Sphere surface area and radius

[Paintjunkie]

I am curious about the relationship between an ever expanding sphere's radius and its surface area.

how would I relate the rate of change in radius to the rate of change in the surface area.

 

[Mark44]

I am curious about the relationship between an ever expanding sphere's radius and its surface area.

how would I relate the rate of change in radius to the rate of change in the surface area.

Start by writing the equation for the surface area of a sphere, which is A = $4\pi r^2$. Now differentiate both sides with respect to t, assuming that both A and t are differentiable functions of time.

 

[Paintjunkie]

so its just as simple as dA/dt=8(pi)dr/dt

 

[Mark44]

so its just as simple as dA/dt=8(pi)dr/dt

It's pretty simple, but what you have isn't correct. Check your differentiation.

 

[Paintjunkie]

dA/dt=8(pi)rdr/dt.

I guess that's right.. and that would be the speed that the surface area is increasing? If the Surface area was increasing at an increasing rate i would just differentiate again?

 

[Mark44]

Yes, that's right now. The rate of change of dA/dt (or A'(t)) would be d²A/dt² or A''(t). To get that, you would differentiate again. You would need to use the product rule, though.

 

[Paintjunkie]

ok so d²A/dt²=8(pi)dr/dt+8(pi)rd²r/dt²

if that's right... could it be said that the surface area is increasing at an increasing rate but the radius is increasing at a constant rate?

i suppose that under those conditions the second term would be 0? resulting in d²A/dt²=8(pi)dr/dt?

could we create a situation where the radius was increasing at a decreasing rate but the surface area, at least for a time, was increasing at an increasing rate?
I guess that would be under the conditions that the first term was greater than the second term, or when the radius velocity term was the dominating term instead of the acceleration term?

kinda throwing all these questions in here at once..

 

[Mark44]

ok so d²A/dt²=8(pi)dr/dt+8(pi)rd²r/dt²

if that's right... could it be said that the surface area is increasing at an increasing rate but the radius is increasing at a constant rate?

Let's look at it the other way around. If r is increasing at a constant rate, that means that dr/dt is constant, and d²r/dt² is zero. So dA²/dt² is proportional to dr/dt.

i suppose that under those conditions the second term would be 0? resulting in d²A/dt²=8(pi)dr/dt?

Yes.

could we create a situation where the radius was increasing at a decreasing rate but the surface area, at least for a time, was increasing at an increasing rate?

Maybe. If r is increasing, but at a decreasing rate, then dr/dt > 0 and d²r/dt² < 0.

If the surface area is increasing at an increasing rate, then d²A/dt² > 0. For this to happen, you would need a situation in which 8(pi)dr/dt+8(pi)rd²r/dt² > 0. IOW, it would have to be true that dr/dt + rd²r/dt² > 0. dr/dt would have to be pretty large in comparison to r.

I guess that would be under the conditions that the first term was greater than the second term, or when the radius velocity term was the dominating term instead of the acceleration term?

kinda throwing all these questions in here at once..

 

[Paintjunkie]

Iow?

 

[Mark44]

IOW = In other words

 

[Paintjunkie]

I would like to start thinking about this with two spheres both starting with the same radius right on top of each other. the outer one obeying our first situation and the inner one obeying our second one. I feel like there should be a relation one could find describing a phase shift, for lack of a better word or maybe that is the right word idk, between the two circles. I am thinking that an angle theta and phi would cut out an area on the surface of the inner sphere that would be smaller than the area cut out on the outer sphere. and that over time, if the spheres were following their rules respectivly, that area difference would increase.

I hope i am explaining this in a way it makes sense. I wish i could draw a picture.

I probably won't be able to respond to this till friday night, finals are killer.