A spherical balloon is inflated with gas at the rate of 20cm^3per mint. How fast the radius of the balloon changing at the instant when radius is 2cm?
We can use the formula for the volume of a sphere:
V = (4/3)πr^3
Taking the derivative with respect to time t,
dV/dt = 4πr^2 (dr/dt)
We are given that dV/dt = 20 cm^3/min. When r = 2 cm, we can solve for dr/dt:
20 = 4π(2^2)(dr/dt)
dr/dt = 20/(16π) = 0.397 cm/min (rounded to 3 decimal places)
Therefore, the radius is changing at a rate of approximately 0.397 cm/min when r = 2 cm.
To find how fast the radius of the balloon is changing, we can apply the volume formula of a sphere, which is given by:
V = (4/3)πr^3
Where V is the volume of the sphere and r is the radius.
We are given that the volume of the balloon is changing at a rate of 20 cm^3 per minute. Therefore, the derivative of the volume with respect to time is:
dV/dt = 20 cm^3/min
To find how the radius is changing, we need to find the derivative of the volume formula with respect to time, using implicit differentiation:
dV/dt = d/dt[(4/3)πr^3]
dV/dt = (4/3)π(3r^2)(dr/dt) (using chain rule)
We are interested in finding the rate of change of the radius (dr/dt) when the radius is 2 cm.
Let's substitute the values we have into the formula:
20 = (4/3)π(3(2^2))(dr/dt)
Simplifying this equation, we get:
20 = 16π(dr/dt)
Now we can solve for dr/dt:
dr/dt = 20/(16π)
dr/dt ≈ 0.398 cm/min
Therefore, the rate at which the radius is changing at the instant when the radius is 2 cm is approximately 0.398 cm/min.