Rate of Change problem

[Drahcir87]

Note: This looks like a lot, but I just ramble on so please take the time to read it and see what you can do.

Problem: A spherical tank is coated uniformly with a 2-inch layer of ice. If the volume of ice is melting at a rate directly proportional to its surface area, show that the outside radius is decreasing at a constant rate.

Now, before I came to college and entered an Honors Calculus class I considered myself fairly proficient in mathematics, but now I'm starting to doubt myself.

Any ways I can't figure out exactly what I'm supposed to do in this problem. I know the Volume of the tank [V(t)] and the Surface Area of the tank [S(t)]. I also know the Volume of the ice [V(i)] and Surface Area of the ice [S(i)].

*P=pi; r=radius; x=thickness of ice; ^=power sign(ex. x^2 is x squared;
x^3 is x cubed).

*V(t) = (4/3)Pr^3

*S(t) = 4Pr^2

*V(i) = (4/3)P(3xr^2 + 3x^2r + x^3)
= 4Pxr^2 + 4Px^2r + (4/3)Px^3

*S(i) = 4P(r^2 + 2xr + x^2)

Now I also know that Surface Area is the derivative of Volume (in a sphere at least), but I have no idea how that will help me.

Thanks for any help.

 

[XtremeChic04]

What you want to do is take the derivative of the volume(Surface area). Plug in a number for x in the derivative, say 5. Then, plug in say 4.9 to show the surface area is decreasing. Take the difference of the answer at 5 & 4.9. as x changes .1, the volume is changing (difference) times more.

 

[Drahcir87]

Sounds pretty simple when you put it that way. Thanks

 

[Gene]

Nice presentation.
I would look at it this way.
r=radius with the ice.
k = a constant.

...the volume of ice is melting at a rate directly proportional to its surface area,

From the definition of directly proportional that gives
dV/dt = kS
dV/dr*dr/dt =
d(4/3Pr^3)/dr*dr/dt =
4Pr^2*dr/dt=k*4Pr^2
dr/dt=k
So the radius is changing at a constant rate k.