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: $$ \mathrm{V}=\frac{4}{3} \pi \mathrm{r}^{3} $$ ) 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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We start with the volume of a sphere

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

Since helium is pumped at a constant rate

$$
\frac{dV}{dt}=2 \text{ ft}^3/\text{s},
$$

differentiate both sides with respect to time $$t$$:

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

Thus,

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

**1. Finding $$\frac{dr}{dt}$$ after 3 minutes**

Three minutes is $$180$$ sec. The volume at that time is

$$
V=2t=2(180)=360 \text{ ft}^3.
$$

Using the volume formula, we have

$$
360=\frac{4}{3}\pi r^3.
$$

Solve for $$r^3$$:

$$
r^3=\frac{3\cdot360}{4\pi}=\frac{1080}{4\pi}=\frac{270}{\pi}.
$$

So,

$$
r=\sqrt[3]{\frac{270}{\pi}}.
$$

Now, substituting into the expression for $$\frac{dr}{dt}$$:

$$
\frac{dr}{dt}=\frac{1}{2\pi \left(\sqrt[3]{\frac{270}{\pi}}\right)^2}=\frac{1}{2\pi\left(\frac{270}{\pi}\right)^{2/3}}.
$$

Numerically, this gives approximately

$$
\frac{dr}{dt}\approx0.0082 \text{ ft/s}.
$$

**2. Finding the time when $$\frac{dr}{dt}=120$$ ft/s**

Set

$$
\frac{1}{2\pi r^2}=120.
$$

Solving for $$r^2$$:

$$
r^2=\frac{1}{240\pi}.
$$

Thus,

$$
r=\frac{1}{\sqrt{240\pi}}.
$$

Now, using the volume formula $$V=\frac{4}{3}\pi r^3$$ and the fact that $$V=2t$$, we have

$$
2t=\frac{4}{3}\pi r^3.
$$

Substitute $$r^3=r\cdot r^2=\frac{1}{\sqrt{240\pi}}\cdot\frac{1}{240\pi}=\frac{1}{(240\pi)^{3/2}}$$:

$$
2t=\frac{4}{3}\pi\cdot\frac{1}{(240\pi)^{3/2}}.
$$

Solve for $$t$$:

$$
t=\frac{1}{2}\cdot\frac{4\pi}{3(240\pi)^{3/2}}=\frac{2\pi}{3(240\pi)^{3/2}}.
$$

Notice that

$$
(240\pi)^{3/2}=(240\pi)\sqrt{240\pi}.
$$

So,

$$
t=\frac{2\pi}{3\cdot240\pi\sqrt{240\pi}}=\frac{2}{720\sqrt{240\pi}}=\frac{1}{360\sqrt{240\pi}}.
$$

Now, calculating numerically:

- Compute $$240\pi$$:

$$
240\pi\approx240\times3.1416\approx753.982.
$$

- Then, $$\sqrt{240\pi}\approx\sqrt{753.982}\approx27.465$$.

Thus,

$$
t\approx\frac{1}{360\times27.465}\approx\frac{1}{9887.4}\approx0.0001011 \text{ sec}.
$$

Rounding to five decimal places, we have

$$
t\approx0.00010 \text{ sec}.
$$

**Final Answers:**

- After 3 minutes, the radius is increasing at a rate of $$0.0082$$ ft/s.
- The radius increases at a rate of 120 ft/s at

$$
\textbf{approximately }0.00010 \text{ sec}.
$$

Thus, the correct choice is:

A. The time at which the radius increases at a rate of 120 feet per second is $$0.00010$$ sec.
After 3 minutes, the radius is increasing at a rate of 0.0082 ft/s. The radius increases at a rate of 120 ft/s at approximately 0.00010 seconds.

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