A piece of ice stays in the shape of a sphere as it melts. The volume of the ice is decreasing at a constant rate of 3π/2 cubic feet per hour. What is the rate of change of the surface area of the piece of ice when the radius is 10 feet?
I found the rate of change of the radius to be -3/800 feet per hour and the rate of change of the surface area to be -3π/10 square feet per hour.
A = 4πr^2
dA/dt = 8πr dr/dt
v = 4/3 πr^3
dv/dt = 4πr^2 dr/dt
when r=10, we have
-3π/2 = 400π dr/dt
dr/dt = -3/800
dA/dt = 80π * -3/800 = -3π/10
You are correct
Well, well, well, we have a melting ice problem here. Now, don't worry, I'll clown around and help you out!
You did a great job finding the rate of change of the radius and the rate of change of the surface area. Now, let's put on our clown shoes and calculate the rate of change of the surface area.
We know that the volume of the ice is decreasing at a constant rate. And we also know that the volume of a sphere is given by the formula V = (4/3)πr^3, where r is the radius.
If the volume is decreasing at a rate of 3π/2 cubic feet per hour, then we have:
(4/3)π(10^3) * (dr/dt) = -3π/2
Simplifying this equation, we have:
(4/3) * 1000 * (dr/dt) = -3/2
(4/3) * 1000 * (dr/dt) = -3/2
Now, let's solve for (dr/dt):
(dr/dt) = -3/2 * (3/4) * (1/1000)
(dr/dt) = -9/8 * (1/1000)
(dr/dt) = -9/8000
So, the rate of change of the radius is -9/8000 feet per hour.
Now, to find the rate of change of the surface area, we can use the formula for the surface area of a sphere: A = 4πr^2.
Differentiating both sides of the equation with respect to time (t), we get:
dA/dt = 8πr * (dr/dt)
Substituting the values we know, we have:
dA/dt = 8π(10) * (-9/8000)
dA/dt = -90π/200
dA/dt = -9π/20
So, the rate of change of the surface area when the radius is 10 feet is -9π/20 square feet per hour.
There you have it! I hope my clowning around made this problem a bit more entertaining.
To find the rate of change of the surface area, we can use the formula for the surface area of a sphere, which is given by:
S = 4πr^2
where S is the surface area and r is the radius.
First, let's differentiate both sides of the equation with respect to time (t):
dS/dt = d/dt (4πr^2)
Using the chain rule, we can rewrite this as:
dS/dt = 8πr (dr/dt)
Given that the volume of the ice is decreasing at a constant rate of 3π/2 cubic feet per hour (which is equivalent to -3/2 cubic feet per hour), we can relate the change in volume (dV/dt) to the change in radius (dr/dt) using the formula for the volume of a sphere:
V = (4π/3)r^3
Differentiating both sides with respect to time (t), we have:
dV/dt = d/dt ((4π/3)r^3)
dV/dt = 4πr^2 (dr/dt)
Since we are given that dV/dt = -3π/2 cubic feet per hour, we can substitute this into the equation above:
-3π/2 = 4πr^2 (dr/dt)
Simplifying, we have:
-3/2 = 4r^2 (dr/dt)
Now, we can solve for dr/dt:
dr/dt = (-3/2) / (4r^2)
Substituting r = 10 feet (since we are interested in finding the rate of change when the radius is 10 feet), we have:
dr/dt = (-3/2) / (4(10)^2)
dr/dt = (-3/2) / 400
dr/dt = -3/800 feet per hour
Now, we can substitute this value of dr/dt into the earlier equation we derived for dS/dt:
dS/dt = 8πr (dr/dt)
dS/dt = 8π(10)(-3/800)
dS/dt = -3π/10 square feet per hour
Therefore, the rate of change of the surface area of the piece of ice when the radius is 10 feet is -3π/10 square feet per hour.
To find the rate of change of the surface area, we need to differentiate the surface area formula of a sphere with respect to time. The formula for the surface area of a sphere is:
A = 4πr^2
Where A is the surface area and r is the radius.
Given that the rate of change of the volume is constant at -3π/2 cubic feet per hour, we can use the volume formula of a sphere to find the rate of change of the radius. The formula for the volume of a sphere is:
V = (4/3)πr^3
Differentiating the above equation with respect to time, we have:
dV/dt = 4πr^2 dr/dt
Where dV/dt is the rate of change of volume, and dr/dt is the rate of change of the radius.
Plugging in the given rate of change of the volume (-3π/2 cubic feet per hour), and the radius (r = 10 feet), we can solve for dr/dt.
-3π/2 = 4π(10^2) dr/dt
-3/2 = 400 dr/dt
dr/dt = -3/800 feet per hour
So, the rate of change of the radius is -3/800 feet per hour.
To find the rate of change of the surface area, we differentiate the surface area formula with respect to time:
dA/dt = 8πr dr/dt
Plugging in the given radius (r = 10 feet) and the rate of change of the radius (dr/dt = -3/800 feet per hour), we can solve for dA/dt.
dA/dt = 8π(10)(-3/800) = -3π/10 square feet per hour
So, the rate of change of the surface area is -3π/10 square feet per hour.