A circular balloon is inflated with air flowing at a rate of 10cm3/s. How fast is the radius of the ballon increasing when the radius is 2cm?
The volume (V) of the balloon with radius (r) is:
V = 4/3πr^3
Differentiating with respect to t, we get:
dV/dt = 4/3π * 3r^2 * dr/dt
dV/dt = 4πr^2 * dr/dt
The rate of change of the volume is:
dV/dt = 10cm^3s^−1
So, we find dr/dt:
dr/dt = 1/4πr^2 * dV/dt
dr/dt = 10/4πr^2
dr/dt = 5/2πr^2
Therefore, dr/dt(2cm) is:
dr/dt(2cm) = 5/2π(2)^2
dr/dt(2cm) = 5/2π4
dr/dt(2cm) = 5π/8
The volume (V) of the balloon with radius (r) is:
V = 4/3πr^3
Differentiating with respect to t, we get:
dV/dt = 4/3π * 3r^2 * dr/dt
dV/dt = 4πr^2 * dr/dt
The rate of change of the volume is:
dV/dt = 10cm^3s^−1
So, we find dr/dt:
dr/dt = 1/4πr^2 * dV/dt
dr/dt = 10/4πr^2
dr/dt = 5/2πr^2
Therefore, dr/dt(2cm) is:
dr/dt(2cm) = 5/2π(2)^2
dr/dt(2cm) = 5/2π4
dr/dt(2cm) = 5π/8