Consider the following: $x = t - 2\sin(t), \quad y = 1 - 2\cos(t), \quad 0 \leq t \leq 4\pi$

1. Write an integral expression that represents the length of the curve described by the parametric equations.

2. Use technology to find the length of the curve. (Round your answer to four decimal places.)

Neetesh Kumar | January 3, 2025

Calculus Homework Help

This is the solution to Math 1c
Assignment: 10.2 Question Number 12
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Step-by-step solution:

Step 1: Formula for arc length of a parametric curve

The arc length of a parametric curve is given by:

$L = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt$

Here, $x = t - 2\sin(t)$ and $y = 1 - 2\cos(t)$ with $t \in [0, 4\pi]$.

Step 2: Compute $\frac{dx}{dt}$ and $\frac{dy}{dt}$

Differentiate $x = t - 2\sin(t)$:

$\frac{dx}{dt} = 1 - 2\cos(t)$

Differentiate $y = 1 - 2\cos(t)$:

$\frac{dy}{dt} = 2\sin(t)$

Step 3: Write the integral for arc length

Substitute $\frac{dx}{dt}$ and $\frac{dy}{dt}$ into the formula for $L$:

$L = \int_0^{4\pi} \sqrt{\left(1 - 2\cos(t)\right)^2 + \left(2\sin(t)\right)^2} \, dt$

Simplify the terms inside the square root:

1. Expand $\left(1 - 2\cos(t)\right)^2$:

   $\left(1 - 2\cos(t)\right)^2 = 1 - 4\cos(t) + 4\cos^2(t)$

2. Expand $\left(2\sin(t)\right)^2$:

   $\left(2\sin(t)\right)^2 = 4\sin^2(t)$

3. Combine the terms:

   $\left(1 - 2\cos(t)\right)^2 + \left(2\sin(t)\right)^2 = 1 - 4\cos(t) + 4\cos^2(t) + 4\sin^2(t)$

Use the Pythagorean identity $\sin^2(t) + \cos^2(t) = 1$:

$\left(1 - 2\cos(t)\right)^2 + \left(2\sin(t)\right)^2 = 1 - 4\cos(t) + 4(1)$

Simplify further:

$\left(1 - 2\cos(t)\right)^2 + \left(2\sin(t)\right)^2 = 5 - 4\cos(t)$

Thus, the arc length is:

$L = \int_0^{4\pi} \sqrt{5 - 4\cos(t)} \, dt$

Step 4: Use technology to evaluate the integral

Using numerical integration, evaluate:

$L = \int_0^{4\pi} \sqrt{5 - 4\cos(t)} \, dt$

After computation:

$L \approx 26.7298$

Final Answer:

1. The integral expression for the arc length is:

$L = \int_0^{4\pi} \boxed{\sqrt{5 - 4\cos(t)} \, dt}$

2. The length of the curve is approximately:

$L \approx \boxed{26.7298}$

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