A curve is given by the following parametric equations. X(T)=4t-4sin(t), y(T)=4-4cos(t). Find the (x,y) point, dx/dt, dy/dt, and the speed at T=pi/6

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.

A curve is given by the following parametric equations. X(T)=4t-4sin(t), y(T)=4-4cos(t). Find the (x,y) point, dx/dt, dy/dt, and the speed at T=pi/6.

Explore how to find the coordinates (x, y), derivatives dx/dt and dy/dt, and the speed of a parametric curve defined by x(t) = 4t - 4sin(t) and y(t) = 4 - 4cos(t) at t = pi/6. Detailed step-by-step solution with trigonometry and calculus for high school students.

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