what is the integral that represents the arc-length of the curve y = 1/x on the interval (2,4)

The formula for the arc length of a curve \( y = f(x) \) over the interval \([a, b]\) is:

\[
L = \int_a^b \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx
\]

For the curve \( y = \frac{1}{x} \), we first need to compute the derivative:

\[
\frac{dy}{dx} = -\frac{1}{x^2}
\]

Now, we substitute this into the arc length formula:

\[
L = \int_2^4 \sqrt{1 + \left( -\frac{1}{x^2} \right)^2} \, dx
\]

Simplifying the expression under the square root:

\[
L = \int_2^4 \sqrt{1 + \frac{1}{x^4}} \, dx
\]

This is the integral that represents the arc length of the curve \( y = \frac{1}{x} \) over the interval \((2,4)\).

Would you like details on how to solve this integral?

Here are 5 related questions to deepen understanding:
1. What is the geometric interpretation of arc length in calculus?
2. How does the formula for arc length change in parametric form?
3. How would you compute the arc length of a circle using this method?
4. Can this method be used to find the arc length of a curve in three dimensions?
5. What happens to the arc length integral if the curve \( y = 1/x \) is over an infinite interval?

**Tip:** The arc length formula can be generalized to find the length of curves in polar and parametric coordinates as well!

What is the integral that represents the arc-length of the curve y = 1/x on the interval (2,4)?

This detailed guide explains how to compute the arc length of the curve y = 1/x on the interval (2,4). Using the arc length formula and integration techniques, this solution breaks down each step. Suitable for college-level calculus students.

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