91Ó°ÊÓ

A cycloid is a type of curve generated by a point on the rim of a circular wheel as the wheel rolls along a straight line. This curve is unique and has various interesting properties. For the given parametric equations, the cycloid is defined as:

- \( x(t) = 2(t - \text{sin}(t)) \)
- \( y(t) = 2(1 - \text{cos}(t)) \)
The parameter `t` usually represents the angle through which the circle has rotated. At `t=0`, the point is at the bottom of the wheel, and as `t` increases, the point traces out the cycloid shape.

To find the arc length of a parametric curve like our cycloid, we first need the derivatives of the parametric functions with respect to the parameter `t`. For the cycloid equations:

- \( x(t) = 2(t - \text{sin}(t)) \)
- \( y(t) = 2(1 - \text{cos}(t)) \)
we calculate \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \):

• For \( x(t) \): \( \frac{dx}{dt} = 2(1 - \text{cos}(t)) \)
• For \( y(t) \): \( \frac{dy}{dt} = 2 \text{sin}(t) \)

These derivatives describe the rate of change of `x` and `y` with respect to `t`, and are vital for calculating the arc length of the cycloid.

The arc length of a curve given by parametric equations \( x(t) \) and \( y(t) \) can be found using the following formula:

\[ s = \int_{a}^{b} \sqrt{ \( \frac{dx}{dt} \)^{2} + \( \frac{dy}{dt} \)^{2}} \, dt \]

For our cycloid, we substitute the derivatives we calculated earlier:
\[ s = \int_{0}^{t} \sqrt{ \( 2(1 - \text{cos}(t)) \)^{2} + \( 2 \text{sin}(t) \)^{2} } \, dt \]

Next, simplify the expression inside the integral. We simplify:
\[ 4(1 - \cos(t))^{2} + 4 \sin^{2}(t) = 4(2 - 2 \cos(t)) = 8(1 - \cos(t)) \]

Which further simplifies the integral to:
\[ s = 2 \sqrt{2} \int_\{0}^\{t} \sqrt{1 - \text{cos}(t)} \, dt \]

To further simplify our integral involving the cycloid's arc length, we utilize a trigonometric identity. Specifically, we use:
\[ 1 - \cos(t) = 2 \sin^{2}( t/2 ) \]

Using this identity, our integral becomes:
\[ s = 2 \sqrt{2} \int_\{0}^\{t} \sqrt{2} \sin(t/2) \, dt \]

Which can be simplified to:
\[ s = 4 \int_\{0}^\{t} \sin(t/2) \, dt \]

From here, we use substitution to solve the integral. Let `u = t/2`, and `du = dt/2`, so `dt = 2du`. Changing the limits of integration accordingly remains 0 and `t/2`:

\[ s = 4 \int_\{0}^\{t/2} \sin(u) \, 2 du \]

Which simplifies to:
\[ s = 8( - \cos(u) ) \bigg|_\{0}^\{t/2} \]

Thus, evaluating the integral gives us:
\[ s = 8(1 - \cos(t/2)) \]

This result elegantly ties the concepts of parametric equations, derivatives, and trigonometric identities together to find the arc length of our cycloid.