To find the parametric equations of the tangent line to the curve defined by the given parametric equations at the specified point $(2\sqrt{3}, 2, 2)$, let’s proceed step by step:

Step 1: Understand the curve equations

The curve is given parametrically as: $x = 4\cos(t), \quad y = 4\sin(t), \quad z = 4\cos(2t).$

The goal is to determine the tangent line at the point $(2\sqrt{3}, 2, 2)$.

Step 2: Find the parameter $t$ corresponding to the given point

1. From $x = 4\cos(t)$, we know: $4\cos(t) = 2\sqrt{3} \implies \cos(t) = \frac{\sqrt{3}}{2}.$ This occurs when $t = \frac{\pi}{6}$ (or any equivalent angle).

2. Check $y$ to confirm this value of $t$: $y = 4\sin(t) \implies 4\sin\left(\frac{\pi}{6}\right) = 4 \cdot \frac{1}{2} = 2.$ This is correct.

3. Verify $z$ at $t = \frac{\pi}{6}$: $z = 4\cos(2t) \implies z = 4\cos\left(2 \cdot \frac{\pi}{6}\right) = 4\cos\left(\frac{\pi}{3}\right).$ Since $\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$, we get: $z = 4 \cdot \frac{1}{2} = 2.$ Therefore, $t = \frac{\pi}{6}$ corresponds to the given point $(2\sqrt{3}, 2, 2)$.

Step 3: Compute the velocity vector (tangent vector)

To find the direction of the tangent line, differentiate the parametric equations with respect to $t$:

1. For $x = 4\cos(t)$: $\frac{dx}{dt} = -4\sin(t).$

2. For $y = 4\sin(t)$: $\frac{dy}{dt} = 4\cos(t).$

3. For $z = 4\cos(2t)$: $\frac{dz}{dt} = -8\sin(2t) \quad \text{(using the chain rule)}.$

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

• $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$ and $\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$,
• $\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$.

Thus:

1. $\frac{dx}{dt} = -4\sin\left(\frac{\pi}{6}\right) = -4 \cdot \frac{1}{2} = -2$,
2. $\frac{dy}{dt} = 4\cos\left(\frac{\pi}{6}\right) = 4 \cdot \frac{\sqrt{3}}{2} = 2\sqrt{3}$,
3. $\frac{dz}{dt} = -8\sin\left(\frac{\pi}{3}\right) = -8 \cdot \frac{\sqrt{3}}{2} = -4\sqrt{3}$.

Therefore, the tangent vector is: $\mathbf{v} = \left(-2, 2\sqrt{3}, -4\sqrt{3}\right).$

Step 4: Write the parametric equations for the tangent line

The tangent line passes through the point $(2\sqrt{3}, 2, 2)$ and has direction vector $(-2, 2\sqrt{3}, -4\sqrt{3})$. The parametric equations are:

$x(t) = 2\sqrt{3} - 2t, \quad y(t) = 2 + 2\sqrt{3}t, \quad z(t) = 2 - 4\sqrt{3}t.$

Final Answer:

The parametric equations for the tangent line are: $\boxed{(x(t), y(t), z(t)) = (2\sqrt{3} - 2t, \, 2 + 2\sqrt{3}t, \, 2 - 4\sqrt{3}t).}$

Would you like me to clarify any steps or provide additional details? Here are 5 related questions you might find helpful:

1. How do you find the tangent line to a curve given parametric equations?
2. What are some strategies for verifying a parameter $t$ for a given point?
3. How is the velocity vector of a parametric curve calculated?
4. What role does the chain rule play in differentiating parametric equations?
5. How can you write the equation of a line passing through a point with a given direction vector?

Tip: Always confirm the point corresponds to the parameter $t$ before proceeding with tangent calculations.