A spherical snowball is melting in such a way that its diameter is decreasing at rate of 0.2 cm/min. At what rate is the volume of the snowball decreasing when the diameter is 9 cm.
To find the rate at which the volume of the snowball is decreasing, we need to differentiate the volume of the sphere with respect to time.
The volume of a sphere can be calculated using the formula: V = (4/3) * π * r^3, where V is the volume and r is the radius.
Since we are given the rate at which the diameter is decreasing, we need to find the relationship between the diameter and the radius. The radius is half the length of the diameter, so r = d/2.
Let's differentiate this expression with respect to time to find the relationship between the rate of change of the diameter (dd/dt) and the rate of change of the radius (dr/dt):
dr/dt = (1/2) * (dd/dt)
Now, we have the relationship between the rate of change of the diameter and the rate of change of the radius. Next, we can differentiate the volume formula:
V = (4/3) * π * (d/2)^3
= (4/3) * π * (1/8) * d^3
= (1/6) * π * d^3
Now, differentiate both sides of the equation with respect to time (t):
dV/dt = (1/6) * 3π * d^2 * (dd/dt)
dV/dt = (1/2) * π * d^2 * (dd/dt)
We are given that the diameter is decreasing at a rate of 0.2 cm/min (dd/dt = -0.2 cm/min), and we want to find the rate at which the volume is decreasing when the diameter is 9 cm.
Substituting the given values into the equation, we have:
dd/dt = -0.2 cm/min
d = 9 cm
dV/dt = (1/2) * π * (9 cm)^2 * (-0.2 cm/min)
dV/dt = 81π * (-0.2) cm^3/min
dV/dt ≈ -50.96π cm^3/min
Therefore, when the diameter is 9 cm, the volume of the snowball is decreasing at a rate of approximately -50.96π cm^3/min.
To find the rate at which the volume of the snowball is decreasing, we can use the formula for the volume of a sphere:
V = (4/3)πr^3
Where V is the volume and r is the radius of the sphere.
Since the diameter is decreasing at a rate of 0.2 cm/min, the radius is also decreasing at a rate of 0.1 cm/min (since the radius is half the diameter).
We are given that the diameter is 9 cm, so the radius is 4.5 cm.
Now, let's find the rate at which the volume is changing when the radius is 4.5 cm:
First, differentiate the volume equation with respect to time (t):
dV/dt = 4πr^2(dr/dt)
The first term, dV/dt, represents the rate of change of volume with respect to time, and the second term, dr/dt, represents the rate of change of the radius with respect to time.
Substituting the values we have:
dV/dt = 4π(4.5)^2(0.1)
dV/dt = 4π(20)(0.1)
dV/dt = 8π
So, when the diameter is 9 cm, the volume of the snowball is decreasing at a rate of 8π cubic cm per minute.
"..diameter is decreasing at rate of 0.2 cm/min"
---> dd/dt = .2, so dr/dt = .1 cm/min
V= (4/3)π r^3
dV/dt = 3π r^2 dr/dt
insert your given data