a spherical snowball with diameter 4 inches is removed from the freezer in June and begins melthing uniformly such that it is shrinking 2 cubic inches per minute. How fast ( in square inches per minute) is the surface area decreasing when the radius is 1 inch?
..it is shrinking 2 cubic inches per minute ----> dV/dt = -2 cubic inches/min
V = (4/3)pi(r^3)
dV/dt = 4pi(r^2)dr/dt
when r - 1 and dV/dt = -2
-2 = 4pi(1)dr/dt
dr/dt = -1/(2pi)
but A = 4pi(r^2)
then dA/dt = 8pi(r)dr/dt
so again when r=1
dA/dt = 8pi(1)(-1/2pi)
=-4 square inches/min
Well, well, looks like our spherical snowball is having a meltdown! Let's crunch some numbers and find out how fast the surface area is decreasing when the radius is 1 inch, shall we?
We're given that the sphere has a diameter of 4 inches, which means the radius is 2 inches (diameter = 2 * radius, basic math right there!). Now, we have the formula to find the surface area of a sphere, which is A = 4πr². Since we're interested in how fast the surface area is changing, we need to take the derivative of this equation.
But before we jump into that, let's determine how fast the radius is changing. We know that the snowball is shrinking uniformly at a rate of 2 cubic inches per minute, so the rate of change of the volume with respect to time (dv/dt) is -2 cubic inches per minute (negative because it's shrinking).
Now, we can relate the volume of the sphere to its radius using the formula V = (4/3)πr³. If we differentiate this equation with respect to time, we get dv/dt = 4πr²(dr/dt), where dr/dt is the rate of change of the radius with respect to time.
Since we're interested in finding dr/dt, we can rearrange the equation to solve for it. dr/dt = (dv/dt) / (4πr²). Plugging in the given values, dr/dt = -2 / (4π(2)²), which simplifies to dr/dt = -1 / 8π.
Now that we have dr/dt, we can plug it into the formula for the surface area A = 4πr² and differentiate it with respect to time. dA/dt = 8πr(dr/dt). Plugging in the values, dA/dt = 8π(1)(-1 / 8π), which simplifies to dA/dt = -1 square inch per minute.
So, when the radius is 1 inch, the surface area is decreasing at a rate of 1 square inch per minute. That snowball better watch out, or it'll disappear faster than an ice cream on a summer day!
To find how fast the surface area is decreasing when the radius is 1 inch, we need to differentiate the surface area formula with respect to time.
The formula for the surface area of a sphere is given by:
A = 4πr^2
Where A is the surface area and r is the radius.
Differentiating both sides of the equation with respect to time (t), we get:
dA/dt = d/dt (4πr^2)
To find how fast the surface area is decreasing at a specific radius, we need to plug in the values into the equation and solve for dA/dt.
Given:
diameter = 4 inches
radius = 1 inch
rate of change of volume (V) = -2 cubic inches per minute
The volume of a sphere is given by:
V = (4/3)πr^3
The rate of change of volume with respect to time (t) is:
dV/dt = d/dt ((4/3)πr^3)
The problem states that the snowball is melting uniformly, meaning the rate of change of volume is constant. Therefore,
dV/dt = -2 cubic inches per minute
We can differentiate the volume equation to find an expression for dV/dt in terms of the radius:
dV/dt = d/dt ((4/3)πr^3)
dV/dt = 4πr^2 * (dr/dt)
Since we are given that the rate of change of volume is -2 cubic inches per minute, we can substitute:
-2 = 4π(1)^2 * (dr/dt)
Simplifying, we find:
-2 = 4π * (dr/dt)
Now we can solve for dr/dt:
dr/dt = -2 / (4π)
dr/dt = -1 / (2π)
Now, substituting this value for dr/dt into our expression for dA/dt:
dA/dt = 4π(1)^2 * (dr/dt)
dA/dt = 4π * (-1 / (2π))
dA/dt = -2 square inches per minute
Therefore, when the radius is 1 inch, the surface area is decreasing at a rate of -2 square inches per minute.
To find the rate at which the surface area is decreasing, we need to differentiate the surface area formula with respect to time and then substitute the given values.
The formula for the surface area of a sphere is:
A = 4πr^2
Where A is the surface area and r is the radius.
To differentiate the equation with respect to time, we will apply the chain rule. Let's denote the rate of change of the surface area with respect to time as dA/dt and the rate of change of the radius with respect to time as dr/dt.
Differentiating the surface area formula with respect to time gives us:
dA/dt = d(4πr^2)/dt
= 8πr(dr/dt) -- Applying the chain rule
We are given that the diameter of the snowball is 4 inches, so the radius (r) would be half the diameter, which is 2 inches. However, we need to find the rate at which the surface area is decreasing when the radius is 1 inch.
To find the value of dr/dt, we are given that the snowball is shrinking 2 cubic inches per minute. Since we know that the snowball is a sphere, we can find the rate of change of the volume with respect to time (dv/dt) using the volume formula for a sphere:
V = (4/3)πr^3
Differentiating the volume formula with respect to time gives us:
dv/dt = d[(4/3)πr^3]/dt
= 4πr^2(dr/dt) -- Applying the chain rule
We know that dv/dt is -2 cubic inches per minute since the snowball is shrinking uniformly.
Now, we can substitute the values into the equation to find dr/dt:
-2 = 4π(2^2)(dr/dt)
-2 = 16π(dr/dt)
Simplifying the equation, we have:
dr/dt = -2/(16π)
dr/dt = -1/(8π) cubic inches per minute
Now we have the value of dr/dt, which is the rate at which the radius is decreasing. To find the rate at which the surface area is decreasing when the radius is 1 inch, we substitute the values into the equation:
dA/dt = 8πr(dr/dt)
= 8π(1)(-1/(8π))
= -1 square inch per minute
Therefore, when the radius is 1 inch, the surface area is decreasing at a rate of -1 square inch per minute. The negative sign indicates a decrease in the surface area.