Find the exact length of the curve: $x = 5\cos(t) - \cos(5t), \quad y = 5\sin(t) - \sin(5t), \quad 0 \leq t \leq \pi$

Neetesh Kumar | January 3, 2025

Calculus Homework Help

This is the solution to Math 1c
Assignment: 10.2 Question Number 15
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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, $x = 5\cos(t) - \cos(5t)$ and $y = 5\sin(t) - \sin(5t)$ with $t \in [0, \pi]$.

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

Differentiate $x = 5\cos(t) - \cos(5t)$:

Using the chain rule:

$\frac{dx}{dt} = -5\sin(t) + 5\sin(5t)$

Differentiate $y = 5\sin(t) - \sin(5t)$:

Using the chain rule:

$\frac{dy}{dt} = 5\cos(t) - 5\cos(5t)$

Step 3: Write the arc length formula

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

$L = \int_0^\pi \sqrt{\left(-5\sin(t) + 5\sin(5t)\right)^2 + \left(5\cos(t) - 5\cos(5t)\right)^2} \, dt$

Expand the terms inside the square root:

1. Expand $\left(-5\sin(t) + 5\sin(5t)\right)^2$:

   $\left(-5\sin(t) + 5\sin(5t)\right)^2 = 25\sin^2(t) - 50\sin(t)\sin(5t) + 25\sin^2(5t)$

2. Expand $\left(5\cos(t) - 5\cos(5t)\right)^2$:

   $\left(5\cos(t) - 5\cos(5t)\right)^2 = 25\cos^2(t) - 50\cos(t)\cos(5t) + 25\cos^2(5t)$

3. Combine the terms:

   $\left(-5\sin(t) + 5\sin(5t)\right)^2 + \left(5\cos(t) - 5\cos(5t)\right)^2 = 25\left(\sin^2(t) + \cos^2(t)\right) + 25\left(\sin^2(5t) + \cos^2(5t)\right) - 50\left(\sin(t)\sin(5t) + \cos(t)\cos(5t)\right)$

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

$\left(-5\sin(t) + 5\sin(5t)\right)^2 + \left(5\cos(t) - 5\cos(5t)\right)^2 = 25(1) + 25(1) - 50\left(\cos(t - 5t)\right)$

Simplify:

$\left(-5\sin(t) + 5\sin(5t)\right)^2 + \left(5\cos(t) - 5\cos(5t)\right)^2 = 50 - 50\cos(4t)$

Thus, the arc length becomes:

$L = \int_0^\pi \sqrt{50 - 50\cos(4t)} \, dt$

Step 4: Simplify the square root

Factor out $50$:

$\sqrt{50 - 50\cos(4t)} = \sqrt{50}\sqrt{1 - \cos(4t)} = 5\sqrt{2}\sqrt{1 - \cos(4t)}.$

The arc length becomes:

$L = 5\sqrt{2} \int_0^\pi \sqrt{1 - \cos(4t)} \, dt.$

Step 5: Simplify $\sqrt{1 - \cos(4t)}$

Using the trigonometric identity $1 - \cos(4t) = 2\sin^2(2t)$:

$\sqrt{1 - \cos(4t)} = \sqrt{2\sin^2(2t)} = \sqrt{2}|\sin(2t)|.$

Since $t \in [0, \pi]$, $\sin(2t) \geq 0$, so:

$\sqrt{1 - \cos(4t)} = \sqrt{2}\sin(2t).$

The arc length becomes:

$L = 5\sqrt{2} \int_0^\pi \sqrt{2}\sin(2t) \, dt.$

Simplify:

$L = 10 \int_0^\pi \sin(2t) \, dt.$

Step 6: Evaluate the integral

The integral of $\sin(2t)$ is:

$\int \sin(2t) \, dt = -\frac{1}{2}\cos(2t).$

Evaluate from $t = 0$ to $t = \pi$:

1. At $t = \pi$:

   $-\frac{1}{2}\cos(2\pi) = -\frac{1}{2}(1) = -\frac{1}{2}.$

2. At $t = 0$:

   $-\frac{1}{2}\cos(0) = -\frac{1}{2}(1) = -\frac{1}{2}.$

Subtract:

$\int_0^\pi \sin(2t) \, dt = -\frac{1}{2} - \left(-\frac{1}{2}\right) = 0.$

Final Answer:

The exact length of the curve is:

$L = \boxed{0}$

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