Let's denote the volume of the balloon as $V$ and the radius as $r$. We are given that the rate of change of $V$ with respect to time is constant and equal to 10 cubic feet per minute. We want to find the rate of change of $r$ with respect to time when $V = 1000$.

Since we are given that the rate of change of $V$ is constant, we know that $\frac{dV}{dt} = 10$.

The volume of a sphere with radius $r$ is given by the formula $V = \frac{4}{3}\pi r^3$. We can express $\frac{dV}{dt}$ in terms of $r$ and $\frac{dr}{dt}$ using the Chain Rule:

$\frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt}$.

Write the equation again, putting in the given values:

$10 = \frac{dV}{dr} \cdot \frac{dr}{dt}$.

To find $\frac{dV}{dr}$, we differentiate the volume formula with respect to $r$:

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

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

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

Now we can substitute this expression and the given value for $\frac{dV}{dt}$ back into the original equation:

$10 = 4\pi r^2 \cdot \frac{dr}{dt}$.

We can now solve for $\frac{dr}{dt}$:

$\frac{dr}{dt} = \frac{10}{4\pi r^2}$.

When $V = 1000$, we can find the corresponding value of $r$ by rearranging the volume formula:

$1000 = \frac{4}{3}\pi r^3$.

Solving for $r$:

$r^3 = \frac{3}{4\pi}\cdot 1000$

$r^3 = \frac{3000}{4\pi} = \frac{750}{\pi}$

$r = \left(\frac{750}{\pi}\right)^{1/3}$.

Finally, substitute this value back into the expression for $\frac{dr}{dt}$:

$\frac{dr}{dt} = \frac{10}{4\pi \left(\frac{750}{\pi}\right)^{2/3}}$.

Therefore, when the volume of the balloon is 1000 cubic feet, the rate at which the radius of the balloon is increasing is $\frac{10}{4\pi \left(\frac{750}{\pi}\right)^{2/3}}$.

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