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 10 cm. (Note the answer is a positive number).
V=4 π r3 / 3
dV/dt = dV/dr . dr/dt
It's very similar to the triangle problem.
diameter is decreasing at a rate of 0.4 cm/min
---> dd/dt = -.4 cm/min or dr/dt = -.2 cm/min
V = (4/3)pi(r^3)
dV/dt = 4pi(r^2)dr/dt
so when d=10, r=5 and
dV/dt = 4pi(25)(-.2) cm^3/min
= -20pi cm^3/min
The negative rate for dV/dt tells me that the volume is decreasing, so if the word "decreasing" is used in the concluding sentence you would use 20pi.
I see it now.
To find the rate at which the volume of the snowball is decreasing, we'll need to use the chain rule from calculus.
Let's start by finding the equation for the volume of a sphere:
V = (4/3) * π * r^3
Here, V represents the volume and r represents the radius of the snowball. Given that the diameter is decreasing at a rate of 0.4 cm/min, we can use this information to find the rate at which the radius is changing.
Since the diameter is twice the radius, we have:
d = 2r
Differentiating both sides with respect to time (t), we get:
dd/dt = 2dr/dt
Given that dd/dt = -0.4 cm/min (as the diameter is decreasing), we can solve for dr/dt:
-0.4 = 2dr/dt
dr/dt = -0.4 / 2
dr/dt = -0.2 cm/min
So, the rate at which the radius is changing is -0.2 cm/min.
Now, let's differentiate the volume equation with respect to time (t) using the chain rule:
dV/dt = dV/dr * dr/dt
We can find dV/dr by differentiating the volume equation:
dV/dr = 4πr^2
Given that the diameter is 10 cm (which means the radius is 5 cm), we can substitute this value into the expression for dV/dr:
dV/dr = 4π(5)^2
dV/dr = 100π
Now, we can substitute the values we found into the equation for dV/dt:
dV/dt = (100π) * (-0.2)
dV/dt = -20π cm³/min
Since the question asks for the rate at which the volume is decreasing (which should be a positive number), we take the absolute value of the result:
Rate of volume decrease = 20π cm³/min