How do you find the arc length of the curve #y = sqrt( 2 − x^2 )#, #0 ≤ x ≤ 1#?

Question

Ezekiel Alexander · Accepted Answer

I found: #s=sqrt(2)pi/4=1.11#

I plotted the arc you need (a bit of a cirle) and found:

Ezra Adams · Answer

undefined To find the arc length of the curve y = √(2 − x^2) over the interval 0 ≤ x ≤ 1, you use the formula for arc length:

$$ L = \int_{a}^{b} \sqrt{1 + (f'(x))^2} \, dx $$

Where $ f(x) = \sqrt{2 - x^2} $ is the function representing the curve.

1. Find the derivative of $ f(x) $, denoted as $ f'(x) $.
2. Plug the derivative into the arc length formula.
3. Integrate the expression from the lower bound to the upper bound of the given interval.

Here are the steps:

1. Find $ f'(x) $ by differentiating $ f(x) = \sqrt{2 - x^2} $.
   $$ f'(x) = \frac{-x}{\sqrt{2 - x^2}} $$

2. Plug $ f'(x) $ into the arc length formula:
   $$ L = \int_{0}^{1} \sqrt{1 + \left(\frac{-x}{\sqrt{2 - x^2}}\right)^2} \, dx $$

3. Simplify the expression under the square root:
   $$ L = \int_{0}^{1} \sqrt{1 + \frac{x^2}{2 - x^2}} \, dx $$

4. Integrate the expression:
   $$ L = \int_{0}^{1} \sqrt{\frac{2 - x^2 + x^2}{2 - x^2}} \, dx $$
   $$ L = \int_{0}^{1} \sqrt{\frac{2}{2 - x^2}} \, dx $$

5. Use trigonometric substitution or other methods to evaluate the integral.

Once the integral is evaluated, you'll have the arc length of the curve over the specified interval.