A spherical balloon is inflated with gas at the rate of 20 cubic feet per minute. How fast is the radius of the balloon increasing at the instant the radius is (a) 1 foot and (b) 2 feet?
v = 4/3 pi r^3
dv/dt = 4 pi r^2 dr/dt
Now just plug in your numbers.
To determine how fast the radius of the balloon is increasing, we can use the volume of a sphere formula and differentiate it with respect to time. The formula for the volume of a sphere is:
V = (4/3)πr³
Where V is the volume and r is the radius of the sphere.
Differentiating both sides with respect to time (t), we get:
dV/dt = d/dt (4/3)πr³
dV/dt represents the rate at which the volume of the balloon is changing with respect to time, and dr/dt represents the rate at which the radius is changing with respect to time.
Now let's substitute the given information. We are given that dV/dt, the rate at which the volume is changing, is 20 cubic feet per minute. So, we have:
20 = d/dt (4/3)πr³
To find the rate at which the radius is changing, dr/dt, we need to solve for it by rearranging the equation and substituting the given radius values.
(a) When the radius is 1 foot:
Substitute r = 1 into the equation:
20 = d/dt (4/3)π(1)³
Simplify:
20 = d/dt (4/3)π
To solve for dr/dt, we can multiply both sides by 3/(4π) to isolate it:
20 * 3/(4π) = d/dt
Evaluate this expression to find the rate at which the radius is changing.
(b) When the radius is 2 feet:
Substitute r = 2 into the equation:
20 = d/dt (4/3)π(2)³
Simplify:
20 = d/dt (4/3)(8π)
To solve for dr/dt, we can multiply both sides by 3/(32π) to isolate it:
20 * 3/(32π) = d/dt
Evaluate this expression to find the rate at which the radius is changing.
To find how fast the radius of the balloon is increasing, we can use the formula for the volume of a sphere. The formula is V = (4/3)πr^3, where V is the volume and r is the radius.
We are given that the balloon is being inflated at a rate of 20 cubic feet per minute. This means that the volume of the balloon is increasing at a rate of 20 cubic feet per minute.
Let's differentiate the volume equation with respect to time to find an expression for the rate of change of volume with respect to time:
dV/dt = d/dt (4/3)πr^3
= 4πr^2 (dr/dt)
Now we can substitute the given rate of change of volume and solve for dr/dt:
20 = 4πr^2 (dr/dt)
(a) When the radius is 1 foot, we can substitute r=1 into the equation:
20 = 4π(1)^2 (dr/dt)
20 = 4π (dr/dt)
(dr/dt) = 20/(4π)
(dr/dt) ≈ 1.59 feet per minute
Hence, when the radius is 1 foot, the radius of the balloon is increasing at a rate of approximately 1.59 feet per minute.
(b) When the radius is 2 feet, we can substitute r=2 into the equation:
20 = 4π(2)^2 (dr/dt)
20 = 4π(4) (dr/dt)
(dr/dt) = 20/(4π*4)
(dr/dt) ≈ 0.40 feet per minute
Therefore, when the radius is 2 feet, the radius of the balloon is increasing at a rate of approximately 0.40 feet per minute.