The sun is shining and a spherical snowball of volume 210 ft3 is melting at a rate of 14 cubic feet per hour. As it melts, it remains spherical. At what rate is the radius changing after 3 hours?
Volume= 4/3 PI r^3
dV/dt=4PI r^2 dR/dt
you are given dV/dt as 14ft^2/hr
you are looking for dr/dt
so what is the radius at t=3?
210-14*3=4/3 PI r^3, solve for r.
put that r into the dV/dt equation, and solve for dR/dt
I followed these steps, but the answer is wrong. Is there a typo here?
To find the rate at which the radius is changing after 3 hours, we need to use the given information about the volume and the rate at which it is changing.
Let's start by noting 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 volume of the snowball is 210 ft^3. So, we can write the equation as:
210 = (4/3) * π * r^3
To find the rate at which the radius is changing, we need to take the derivative of this equation with respect to time. Let's call the rate at which the radius is changing dr/dt:
dV/dt = d((4/3) * π * r^3)/dt
Since the volume is changing at a rate of 14 ft^3/hour, we can substitute dV/dt with 14:
14 = d((4/3) * π * r^3)/dt
Now, let's differentiate with respect to time using the chain rule:
14 = (4/3) * π * 3r^2 * (dr/dt)
We can simplify this equation further:
14 = 4πr^2 * (dr/dt)
Now, we need to solve for dr/dt, the rate at which the radius is changing.
Divide both sides of the equation by 4πr^2:
14 / (4πr^2) = dr/dt
After simplifying, we get:
dr/dt = 14 / (4πr^2)
Now, we can substitute the value of r after 3 hours into the equation to find the rate at which the radius is changing:
dr/dt = 14 / (4π * r^2)
To determine r, we need to use the volume formula again:
210 = (4/3) * π * r^3
Solve the equation for r:
r^3 = (3/4π) * 210
r^3 = 70/π
r = (70/π)^(1/3)
Substitute this value of r into the equation for dr/dt:
dr/dt = 14 / (4π * ((70/π)^(1/3))^2)
dr/dt ≈ 0.108 ft/hour
Therefore, the rate at which the radius is changing after 3 hours is approximately 0.108 ft/hour.