Answer
$$L = \int_0^{4\pi} \sqrt{(5 -4cos(t))} \space dt \approx 26.7298$$
Work Step by Step
1. Write the formula for the length "$L$"of a curve:
$$L = \int_{\alpha}^{\beta} \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} \space dt$$
2. Find $\frac{dx}{dt}$ and $\frac{dy}{dt}$
$\frac{dx}{dt} = \frac{d(t - 2sin(t))}{dt} = 1 - 2cos(t)$
$\frac{dy}{dt} = \frac{d(1 - 2cos(t))}{dt} = 0 - 2(-sin(t)) = 2sin(t)$
3. Substitute the values into the definite integral:
$L = \int_{\alpha}^{\beta} \sqrt{(1 - 2cos(t))^2 + (2sin(t))^2} \space dt$
$L = \int_{\alpha}^{\beta} \sqrt{(1 -4cos(t) + 4cos^2(t)) + (4sin^2(t))} \space dt$
$L = \int_{\alpha}^{\beta} \sqrt{(1 -4cos(t) + 4 (cos^2(t) + sin^2(t)))} \space dt$
$L = \int_{\alpha}^{\beta} \sqrt{(1 -4cos(t) + 4)} \space dt$
$L = \int_{\alpha}^{\beta} \sqrt{(5 -4cos(t))} \space dt$
- Since $t$ goes from 0 to 4$\pi$:
$L = \int_0^{4\pi} \sqrt{(5 -4cos(t))} \space dt$
4. Write this integral on your calculator (you can use online integral calculators), and get the result.
$L \approx 26.7298$
