Helium is pumped into a spherical balloon at a constant rate of 2 cubic feet per second. How fast is the radius increasing after 3 minutes?. At what time (if any) is the radius increasing at a rate of 120 feet per second? (Volume of a sphere: $$ V=\frac{4}{3} \pi r^{3} $$ ) The radius is increasing at a rate of $$ 0.0082 \mathrm{ft} / \mathrm{sec} $$. (Type an integer or a decimal. Do not round until the final answer. Then round to four decimal places as needed.) Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The time at which the radius increases at a rate of 120 feet per second is $$ \square $$ sec (Type an integer or a decimal. Do not round until the final answer. Then round to five decimal places as needed.) B. There is no solution.

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Alright, I need to solve this problem about the rate at which the radius of a spherical balloon is increasing as helium is pumped into it. The balloon is being inflated at a constant rate of 2 cubic feet per second. I need to find two things:

1. The rate at which the radius is increasing after 3 minutes.
2. The time at which the radius is increasing at a rate of 120 feet per second.

First, let's recall the formula for the volume of a sphere:

$$ V = \frac{4}{3} \pi r^3 $$

Given that the volume is increasing at a constant rate of 2 cubic feet per second, I can denote this rate as $$ \frac{dV}{dt} = 2 $$ cubic feet per second.

To find the rate at which the radius is increasing, $$ \frac{dr}{dt} $$, I'll need to differentiate the volume formula with respect to time $$ t $$:

$$ \frac{dV}{dt} = \frac{d}{dt} \left( \frac{4}{3} \pi r^3 \right) $$
$$ \frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt} $$

Now, I can solve for $$ \frac{dr}{dt} $$:

$$ \frac{dr}{dt} = \frac{\frac{dV}{dt}}{4 \pi r^2} $$
$$ \frac{dr}{dt} = \frac{2}{4 \pi r^2} $$
$$ \frac{dr}{dt} = \frac{1}{2 \pi r^2} $$

So, the rate at which the radius is increasing depends on the current radius $$ r $$.

**Part 1: Rate after 3 minutes**

First, convert 3 minutes to seconds:

$$ 3 \text{ minutes} = 3 \times 60 = 180 \text{ seconds} $$

Now, I need to find the radius $$ r $$ after 180 seconds. To do this, I'll integrate the rate of change of volume to find the total volume after 180 seconds:

$$ V = \int \frac{dV}{dt} dt = \int 2 dt = 2t + C $$

Assuming the balloon starts with zero volume, $$ C = 0 $$:

$$ V = 2t $$

After 180 seconds:

$$ V = 2 \times 180 = 360 \text{ cubic feet} $$

Now, use the volume formula to find $$ r $$:

$$ 360 = \frac{4}{3} \pi r^3 $$
$$ r^3 = \frac{360 \times 3}{4 \pi} $$
$$ r^3 = \frac{1080}{4 \pi} $$
$$ r^3 = \frac{270}{\pi} $$
$$ r = \left( \frac{270}{\pi} \right)^{1/3} $$

Calculating this:

$$ r \approx \left( \frac{270}{3.1416} \right)^{1/3} $$
$$ r \approx (85.73)^{1/3} $$
$$ r \approx 4.42 \text{ feet} $$

Now, plug this back into the rate equation:

$$ \frac{dr}{dt} = \frac{1}{2 \pi (4.42)^2} $$
$$ \frac{dr}{dt} = \frac{1}{2 \times 3.1416 \times 19.5364} $$
$$ \frac{dr}{dt} = \frac{1}{123.10} $$
$$ \frac{dr}{dt} \approx 0.0081 \text{ feet per second} $$

So, the radius is increasing at approximately 0.0082 feet per second after 3 minutes.

**Part 2: Time when the radius is increasing at 120 feet per second**

Set $$ \frac{dr}{dt} = 120 $$ feet per second and solve for $$ t $$:

$$ 120 = \frac{1}{2 \pi r^2} $$
$$ r^2 = \frac{1}{2 \pi \times 120} $$
$$ r^2 = \frac{1}{240 \pi} $$
$$ r = \left( \frac{1}{240 \pi} \right)^{1/2} $$
$$ r = \frac{1}{\sqrt{240 \pi}} $$
$$ r \approx \frac{1}{\sqrt{753.98}} $$
$$ r \approx \frac{1}{27.45} $$
$$ r \approx 0.0364 \text{ feet} $$

Now, use the volume formula to find the time $$ t $$:

$$ V = \frac{4}{3} \pi r^3 $$
$$ V = \frac{4}{3} \pi (0.0364)^3 $$
$$ V = \frac{4}{3} \pi (0.000047) $$
$$ V \approx \frac{4}{3} \times 3.1416 \times 0.000047 $$
$$ V \approx 0.000196 \text{ cubic feet} $$

Since $$ V = 2t $$:

$$ 0.000196 = 2t $$
$$ t = \frac{0.000196}{2} $$
$$ t = 0.000098 \text{ seconds} $$

So, the radius is increasing at 120 feet per second after approximately 0.000098 seconds.

The radius is increasing at a rate of 0.0082 feet per second after 3 minutes. A. The time at which the radius increases at a rate of 120 feet per second is 0.000098 seconds.

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