Is the answer 4?

Question

What is the length of the curve with parametric equations x = t - cos(t), y = 1 - sin(t) from t = 0 to

t = π? (20 points)

	8
	4

Answer 1

Here, dx/dt is 1 + sin t; (dx/dt)² is 1 + 2sin t + sin²t.

Then take dy/dt as -cos t and (dy/dt)² as cos²t.

Next construct curve length or s equal to ∫_{(from 0 to π)}√[(dx/dt)² + (dy/dt)²]dt.

Place results above and put down s = ∫_{(from 0 to π)}√[1 + 2sin t + sin²t + cos²t]dt.

Rewrite this last integral as s = ∫_{(from 0 to π)}√[2 + 2sin t]dt or √2∫_{(from 0 to π)}√[1 + sin t]dt.

Using sin²θ + cos²θ = 1 and sin 2θ = 2sin θcos θ, develop s = √2∫_{(from 0 to π)}√[1 + sin 2(t/2)]dt.

Express s again as √2∫_{(from 0 to π)}√[sin²(t/2)+ 2sin (t/2)cos (t/2) + cos²(t/2)]dt or

√2∫_{(from 0 to π)}√{ [sin (t/2)+ cos (t/2)]² }dt which reduces to √2∫_{(from 0 to π)}{ cos (t/2)+ sin (t/2) }dt.

Integrate this last to √2[2sin (t/2) − 2cos (t/2) | _{(from 0 to π)}].

Apply limits of integration to gain [2√2sin (π/2) − 2√2cos (π/2) − 2√2sin (0/2) + 2√2cos (0/2)]

which calculates to 2√2[1 − 0 − 0 + 1] or 4√2 units of length.