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 0.02 inches per second, and that r = 36 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 = 360 seconds.
V(t) = 4π(0.02t)2; 651.44 in3

V(t) = 4π(36 + 0.02t)2; 1,694,397.14 in3

V(t) = four pi times the product of zero point zero two and t to the third power divided by three.; 4,690.37 in3

V(t) = four pi times the quantity of thirty six plus zero point zero two t to the third power divided by three.; 337,706.83 in3

Answer 1

Answer:

V(t)= $\frac{4\pi }{3} (36+0.02t)^{3}$

337 706, 83 $in^{3}$

Step-by-step explanation:

The volume of a sphere is $V=\frac{4\pi }{3} r^{3}$

So like the initial radius is 36 and is increasing of 0.02 every second, we define the radius:

r= 36+0.02t

So, we replace in the formula of the volume and we obtain

$V(t)= \frac{4\pi }{3} (36+0.02t)^{3}$

and if we replace t= 360 seconds in the equation of volume, we get

$V(360)= \frac{4\pi }{3} (36+(0.02*360))^{3}$

$V(360)=337 706, 83 in^{3}$