October 24th, 2007, 08:38 PM  #1 
Newbie Joined: Oct 2007 Posts: 12 Thanks: 0  Confusing problem
Hey everyone, this is a problem I have been assigned and I am stuck on it. Physics tells us that a drop of mist light enough to float in the air and not fall to the ground as a raindrop is shaped like a perfect sphere, and water molecules can join this drop from any direction from outside the drop itself meaning that the drop gains moisture at a rate proportional to its surface area. Show that, under these conditions, the radius of the drop is growing at a constant rate, no matter how large the drop currently is. I know that I need to use the formula for the surface area of a sphere and possibly its derivative, but I just am confuesed as where to go next. Any help is greatly appreciated. 
October 24th, 2007, 09:02 PM  #2 
Newbie Joined: Oct 2007 From: Los Angeles Posts: 7 Thanks: 0  try this
water molecules can join this drop from any direction from outside the drop itself meaning that the drop gains moisture at a rate proportional to its surface area. this translates to: dV = k A dV = k 4 (pi) r^2 where dV is the change in volume, k is a constant, and A is the surface area. The formula for the volume of a sphere is this: V = (4/3)(pi)r^3 Now, taking the derivative of the volume of a sphere, we can write dV = 4(pi)r^2 dr Substitute dV from above, and we have k 4 (pi) r^2 = 4(pi)r^2 dr Cancel 4 (pi) r^2 from both sides, and we have k = dr Therefore, the change in radius is constant 
October 24th, 2007, 09:33 PM  #3 
Newbie Joined: Oct 2007 Posts: 12 Thanks: 0 
Ok I don't understand the need for the constant k. I i think i was initially thinking to hard about the problem, this is what I came up with. Surface Area of a sphere a= 4(pi)r^2 So the change in the surface area = da/dt = 8(pi)r dr/dt so the change in the radius dr/dt = 1/8(pi)r x da/dt I believe this is what i'm supposed to show, what do you guys think? 
October 24th, 2007, 10:53 PM  #4 
Newbie Joined: Oct 2007 From: Los Angeles Posts: 7 Thanks: 0 
i think the problem with leaving it as dr/dt = 1/8(pi)r x da/dt is that according to this equation, the change in the radius is dependent on r and the rate of change in the area. they want to see that the rate of change is constant.

October 24th, 2007, 11:19 PM  #5 
Newbie Joined: Oct 2007 Posts: 12 Thanks: 0 
ok that makes sense, but where does the constant k come from, what is it, what does it stand for, im just confused as to where you got the k from.

October 25th, 2007, 12:43 AM  #6 
Newbie Joined: Oct 2007 From: Los Angeles Posts: 7 Thanks: 0 
i originally had dV = kA because the question read, "drop gains moisture at a rate proportional to its surface area" i translated this to mean, "the volume of the drop changes directly in proportion to the surface area", making this a "direct variation" equation (y=kx) i used "k" as the constant of proportionality. 

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