A spherical balloon is inflated at a rate of 10 cubic feet per minute. How fast is the radius of the balloon changing at the instant the radius is 4 feet?
And
The radius of a circle is decreasing at a rate of 2 ft/minute. Find the rate at which the area is decreasing with respect to time when the radius is 4 feet?
V = (4/3)πr^3
dV/dt = 4πr^2 dr/dt
given dV/dt = 10 ft^3/min
so at r = 4
10 = 4π(16) dr/dt
dr/dt = 10/(64π) or appr. .05 ft/min
If A = 4πr^2, do the second question the same way.
Well, well, well, we have some inflatable questions here! Let's dive in.
For the first question, we know that the volume of a sphere is given by V = (4/3)πr^3. So, if the balloon is inflating at a rate of 10 cubic feet per minute, we can say that dV/dt = 10. We want to find dr/dt, the rate at which the radius is changing when r = 4.
To solve this, we need to use the chain rule. It's the superhero of calculus! We know that V = (4/3)πr^3, so we can rewrite it as V = (4/3)πr^3. Now, let's differentiate both sides with respect to time:
dV/dt = d/dt[(4/3)πr^3].
Using the chain rule, we get:
dV/dt = (4/3)(3πr^2)(dr/dt).
Now, we substitute dV/dt = 10 and r = 4:
10 = (4/3)(3π(4)^2)(dr/dt).
Simplifying this becomes a piece of cake:
10 = (4/3)(48π)(dr/dt).
Now we can solve for dr/dt:
(dr/dt) = 10 / [(4/3)(48π)].
And just a pinch more simplification:
(dr/dt) = 10 / (16π).
So, when the radius is 4 feet, dr/dt is equal to 10 / (16π).
Now, moving on to the second question:
We know that the area of a circle is given by A = πr^2. We also know that dr/dt = -2 ft/minute, meaning the radius is decreasing.
Now, we want to find dA/dt, the rate at which the area is changing with respect to time when r = 4.
To find dA/dt, we differentiate both sides of the equation A = πr^2 with respect to time:
dA/dt = d/dt(πr^2).
Using the chain rule, we get:
dA/dt = 2πr(dr/dt).
Substituting r = 4 and dr/dt = -2, we have:
dA/dt = 2π(4)(-2) = -16π.
So, when the radius is 4 feet, the area is decreasing at a rate of -16π square feet per minute.
I hope that added a little bit of humorous math to your day!
To find the first derivative, we use the formula
\( dV/dt = (dV/dr) * (dr/dt) \),
where V is the volume of the balloon, r is the radius of the balloon, and t is time.
We are given that dV/dt = 10 cubic feet/minute.
Since the volume of a sphere is given by the formula \( V = (4/3) * π * r^3 \), we can differentiate both sides with respect to time to get dV/dr:
\( dV/dr = 4 * π * r^2 \).
Substituting the given values, we have:
\( 10 = (4 * π * 4^2) * (dr/dt) \).
To solve for dr/dt, we can divide both sides by \( (4 * π * 4^2) \):
\( dr/dt = 10 / (4 * π * 4^2) \).
Simplifying, we have:
\( dr/dt ≈ 10 / 201 ≈ 0.0498 \) ft/minute.
Therefore, the radius of the balloon is changing at a rate of approximately 0.0498 ft/minute when the radius is 4 feet.
For the second question, we want to find the rate at which the area is decreasing with respect to time. We are given that dr/dt = -2 ft/minute, where the negative sign indicates that the radius is decreasing.
The area of a circle is given by the formula A = π * r^2.
To find dA/dt, we can differentiate both sides with respect to time:
\( dA/dt = 2π * r * (dr/dt) \).
Substituting the given values, we have:
\( dA/dt = 2π * 4 * (-2) \).
Simplifying, we have:
\( dA/dt = -16π \) ft^2/minute.
Therefore, the area is decreasing at a rate of 16π square feet/minute when the radius is 4 feet.
To solve both of these problems, we can use the formulas that relate the variables (radius, volume, and area) of the given shapes (sphere and circle) to find the desired rates of change.
1. For the first problem, we are given the rate of change of volume and asked to find the rate of change of radius. This requires using the formula for the volume of a sphere:
V = (4/3)πr³
Differentiating both sides of the equation with respect to time t, we get:
dV/dt = (4/3)(π)(3r²)(dr/dt)
Since the rate of change of volume is given as 10 cubic feet per minute (dV/dt = 10), and we are asked to find the rate of change of radius (dr/dt) when the radius is 4 feet (r = 4), we can substitute these values into the equation and solve for dr/dt:
10 = (4/3)(π)(3(4²))(dr/dt)
dr/dt = 10 / [(4/3)(π)(3(4²))]
Simplifying the equation gives the value of dr/dt.
2. For the second problem, we are given the rate of change of radius and asked to find the rate of change of area. This requires using the formula for the area of a circle:
A = πr²
Differentiating both sides of the equation with respect to time t, we get:
dA/dt = (π)(2r)(dr/dt)
Since the rate of change of radius is given as 2 ft/minute (dr/dt = -2, as the radius is decreasing), and we are asked to find the rate of change of area (dA/dt) when the radius is 4 feet (r = 4), we can substitute these values into the equation and solve for dA/dt:
dA/dt = (π)(2(4))(-2)
Simplifying the equation gives the value of dA/dt.