We are given the parametric equations:

$x(t) = 4t - 4\sin(t)$ $y(t) = 4 - 4\cos(t)$

Step 1: Find the $(x, y)$ coordinates at $t = \frac{\pi}{6}$

Substitute $t = \frac{\pi}{6}$ into $x(t)$ and $y(t)$:

$x\left(\frac{\pi}{6}\right) = 4\left(\frac{\pi}{6}\right) - 4\sin\left(\frac{\pi}{6}\right)$ Since $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$:

$x\left(\frac{\pi}{6}\right) = \frac{2\pi}{3} - 4\left(\frac{1}{2}\right) = \frac{2\pi}{3} - 2$

$y\left(\frac{\pi}{6}\right) = 4 - 4\cos\left(\frac{\pi}{6}\right)$ Since $\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$:

$y\left(\frac{\pi}{6}\right) = 4 - 4\left(\frac{\sqrt{3}}{2}\right) = 4 - 2\sqrt{3}$

So, the coordinates at $t = \frac{\pi}{6}$ are:

$\left(\frac{2\pi}{3} - 2, 4 - 2\sqrt{3}\right)$

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

Differentiate $x(t)$ and $y(t)$ with respect to $t$:

For $x(t) = 4t - 4\sin(t)$:

$\frac{dx}{dt} = 4 - 4\cos(t)$

For $y(t) = 4 - 4\cos(t)$:

$\frac{dy}{dt} = 4\sin(t)$

Step 3: Evaluate $\frac{dx}{dt}$ and $\frac{dy}{dt}$ at $t = \frac{\pi}{6}$

Substitute $t = \frac{\pi}{6}$:

$\frac{dx}{dt}\Bigg|_{t = \frac{\pi}{6}} = 4 - 4\cos\left(\frac{\pi}{6}\right) = 4 - 4\left(\frac{\sqrt{3}}{2}\right) = 4 - 2\sqrt{3}$

$\frac{dy}{dt}\Bigg|_{t = \frac{\pi}{6}} = 4\sin\left(\frac{\pi}{6}\right) = 4\left(\frac{1}{2}\right) = 2$

Step 4: Find the speed at $t = \frac{\pi}{6}$

The speed is given by:

$\text{Speed} = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}$

Substitute the values:

$\text{Speed} = \sqrt{(4 - 2\sqrt{3})^2 + 2^2}$

First, simplify $(4 - 2\sqrt{3})^2$:

$(4 - 2\sqrt{3})^2 = 16 - 16\sqrt{3} + 12 = 28 - 16\sqrt{3}$

So, the speed becomes:

$\text{Speed} = \sqrt{(28 - 16\sqrt{3}) + 4} = \sqrt{32 - 16\sqrt{3}}$

Summary of results at $t = \frac{\pi}{6}$:

• The point $(x, y)$ is $\left(\frac{2\pi}{3} - 2, 4 - 2\sqrt{3}\right)$
• $\frac{dx}{dt} = 4 - 2\sqrt{3}$
• $\frac{dy}{dt} = 2$
• Speed $= \sqrt{32 - 16\sqrt{3}}$

Would you like further details on any step?

Here are 5 related questions to further explore:

1. How do you interpret the physical meaning of the speed for parametric curves?
2. What happens to the speed at different points $t$?
3. How would you find the acceleration of the curve at $t = \frac{\pi}{6}$?
4. How does the curvature of the curve vary with $t$?
5. Can the speed ever be zero? If so, under what conditions?

Tip: Speed for a parametric curve is always the magnitude of the velocity vector, which combines both the $x$- and $y$-direction changes.