A spherical ballon is leaking air at 2 cubic inches per hour.l How fast is the ballons radisu chancing when the radisu si 3 inches.
v = 4/3 pi r^3
dv/dt = 4pi r^2 dr/dt
-2 = 4pi*9 dr/dt
dr/dt = -2/36pi = -1/18pi = -.0177 in/hr
What is a radisu?
Once I can understand as a typo, but twice ?
V= (4/3)πr^3
dV/dt = 4π r^2 dr/dt
-2 = 4π(9)dr/dt
etc.
Sorry about that Reiny. I have a quick question though, where does the 9 come from?
Notice both Steve and I have the same derivative equation of
dV/dt = 4π r^2 dr/dt
what is the value of r^2 when r = 3 ?
Oh okay. I understand that, i was just not understand where the 9 came from, but i see now. Thank you both
To find how fast the balloon's radius is changing, we can use the rate at which the balloon is losing air. We can relate the change in volume to the change in radius using the formula for the volume of a sphere: V = (4/3)πr^3, where V is the volume and r is the radius.
Since the balloon is leaking air at a rate of 2 cubic inches per hour, the change in volume with respect to time (dv/dt) is -2 cubic inches per hour (negative because the volume is decreasing).
We need to find dr/dt, the rate at which the radius is changing when the radius is 3 inches. We can differentiate the volume equation with respect to time to find an expression relating dv/dt and dr/dt:
dV/dt = (d/dt)((4/3)πr^3)
dv/dt = 4πr^2(dr/dt)
Rearranging the equation and substituting the known values, we get:
-2 = 4π(3^2)(dr/dt)
-2 = 36π(dr/dt)
dr/dt = -2/(36π)
dr/dt ≈ -0.01745
Therefore, when the radius is 3 inches, the balloon's radius is changing at a rate of approximately -0.01745 inches per hour.