A spherical snowball is melting in such a way that its diameter is decreasing at rate of 0.4 cm/min. At what rate is the volume of the snowball decreasing when the diameter is 8 cm. (Note the answer is a positive number).
v = 4π/3 r3
dv/dt = 4πr2 dr/dt
= 4π*42 (-.4)
= -25.6π cm3/min
Rats. dr/dt = -.2, not -.4 (diameter d/dt)
So, take half of that answer.
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 0.4 cm/min, the radius is also decreasing at half that rate, which is 0.2 cm/min.
Given that the diameter is 8 cm, we can find the radius of the snowball:
r = d/2 = 8/2 = 4 cm.
We can now differentiate the volume formula with respect to time (t) and solve for dV/dt, the rate of change of volume:
V = (4/3)πr^3
dV/dt = (4/3)π * 3r^2 * dr/dt.
Substituting the values we know:
dV/dt = (4/3)π * 3 * 4^2 * (0.2).
Calculating this gives us:
dV/dt = (4/3)π * 48 * (0.2) = 32π * 0.2.
The value of dV/dt is therefore 6.4π cm^3/min.
Since the question asks for the answer to be positive, the final answer is 6.4π cm^3/min.
To find the rate at which the volume of the snowball is decreasing, we need to differentiate the volume equation with respect to time.
The volume of a sphere can be calculated using the formula:
V = (4/3)πr^3
Since we are given the rate of change of the diameter, we need to find the relationship between the diameter (d) and the radius (r).
The relationship between diameter and radius is:
r = d/2
If the diameter is decreasing at a rate of 0.4 cm/min, then the rate of change of the radius is:
dr/dt = (d/2)' = 0.4 cm/min
To find the rate at which the volume is decreasing when the diameter is 8 cm, we need to substitute the values into the volume equation and differentiate it with respect to time.
Let's start by finding the radius when the diameter is 8 cm:
d = 8 cm
r = d/2 = 8/2 = 4 cm
Now, we can differentiate the volume equation with respect to time (t):
V = (4/3)πr^3
dV/dt = d/dt [(4/3)πr^3]
Using the chain rule, the right side can be rewritten as:
dV/dt = (dV/dr) * (dr/dt)
We have the value of dr/dt as 0.4 cm/min, and we need to find dV/dt when r = 4 cm.
First, let's differentiate the volume equation with respect to r to find dV/dr:
dV/dr = d/dr [(4/3)πr^3]
dV/dr = (4/3)π * 3r^2
dV/dr = 4πr^2
Now, substitute the value of r = 4 cm into dV/dr:
dV/dr = 4π(4)^2
dV/dr = 4π(16)
dV/dr = 64π
Finally, substitute the values of dV/dr = 64π and dr/dt = 0.4 cm/min into dV/dt:
dV/dt = (dV/dr) * (dr/dt)
dV/dt = (64π) * (0.4)
dV/dt = 25.6π cm^3/min
Therefore, when the diameter is 8 cm, the volume of the snowball is decreasing at a rate of approximately 25.6π cm^3/min.