A spherical balloon is being filled with air in such a way that its radius is increasing at a rate of 2 centimeters per second. At what rate is the volume of the balloon changing at the instant when its surface has an area of 4 pi square centimeters?
The balloon is spherical, so
V = (4/3)(pi)r^3
SA = 4(pi)r^2
Differentiate both with respect to time.
dV/dt = 4(pi)r^2(dr/dt)
dSA/dt = 8(pi)r(dr/dt)
We're given the rate of change in SA and need to find the rate of change in volume. Let's write an equation for it.
(dV/dt)/r = 4(pi)r(dr/dt)
(dSA/dt)/2 = 4(pi)r(dr/dt)
(dV/dt)/r = (dSA/dt)/2
(dV/dt) = (r/2)(dSA/dt)
Edit: actually we're given dr/dt and SA.
Since dV/dt = 4(pi)r^2(dr/dt) and SA = 4(pi)r^2, dV/dt = SA(dr/dt).
To find the rate at which the volume of the balloon is changing, we can use the formula for the volume of a sphere.
The volume (V) of a sphere with radius (r) is given by the formula:
V = (4/3) * π * r^3
Given that the radius of the balloon is increasing at a rate of 2 centimeters per second, we can express the rate of change of the radius as dr/dt = 2 cm/s.
We are asked to find the rate of change of the volume (dV/dt) when the surface area (A) of the balloon is 4π square centimeters.
The formula for the surface area of a sphere is:
A = 4 * π * r^2
Since we know the surface area A, we can find the radius r by rearranging the formula:
r = √(A / (4 * π))
Substituting the given value of A = 4π, we get:
r = √(4π / (4 * π))
= √(1)
= 1 cm
Now, we can differentiate the volume formula with respect to time (t):
V = (4/3) * π * r^3
Differentiating both sides of the equation with respect to time, we get:
dV/dt = (4/3) * π * 3 * r^2 * dr/dt
Substituting the given value of dr/dt = 2 cm/s and r = 1 cm, we can calculate dV/dt:
dV/dt = (4/3) * π * 3 * (1^2) * 2
= 8π cm^3/s
So, the volume of the balloon is changing at a rate of 8π cubic centimeters per second at the instant when its surface has an area of 4π square centimeters.
To find the rate at which the volume of the balloon is changing, we need to differentiate the volume with respect to time. However, before doing that, we need to find an expression for the volume of the balloon in terms of its radius.
The volume of a sphere is given by the formula V = (4/3) * pi * r^3, where V is the volume and r is the radius.
Now, we can differentiate both sides of the equation with respect to time:
dV/dt = (4/3) * pi * 3r^2 * dr/dt
Here, dV/dt represents the rate of change of the volume with respect to time (the quantity we want to find), and dr/dt represents the rate at which the radius is changing, which is given as 2 cm/s in the problem.
Now, let's substitute the given information into the equation. We are told that the surface area of the balloon is 4π square centimeters. The surface area of a sphere is given by the formula A = 4πr^2, where A is the surface area and r is the radius.
So, we have:
4π = 4πr^2
Dividing both sides by 4π, we get:
1 = r^2
Taking the square root of both sides, we find that r = 1 cm.
Now, substitute this value of r into the previous equation:
dV/dt = (4/3) * π * 3(1^2) * 2
Simplifying further:
dV/dt = 8π
Therefore, the rate at which the volume of the balloon is changing at the instant when its surface has an area of 4π square centimeters is 8π cubic centimeters per second.