius. So the radius is the constant of proportionality? I am thinking that we should have something like the following: (d/s/dt)=constant of proportionality*surface area?

    I am not sure how to understand this problem.

    Please explain the problem. There is nothing worse than a teacher that does a bunch of math on a chalkboard, gives the most minute explanation, and calls it teaching (just my opinion)

    Please use words and explain what is happening. Thanks
     
  2. jcsd
  3. Jul 24, 2016 #2

    fresh_42

    Staff: Mentor

    You know that the rate of melting is ##\frac{dV(r,t)}{dt} = c \cdot S(r,t) \sim 4\pi r(t)^2## for some constant ##c##. Do you know how volume ##V## and surface ##S## are related?
     
  4. Jul 24, 2016 #3
    The derivative of volume is equal to surface area?
     
  5. Jul 24, 2016 #4
    Also, I do not understand your notation. Can you please use words?

    It seems that you are saying that the derivative of volume at some arbitrary time for some arbitrary radius (with respect to time) is equal to some constant of proportionality multiplied by the surface area at some arbitrary time for some arbitrary radius. It seems that you are also implying that surface area is a function of r and t and changes as r and t change. I have not taken multivariable calculus....
     
  6. Jul 24, 2016 #5

    fresh_42

    Staff: Mentor

    Yep! And with respect to what? I used the same notation as you did, only an additional ##\sim## to indicate proportionality.
    If you write your answer above in terms of the Leibniz notation as you did with the surface and add my equation which is given by the exercise, then you will only have to rearrange the terms.
     
  7. Jul 24, 2016 #6
    I see. I will research your notation in a bit to try to understand how they are equal to mine.

    I was just working on this again and I think I see where you are going... but it slipped my mind... The volume of the snowball melts at a rate proportional to the surface area. I am still struggling to use this information. So dr/dt is the constant of proportionality? I still do not understand this problem...

    Also, out of curiosity, is the problem true if and only if the surface temperature of the snowball is constant? It seems that the rate of change of volume is going to depend heavily on temperature..

    If you solve for the given information, you get 1/(4pir^2)=dr/dt. But isn't r changing?
     
  8. Jul 24, 2016 #7

    fresh_42

    Staff: Mentor

    ##V## is simply sloppy for ##V(r,t)## or even more precisely ##V(r(t),t)##. The same with ##S##. Since we have two variables to consider, radius and time, it is as saying ##f(x)## which you are probably used to. The ##\sim## sign to state something is proportional might be a bit of a personal taste. I'm not quite sure about that.

    Forget about the ##4\pi r^2##. You know ##\frac{dV}{dt} = \text{ melting rate of snowball } = c \cdot S## as given in the text. Then you know, as you have said, ##\frac{dV}{dr} = S##. If you treat the fractions as normal fractions of numbers, you will find ##\frac{dr}{dt}##. (Assuming you know what ##\frac{d}{dV}V## is.) This is a bit of a hand waving method, but I thought: Hey, it's a physics forum, so why to complicate things. :wink:

    (Hopefully they won't hit me for this remark ....:nb))
     
  9. Jul 24, 2016 #8
    I understand what you are saying; however, multivariable calculus goes beyond the scope of my education and the discussion. Conceptually, I understand what you getting at. Notice that the question is slotted as a beginning level calculus question, not a third semester college calculus question.

    Anyway, we have (dv/dt)=cS

    dv/dt=4pir^2(dr/dt)

    4pir^2(dr/dt)=c(4pir^2)

    Multiply through by (1/4pir^2) and we have dr/dt=c

    Thanks for the help
     
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