A spherical balloon is being inflated so at the rate of 8cm3/sec. determine the rate at which the radius of the balloon is changing when the radius is 10
v = 4/3 pi r^3
dv/dt = 4 pi r^2 dr/dt
Now just plug in your numbers and solve for dr/dt.
To determine the rate at which the radius of the balloon is changing, we can use the formula for the volume of a sphere:
V = (4/3)πr^3
Where V is the volume of the sphere and r is the radius.
Now, let's take the derivative of both sides of the equation with respect to time:
dV/dt = 4πr^2 (dr/dt)
Here, dV/dt represents the rate at which the volume is changing with respect to time, and dr/dt represents the rate at which the radius is changing with respect to time.
Given that the volume is changing at a rate of 8 cm^3/sec, which is equal to dV/dt = 8 cm^3/sec, and the radius is 10 cm, we can substitute these values into the equation:
8 = 4π(10^2)(dr/dt)
Simplifying the equation, we have:
8 = 400π(dr/dt)
Divide both sides of the equation by 400π:
dr/dt = 8 / (400π)
dr/dt ≈ 0.0063662 cm/sec (rounded to 5 decimal places)
Therefore, the rate at which the radius of the balloon is changing when the radius is 10 cm is approximately 0.0063662 cm/sec.