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 10 cm.
Can someone help & explain how to solve this question? Please.
v = 4/3 pi r^3
dv/dt = 4pi r^2 dr/dt
= 4pi * 25cm^2 * -0.3cm/min
= -30pi cm^3/min
small correction to Steve's solution
d(diameter)/dt = -.3
dr/dt = -.15
so dv/dt = 4π(25)(-.15) = -15π 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) * π * r^3
where V is the volume and r is the radius.
We are given that the diameter is decreasing at a rate of 0.3 cm/min. Since the radius is half the diameter, the rate at which the radius is changing can be found by dividing the rate at which the diameter is changing by 2:
dr/dt = -0.3 cm/min / 2 = -0.15 cm/min
Note that the negative sign indicates that the radius is decreasing.
We are asked to find dV/dt, the rate at which the volume is decreasing. To do this, we need to differentiate the volume equation with respect to time:
dV/dt = d/dt[(4/3) * π * r^3]
Using the chain rule, we have:
dV/dt = (4/3) * π * 3r^2 * dr/dt
Since we know the rate at which the radius is changing (dr/dt = -0.15 cm/min), and we are given that the sphere has a diameter of 10 cm, we can substitute these values into the equation:
r = 10 cm / 2 = 5 cm
dV/dt = (4/3) * π * 3(5 cm)^2 * (-0.15 cm/min)
Simplifying, we find:
dV/dt = -0.3π cm^3/min
Therefore, the volume of the snowball is decreasing at a rate of 0.3π cm^3/min when the diameter is 10 cm.
Sure! To solve this problem, we can use the formula for the volume of a sphere, which is V = (4/3) * π * r^3, where V is the volume and r is the radius.
Given that the snowball is melting such that its diameter is decreasing at a rate of 0.3 cm/min, we can find the rate at which the radius is changing by dividing the rate of change of the diameter by 2. So, the rate at which the radius is changing is -0.3 cm/min / 2 = -0.15 cm/min. The negative sign indicates that the radius is decreasing.
We are asked to find at what rate the volume of the snowball is decreasing when the diameter is 10 cm. Since diameter = 2 * radius, when the diameter is 10 cm, the radius will be 10 cm / 2 = 5 cm.
To find the rate at which the volume is decreasing at this point, we need to differentiate the volume formula with respect to time (t) and multiply it by the rate at which the radius is changing (dr/dt):
dV/dt = (dV/dr) * (dr/dt)
Differentiating the volume formula with respect to the radius (r), we get:
dV/dr = 4 * π * r^2
Plugging in the value of the radius when the diameter is 10 cm, which is 5 cm, we get:
dV/dr = 4 * π * (5^2)
= 4 * π * 25
= 100 * π
Now, we can plug this value and the rate at which the radius is changing, which is -0.15 cm/min, into the equation:
dV/dt = (100 * π) * (-0.15)
= -15π cm^3/min
So, the volume of the snowball is decreasing at a rate of -15π cm^3/min when the diameter is 10 cm.