A spherical snowball is uniformly melting art of rate of pi/18 in^3/min. At what rate is its diameter decreasing when its surface area is 3 in^2?
a = pi d^2
when a=3, d^2 = 3/pi
v = pi/6 d^3
dv/dt = pi/2 d^2 dd/dt
So,
-pi/18 = pi/2 (3/pi) dd/dt
dd/dt = -pi/27 in/min
To find the rate at which the diameter of the snowball is decreasing, we'll need to use related rates.
Let's denote the radius of the snowball as r and the diameter as d. The surface area of a sphere is given by the formula S = 4πr^2, and we're given that the surface area is 3 in^2.
To start, let's differentiate the equation S = 4πr^2 with respect to time t (since all variables are changing with respect to time):
dS/dt = d(4πr^2)/dt
The left-hand side represents the rate of change of the surface area with respect to time. We're given that the surface area is decreasing at a rate of dS/dt = -3 in^2/min.
So now we have:
-3 in^2/min = d(4πr^2)/dt
Next, we want to find dr/dt, the rate at which the radius is changing with respect to time. Since the diameter is simply 2r, we can differentiate both sides of the equation d = 2r with respect to time t:
dd/dt = d(2r)/dt
The left-hand side represents the rate of change of the diameter with respect to time, and we're looking for this value. Let's rewrite the equation to solve for dr/dt:
dr/dt = (dd/dt) / 2
Now we have an equation for dr/dt in terms of dd/dt. We need to determine the value of dd/dt.
From the given information, we know that the snowball is melting at a rate of dV/dt = π/18 in^3/min, where V represents the volume of the snowball. The volume of a sphere is given by the formula V = (4/3)πr^3.
Differentiating the equation V = (4/3)πr^3 with respect to t, we get:
dV/dt = d((4/3)πr^3)/dt
The left-hand side represents the rate of change of the volume with respect to time, which we're given as dV/dt = π/18 in^3/min.
Now, let's differentiate the right-hand side of the equation:
π/18 = (4/3)π (d(r^3)/dt)
Using the Chain Rule, we can differentiate r^3 with respect to t:
d(r^3)/dt = 3r^2 (dr/dt)
Plugging this into the equation, we have:
π/18 = (4/3)π (3r^2)(dr/dt)
Canceling out π and simplifying, we get:
1/18 = 4r^2(dr/dt)
Now, we have an equation for dr/dt in terms of r. We want to find the value of dr/dt when the surface area is 3 in^2, so we need to substitute this into the equation for r:
3 = 4πr^2
Solving this equation for r, we get:
r^2 = 3/(4π)
r = sqrt(3/(4π))
Now substitute this value of r back into the equation:
1/18 = 4(sqrt(3/(4π)))^2(dr/dt)
Simplifying,
1/18 = 4(3/(4π))(dr/dt)
dr/dt = (1/18)(π/3)(4/3)
dr/dt = π/81 in/min
Therefore, the rate at which the diameter is decreasing is π/81 in/min when the surface area is 3 in^2.