A spherical snowball is melting in such a way that its diameter is decreasing at rate of 0.3 cm/min. At what rate is the volume of the snowball decreasing when the diameter is 13 cm. (Note the answer is a positive number).
v = π/6 d^3
so, you have
dv/dt = π/2 d^2 dd/dt
and just plug in your numbers.
To find the rate at which the volume of the snowball is decreasing when the diameter is 13 cm, we can use the formula for the volume of a sphere.
The volume of a sphere is given by the formula V = (4/3) * π * r^3, where V is the volume and r is the radius of the sphere.
In this case, we are given that the diameter (D) of the snowball is decreasing at a rate of 0.3 cm/min. We need to find the rate at which the volume (V) is decreasing when D = 13 cm.
To relate the diameter (D) and the radius (r), we use the relationship D = 2r. Therefore, when the diameter is 13 cm, the radius is 6.5 cm.
Now, we can differentiate the volume formula V = (4/3) * π * r^3 with respect to time (t) using the chain rule.
dV/dt = dV/dr * dr/dt
The derivative of the volume V with respect to the radius r is
dV/dr = 4π * r^2.
The rate at which the diameter is decreasing is given as -0.3 cm/min, which means that the rate at which the radius is decreasing, dr/dt, is also -0.15 cm/min.
Now substitute the values:
dV/dt = (4π * r^2) * (dr/dt)
= (4π * (6.5)^2) * (-0.15)
Calculating the expression, we get:
dV/dt ≈ - 196.35π
The negative sign indicates that the volume is decreasing, but since we are asked for the magnitude, the answer is the absolute value:
|dV/dt| ≈ 196.35π
Therefore, the rate at which the volume of the snowball is decreasing when the diameter is 13 cm is approximately 196.35π cubic cm/min.