a spherical hailstone is losing mass by melting uniformly over its surface as it falls. At a certain time, its radius is 2 cm and its volume is decreasing at the rate of o.1 cm^3/s. How fast is the radius decreasing at the time?
V=pi*r^3
dV/dt=3*pi*(r^2)*(dr/dt)
dV/dt=.1
r=2
pi=3.14
dr/dt=?
To find the rate at which the radius is decreasing over time, we can use the related rates formula.
The formula for the volume of a sphere is: V = (4/3)πr^3
Differentiating both sides of the equation with respect to time t, we get:
dV/dt = 4πr^2(dr/dt)
Given that dV/dt = -0.1 cm^3/s and r = 2 cm, we can substitute these values into the equation:
-0.1 = 4π(2^2)(dr/dt)
Simplifying the equation further:
-0.1 = 16π(dr/dt)
Now we can solve for dr/dt, the rate at which the radius is decreasing:
dr/dt = -0.1 / (16π)
dr/dt ≈ -0.0063 cm/s
Therefore, the radius is decreasing at a rate of approximately -0.0063 cm/s at the given time.