A spherical balloon is being inflated so that radius is increasing at a rate of 10 mm/sec.
sounds like spherical trig to me.
To find the rate at which the volume of the balloon is increasing, we can use the formula for the volume of a sphere: V = (4/3) * π * r^3, where V is the volume and r is the radius.
Given that the radius is increasing at a rate of 10 mm/sec, we can substitute this rate into the formula to find the rate at which the volume is increasing.
To do this, we take the derivative of the volume formula with respect to time (t):
dV/dt = (4/3) * π * 3 * r^2 * dr/dt
Here, dV/dt is the rate of change of the volume with respect to time, dr/dt is the rate of change of the radius with respect to time, and r is the radius.
Substituting the given values into the equation, we get:
dV/dt = (4/3) * π * 3 * (10 mm/sec)^2
Simplifying further:
dV/dt = (4/3) * π * 3 * 100 mm^2/sec^2
dV/dt = (4/3) * π * 300 mm^2/sec^2
Therefore, the volume of the balloon is increasing at a rate of (4/3) * π * 300 mm^2/sec^2, or approximately 1256.6 mm^2/sec^2.