Question

A high-altitude spherical weather balloon expands as it rises due to the drop in atmospheric pressure. Suppose that the radius r increases at the rate of 0.16 inches per second and that r = 38 inches at time t = 0. Determine the equation that models the volume V of the balloon at time t and find the volume when t = 280 seconds.



V(t) = four pi times the product of zero point one six and t to the third power divided by three.; 1,129,910.14 in.3

V(t) = four pi times the quantity of thirty eight plus zero point one six t to the third power divided by three.; 2,377,823.53 in.3

Answer 1

Are those your only two choices?

Answer 2

No, my apologies.
V(t) = four pi times the product of zero point one six and t to the third power divided by three.; 1,129,910.14 in.3

V(t) = four pi times the quantity of thirty eight plus zero point one six t to the third power divided by three.; 2,377,823.53 in.3

V(t) = 4π(0.16t)2; 25,221.21 in.3

V(t) = 4π(38 + 0.16t)2; 1,325,689.77 in.3

Answer 3

I chose the second option as my answer, well that is what I got.

Answer 4

OK. The thing that bothers me is that there is an initial volume that would be added to the volume of expansion. So, you would start out with Vo=(4/3)*pi*(38 in.)^3=229847.296 cubic inches. Then you would add the expansion V(t)= (4/3)*pi*((.16)^3)*280.

Answer 5

Or V(t) = Vo + V(280)

Answer 6

That is what I did

r(t)=0.16*t+38

V=(4/3)(pi)(r(t))^3

Answer 7

Thank you for your help, I got my answer verified.

Answer 8

Ah, I see the 38 value in your second option in the first part of your question. I didn't see that at first. It's difficult to read this at times. I don't know if it's the same for other users, but my screen display bounces around a bit while I'm typing on here.