A melting snowball has a radius that decreases 1/4 inch per hour.?Find the rate of change of the volume with respect to time when the radius is 4in ?
Calculus solution
V=4/3 PI r^3
dv/dt= 4 PI r^2 dr/dt
dr/dt is given as .25 in/hr
then solve for dv/dt when r=4
To find the rate of change of the volume with respect to time, we need to 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 decreasing 1/4 inch per hour, we can express the rate of change of the radius with respect to time as:
dr/dt = -1/4 inch/hour
To find the rate of change of the volume with respect to time, we need to differentiate the volume formula with respect to time:
dV/dt = (dV/dr) * (dr/dt)
First, let's find dV/dr by differentiating the volume formula with respect to r:
dV/dr = (4/3) * π * (3r^2)
Now, we substitute the given value of the radius (r = 4 inches) into the expression for dV/dr:
dV/dr = (4/3) * π * (3(4)^2)
= (4/3) * π * 48
= 64π
Finally, substitute the values of dV/dr and dr/dt into the expression for dV/dt:
dV/dt = (64π) * (-1/4)
= -16π
Therefore, when the radius is 4 inches, the rate of change of the volume with respect to time is -16π cubic inches per hour.