Find the length of one arch of the cycloid.

Steps to Solve

1. Identify the parametric equations of the cycloid

The parametric equations for a cycloid, where a circle of radius $r$ rolls along a straight line, are given by: $$ x(t) = r(t - \sin(t)) $$ $$ y(t) = r(1 - \cos(t)) $$ Here, $t$ is the parameter that usually represents the angle in radians.

2. Determine the limits of integration

For one complete arch of the cycloid, $t$ varies from $0$ to $2\pi$. These limits represent the full rotation of the circle.

3. Calculate the derivatives of x and y

Next, we need to find the derivatives of $x$ and $y$ with respect to $t$: $$ \frac{dx}{dt} = r(1 - \cos(t)) $$ $$ \frac{dy}{dt} = r \sin(t) $$

4. Set up the length of the arch formula

The length (L) of one arch of the cycloid can be calculated using the formula: $$ L = \int_{0}^{2\pi} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} , dt $$

5. Substitute the derivatives into the length formula

Substituting the derivatives into the integral: $$ L = \int_{0}^{2\pi} \sqrt{(r(1 - \cos(t)))^2 + (r \sin(t))^2} , dt $$

6. Simplify the integrand

The expression inside the square root simplifies as follows: $$ L = \int_{0}^{2\pi} \sqrt{r^2(1 - \cos(t))^2 + r^2 \sin^2(t)} , dt $$ Factor out $r^2$ from the square root: $$ L = r \int_{0}^{2\pi} \sqrt{(1 - \cos(t))^2 + \sin^2(t)} , dt $$

7. Use trigonometric identities

Since $1 - \cos(t) = 2 \sin^2(t/2)$ and $\sin^2(t) = 1 - \cos^2(t)$, we can simplify the expression further using these identities.

8. Evaluate the integral

After further simplifications and integration, the result of the integral evaluates to: $$ L = 8r $$ This represents the length of one arch of the cycloid.