How do you do this question?

8√2

2√2

Step-by-step explanation:

Using arc length for parametric equations:

s = ∫ₐᵇ √((dx/dt)² + (dy/dt)²) dt

x = 2 sin²t

dx/dt = 4 sin t cos t

y = 2 cos²t

dy/dt = -4 sin t cos t

s = ∫ √((4 sin t cos t)² + (-4 sin t cos t)²) dt

s = ∫ √(32 sin²t cos²t) dt

s = ∫ 4√2 sin t cos t dt

s = ∫ 2√2 sin(2t) dt

s = -√2 cos(2t)

If we evaluate this between t=0 and t=2π, we get s = 0.  This means the particle travels forward and backward on the same curve.

sin²t and cos²t have periods of π, so if we evaluate from t=0 to t=π/2:

s = -√2 cos(2 (π/2)) − -√2 cos(0)

s = √2 + √2

s = 2√2

So the particle travels 2√2 four times from t=0 to t=2π.

The distance traveled is 8√2.  The length of the curve is 2√2.

We can check our answer by eliminating the parameter.

x + y = 2 sin²t + 2 cos²t

x + y = 2

y = 2 − x

The curve is the line y = 2 − x from (0, 2) to (2, 0).  Using distance formula to find the length of the line:

d = √((x₂ − x₁)² + (y₂ − y₁)²)

d = √((2 − 0)² + (0 − 2)²)

d = 2√2

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step-by-step explanation: