its about a weather balloon. if the radius increases at rate of 0.03 inches per second and the radius=48 at T=0. find the equation for volume of balloon at time T and find the volume when T=300seconds. can you please explain how to do this .
2007-08-26 06:06:19 · 4 answers · asked by shelykid 2 in Science & Mathematics ➔ Mathematics
can someone break it down it simple terms?
2007-08-26 06:22:11 · update #1
Hi,
At time 0 the radius starts at 48 inches. Since it increases by .03 inches per second, the radius is increasing at a constant rate, so a linear equation would represent the increase in the radius. This equation would be r = 48 + .03T, where T is the time in seconds after when T = 0.
Since the volume of a sphere is found by 4/3πr³, then the volume of the balloon will be V = 4/3π (48 + .03T)³.
When T = 300, this formula would become:
V = 4/3π (48 + .03(300))³ which simplifies to:
V = 4/3π (48 + 9)³ =
V = 4/3π (57)³ = 246,924π or 775734.6 cubic inches
This is the volume after 300 seconds.
I hope that helps!! :-)
2007-08-26 06:20:39 · answer #1 · answered by Pi R Squared 7 · 0⤊ 0⤋
v = (4/3)pi r^3
Differentiate with respect to time t,
v' = 4pi r^2 r' = 0.12pi r^2, where r = 48+0.03t
Integrating gives,
v = (4/3)pi (48)^3 + â«v' dt , t from 0 to T
v(300) = (4/3)pi (48)^3 + â«v' dt, t from 0 to 300 = 7.757*10^5 in^3
2007-08-26 13:14:55 · answer #2 · answered by sahsjing 7 · 0⤊ 0⤋
V= 4/3 pi(48+.03t)^3
When t = 300,
V= 4/3*pi(57)^3 = 775,732.7 in^3
2007-08-26 13:46:48 · answer #3 · answered by ironduke8159 7 · 0⤊ 0⤋
r(t)=48+0.03t, t in seconds
V(t)=4pi/3r^3(t)
=4pi/3[48+0.03t]^3 ANS.
when t=300s,
V(300)=4pi/3[]48+9]^3
=4pi/3(57)^3 ANS
2007-08-26 13:15:04 · answer #4 · answered by Anonymous · 0⤊ 0⤋