Find the exact length of the curve.
x =
y4/8 +  1/4y2
,   1 ≤ y ≤ 3

To find the exact length of the curve given by the equation \( x = \frac{y^4}{8} + \frac{1}{4}y^2 \) for \( 1 \leq y \leq 3 \), we will use the formula for the arc length of a curve \( x = f(y) \), which is:

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

### Step 1: Find \(\frac{dx}{dy}\)
The function for \( x \) is given as:
\[
x = \frac{y^4}{8} + \frac{1}{4}y^2
\]

We first compute the derivative \(\frac{dx}{dy}\):

\[
\frac{dx}{dy} = \frac{d}{dy} \left( \frac{y^4}{8} + \frac{1}{4}y^2 \right) = \frac{4y^3}{8} + \frac{2y}{4} = \frac{y^3}{2} + \frac{y}{2}
\]

### Step 2: Square \(\frac{dx}{dy}\)
Next, we square this derivative:

\[
\left(\frac{dx}{dy}\right)^2 = \left( \frac{y^3}{2} + \frac{y}{2} \right)^2
\]
Expanding this:

\[
\left( \frac{y^3}{2} + \frac{y}{2} \right)^2 = \frac{y^6}{4} + \frac{y^4}{2} + \frac{y^2}{4}
\]

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

\[
L = \int_1^3 \sqrt{1 + \frac{y^6}{4} + \frac{y^4}{2} + \frac{y^2}{4}} \, dy
\]

### Step 4: Simplify the integrand
Inside the square root, we can combine the terms:

\[
1 + \frac{y^6}{4} + \frac{y^4}{2} + \frac{y^2}{4}
\]

So the integrand becomes:

\[
\sqrt{\frac{y^6}{4} + \frac{y^4}{2} + \frac{y^2}{4} + 1}
\]

### Step 5: Solve the integral
This integral doesn't have a simple elementary form, so it needs to be evaluated either numerically or with more advanced integration techniques.

Let me calculate the numerical value of this integral for the exact length.The exact length of the curve for \( x = \frac{y^4}{8} + \frac{1}{4}y^2 \) over the interval \( 1 \leq y \leq 3 \) is approximately \( 12.276 \) units.

Would you like further clarification on any step or details?

Here are some related questions to explore:
1. How is arc length derived for curves in parametric form?
2. How would the calculation differ if the curve was defined in polar coordinates?
3. What is the geometric interpretation of arc length in three dimensions?
4. Can you apply this arc length formula for curves defined by \( y = f(x) \)?
5. How would the presence of a trigonometric function in \( f(y) \) affect the integration?

**Tip:** For functions that are difficult to integrate analytically, numerical methods such as Simpson's rule or trapezoidal rule can be helpful.

Find the exact length of the curve. x = y^4/8 + 1/4y^2, 1 ≤ y ≤ 3

Find the exact length of the curve x = y^4/8 + 1/4y^2 over the interval 1 ≤ y ≤ 3 using the arc length formula. This solution involves differentiation and integral calculus. Suitable for first-year calculus students.

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