Consider the following. x = t − 2 sin(t), y = 1 − 2 cos(t), 0 ≤ t ≤ 2π Set up an integral that represents the length of the curve. 2π 0 dt Use your calculator to find the length correct to four decimal places.

L = 13.3649

Step-by-step explanation:

We are given;

x = t − 2 sin(t)

dx/dt = 1 - 2 cos(t)

Also, y = 1 − 2 cos(t)

dy/dt = 2 sin(t)

0 ≤ t ≤ 2π

The arc length formula is;

L = (α,β)∫√[(dx/dt)² + (dy/dt)²]dt

Where α and β are the boundary points. Thus, applying this to our question, we have;

L = (0,2π)∫√((1 - 2 cos(t))² + (2 sin(t))²)dt

L = (0,2π)∫√(1 - 4cos(t) + 4cos²(t) + 4sin²(t))dt

L = (0,2π)∫√(1 - 4cos(t) + 4(cos²(t) + sin²(t)))dt

From trigonometry, we know that;

cos²t + sin²t = 1.

Thus;

L = (0,2π)∫√(1 - 4cos(t) + 4)dt

L = (0,2π)∫√(5 - 4cos(t))dt

Using online integral calculator, we have;

L = 13.3649

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