A balloon is being inflated,the radius of the balloon,initially zero,increases at the rate of 9 ft/sec.How fast is the volume increasing after 1/2 second?
V = (4/3) π r^3
dV/dt = 4πr^2 dr/dt
after 1/2 second, the radius is 4.5 ft
back in dV/dt
= 4π(4.5)^2 (9) = ....
To find out how fast the volume is increasing after 1/2 second, we need to use the formula for the volume of a balloon, which is V = (4/3)πr³, where V is the volume and r is the radius.
We are given that the radius is increasing at a rate of 9 ft/sec. So, the rate at which the radius is changing is dr/dt = 9 ft/sec.
We need to find dV/dt, the rate at which the volume is changing with respect to time. To do this, we can use the chain rule of differentiation.
First, differentiate the volume equation with respect to time:
dV/dt = (dV/dr)(dr/dt)
Next, substitute the values we know into the equation:
dv/dt = (4π(1/2)(9))(9) = (4π)(9)(9/2) = 162π ft³/sec
So, the volume is increasing at a rate of 162π ft³/sec after 1/2 second.