Let's work through each part of the problem step by step.

Given:

• The radius of the original circular piece of paper is $R = 4$ inches.
• A sector with an arc length of $x$ is removed, and the remaining paper is joined to form a cone.

(a) Why is the circumference of the base of the cone $8\pi - x$?

1. The entire circumference of the original circular piece of paper is given by: $C = 2\pi R = 2\pi \cdot 4 = 8\pi$
2. When a sector with an arc length of $x$ is cut out, the remaining arc length (which becomes the circumference of the cone's base) is: $8\pi - x$

Therefore, the circumference of the base of the cone is $8\pi - x$.

(b) Express the radius $r$ as a function of $x$

1. The circumference of the cone's base is $2\pi r$.
2. Since the circumference of the cone's base is equal to the remaining arc length after cutting, we have: $2\pi r = 8\pi - x$
3. Solving for $r$: $r = \frac{8\pi - x}{2\pi} = 4 - \frac{x}{2\pi}$

Thus, the radius $r$ of the cone is: $r(x) = 4 - \frac{x}{2\pi}$

(c) Express the height $h$ as a function of $x$

1. The slant height of the cone (the distance from the tip of the cone to the edge of the base) is equal to the radius of the original circular piece, $R = 4$.
2. We can use the Pythagorean theorem to relate the height $h$, radius $r$, and slant height $R$ in the cone: $h = \sqrt{R^2 - r^2}$
3. Substitute $R = 4$ and $r = 4 - \frac{x}{2\pi}$ (from part (b)): $h(x) = \sqrt{4^2 - \left(4 - \frac{x}{2\pi}\right)^2}$ Simplifying further: $h(x) = \sqrt{16 - \left(16 - 8\frac{x}{2\pi} + \frac{x^2}{4\pi^2}\right)}$ $h(x) = \sqrt{8\frac{x}{\pi} - \frac{x^2}{4\pi^2}}$

So, the height $h$ in terms of $x$ is: $h(x) = \sqrt{8\frac{x}{\pi} - \frac{x^2}{4\pi^2}}$

(d) Express the volume $V$ of the cone as a function of $x$

The volume $V$ of a cone is given by: $V = \frac{1}{3} \pi r^2 h$

Substitute $r = 4 - \frac{x}{2\pi}$ and $h = \sqrt{8\frac{x}{\pi} - \frac{x^2}{4\pi^2}}$:

1. First, find $r^2$: $r^2 = \left(4 - \frac{x}{2\pi}\right)^2 = 16 - 8\frac{x}{\pi} + \frac{x^2}{4\pi^2}$
2. Now, substitute $r^2$ and $h$ into the volume formula: $V(x) = \frac{1}{3} \pi \left(16 - 8\frac{x}{\pi} + \frac{x^2}{4\pi^2}\right) \sqrt{8\frac{x}{\pi} - \frac{x^2}{4\pi^2}}$

This is the volume $V$ as a function of $x$. It can be simplified further, but this form accurately represents the volume of the cone.

Would you like a further breakdown of any of these steps or simplifications?

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Here are some related questions to explore further:

1. What happens to the cone's volume as $x$ approaches 0?
2. How does the height $h$ change as $x$ increases?
3. Could you solve for the maximum possible volume of the cone given a range for $x$?
4. How does the shape of the cone (height and radius) change as $x$ varies?
5. How would the problem change if the initial radius of the paper were different?

Tip: Always double-check the physical dimensions of derived formulas to ensure they make sense in practical applications!