A certain calculus student hit Mrs. Evans in the head with a snowball. If the snowball is melting at the rate of 10 cubic feet per minute, at what rate is the radius of the snowball changing when the snowball is 1 foot in radius?
That is a pretty big snowball !
V = (4/3) pi r^3
dV/dt = (4/3) pi 3 r^2 dr/dt
10 = 4 pi r^2 dr/dt
since r = 1
dr/dt = 10/(4 pi)
note : using common sense we knew that. The rate of change of the volume is the surface area 4 pi r^2 times the rate of radius change
To find the rate at which the radius of the snowball is changing, we can use related rates, which involves applying the chain rule from calculus.
Let's denote the radius of the snowball as r (in feet) and the volume of the snowball as V (in cubic feet). We are given that the volume of the snowball is decreasing at a constant rate of 10 cubic feet per minute.
The volume of a sphere can be expressed as V = (4/3)πr^3, where π is a mathematical constant (approximately 3.14). Differentiating this equation implicitly with respect to time t, we get:
dV/dt = (4/3)π(3r^2)(dr/dt)
Where dV/dt represents the rate of change of volume with respect to time, and dr/dt represents the rate of change of the radius with respect to time.
We are given that dV/dt = -10, as the volume is decreasing at a rate of 10 cubic feet per minute. At the moment we are interested in, the radius is 1 foot.
So, now we can solve for dr/dt:
-10 = (4/3)π(3(1)^2)(dr/dt)
-10 = 4π(dr/dt)
dr/dt = -10/4π
dr/dt ≈ -2.53 ft/min
Therefore, the rate at which the radius of the snowball is changing is approximately -2.53 feet per minute. Note that the negative sign indicates that the radius is decreasing.