If a spherical balloon is inflated and its volume is increasing at a rate of 6 in^3/min. What is the rate of change of the radius when the radius is 3 inches??
V = (4/3)πr^3
dV/dt = 4πr^2 dr/dt
given: dV/dt = 6, r=3
6 = 4π (9) dr/dt
dr/dt = 6/(36π) inches/min
= 1/(6π) in/min
Ah, the world of inflatable fun! Let's find the rate of change of the radius.
We know that the volume of a sphere is given by V = (4/3)πr³, where V is the volume and r is the radius. To find the rate of change of the radius, we can differentiate both sides with respect to time t:
dV/dt = d((4/3)πr³)/dt
But hold your balloons, we need to find dV/dt! Thankfully, you've provided that information. dV/dt = 6 in³/min.
Now, let's substitute the radius value you've given, r = 3 inches, into the equation so we can solve for d(r)/dt:
6 in³/min = d((4/3)π(3)³)/dt
6 in³/min = d[(4/3)(π)(27)]/dt
6 in³/min = d(36π)/dt
6 in³/min = 0
Wait a minute... it seems that something's not quite right here. According to the equation, the rate of change of the radius when r = 3 inches is... *drumroll*... zero! But that can't be! We can't have a balloon with a radius that isn't changing!
Well, partner, it seems that my balloon math skills might need a little inflating. I apologize for the confusion. Let's try a different approach or double-check those figures.
To find the rate of change of the radius, we need to use the formula for the volume of a sphere:
V = (4/3) * π * r^3
First, let's differentiate both sides of the equation with respect to time (t):
dV/dt = d/dt[(4/3) * π * r^3]
The derivative of 4/3 is 0, and the derivative of π is also 0. Therefore, the equation simplifies to:
dV/dt = 4πr^2 * dr/dt
We are given that the volume is increasing at a rate of 6 in^3/min (dV/dt = 6 in^3/min), and we need to find the rate of change of the radius (dr/dt) when the radius is 3 inches (r = 3). Plugging in these values, we have:
6 in^3/min = 4π(3)^2 * dr/dt
Now, let's solve for dr/dt:
6 in^3/min = 36π * dr/dt
dr/dt = 6 in^3/min / (36π)
Simplifying further:
dr/dt = 1 / (6π) in/min
Therefore, the rate of change of the radius when the radius is 3 inches is approximately 0.053 in/min.
To find the rate of change of the radius, we can use the formula for the volume of a sphere, which is given by:
V = (4/3) * π * r^3
where V represents the volume and r represents the radius. Since the volume is increasing at a rate of 6 in^3/min, we can differentiate the volume equation with respect to time (t) to find the rate of change of the volume:
dV/dt = (d/dt) [(4/3) * π * r^3]
To differentiate the equation, we use the chain rule:
dV/dt = (4/3) * π * (3r^2) * (dr/dt)
We can rearrange the equation to solve for dr/dt:
dr/dt = (dV/dt) / [(4/3) * π * (3r^2)]
Substituting the given values, with dV/dt = 6 in^3/min and r = 3 inches, we can calculate the rate of change of the radius:
dr/dt = (6 in^3/min) / [(4/3) * π * (3(3)^2)]
Simplifying the equation further:
dr/dt = (6 in^3/min) / [(4/3) * π * 27]
dr/dt = (6 in^3/min) / (36π)
dr/dt ≈ 0.053 in/min
Therefore, when the radius is 3 inches, the rate of change of the radius is approximately 0.053 inches per minute.