Find the length of the curve x = (y^3/6) + (1/2y) for 1 <= y <= 2

To find the length of the curve \( x = \frac{y^3}{6} + \frac{1}{2y} \) for \( 1 \leq y \leq 2 \), we use the formula for the arc length of a curve defined parametrically or as a function of \( y \). The general formula for arc length when \( x \) is a function of \( y \) is:

\[
L = \int_{y_1}^{y_2} \sqrt{1 + \left( \frac{dx}{dy} \right)^2} \, dy
\]

### Step 1: Compute \( \frac{dx}{dy} \)
Given:
\[
x = \frac{y^3}{6} + \frac{1}{2y}
\]
Differentiate \( x \) with respect to \( y \):

\[
\frac{dx}{dy} = \frac{d}{dy} \left( \frac{y^3}{6} \right) + \frac{d}{dy} \left( \frac{1}{2y} \right)
\]
\[
\frac{dx}{dy} = \frac{3y^2}{6} - \frac{1}{2y^2} = \frac{y^2}{2} - \frac{1}{2y^2}
\]

### Step 2: Compute \( \left( \frac{dx}{dy} \right)^2 \)
Now square \( \frac{dx}{dy} \):

\[
\left( \frac{dx}{dy} \right)^2 = \left( \frac{y^2}{2} - \frac{1}{2y^2} \right)^2
\]
This expands as:

\[
\left( \frac{dx}{dy} \right)^2 = \left( \frac{y^4}{4} - \frac{1}{2} + \frac{1}{4y^4} \right)
\]

### Step 3: Set up the arc length integral
The arc length formula becomes:

\[
L = \int_1^2 \sqrt{1 + \left( \frac{y^4}{4} - \frac{1}{2} + \frac{1}{4y^4} \right)} \, dy
\]

Simplify the expression inside the square root:

\[
L = \int_1^2 \sqrt{\frac{y^4}{4} + \frac{1}{2} + \frac{1}{4y^4}} \, dy
\]

### Step 4: Evaluate the integral
The integral does not have a simple elementary antiderivative, so it would generally be evaluated using numerical methods.

I can compute this integral numerically to give the final arc length.

Let me compute it.The length of the curve \( x = \frac{y^3}{6} + \frac{1}{2y} \) for \( 1 \leq y \leq 2 \) is approximately \( 1.417 \) units.

Would you like more details on the process or have any other questions?

Here are some related questions for further exploration:
1. How does the arc length formula change if the curve is given as \( y \) as a function of \( x \)?
2. Can you compute the arc length of a straight line using the same formula?
3. What is the arc length of the curve \( x = y^2 \) from \( y = 0 \) to \( y = 1 \)?
4. How would the computation differ if \( x = \frac{y^3}{6} + \frac{1}{2y} \) were parametrized differently?
5. Can the arc length formula be used to find surface areas of revolution?

**Tip:** For curves with complex derivatives, numerical methods such as Simpson's rule or Gaussian quadrature are commonly used to estimate arc lengths.

Learn how to find the arc length of the curve x = (y^3/6) + (1/2y) from y = 1 to y = 2 using the arc length formula. This solution involves differentiation and integral calculus, making it suitable for undergraduate students studying calculus.

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