We are given the parametric equations:

$x = 6 \cos t, \quad y = 6 \sin t, \quad z = 8 \cos(2t)$

Step 1: Find the Parameter $t$ at the Given Point

The given point is $(3, 3, 4)$. We solve for $t$:

1. Solve for $t$ from $x$: $3 = 6 \cos t$ $\cos t = \frac{3}{6} = \frac{1}{2}$ This implies $t = \frac{\pi}{3}$ or $t = -\frac{\pi}{3}$ (choosing the principal value).

2. Verify with $y$: $3 = 6 \sin t$ $\sin t = \frac{3}{6} = \frac{1}{2}$ Since $\sin(\pi/3) = 1/2$, we confirm $t = \pi/3$.

3. Verify with $z$: $4 = 8 \cos(2t)$ $\cos(2t) = \frac{4}{8} = \frac{1}{2}$ Since $\cos(\pi/3) = 1/2$, we confirm that $2t = \pi/3$, so $t = \pi/6$, not $\pi/3$.

Thus, the correct $t$ value for the given point is $t = \pi/6$.

Step 2: Compute the Tangent Vector

The tangent vector is given by the derivative of the position function:

$\frac{dx}{dt} = -6 \sin t, \quad \frac{dy}{dt} = 6 \cos t, \quad \frac{dz}{dt} = -16 \sin(2t)$

At $t = \pi/6$:

1. Compute $\frac{dx}{dt}$: $\frac{dx}{dt} = -6 \sin \left(\frac{\pi}{6}\right) = -6 \times \frac{1}{2} = -3$

2. Compute $\frac{dy}{dt}$: $\frac{dy}{dt} = 6 \cos \left(\frac{\pi}{6}\right) = 6 \times \frac{\sqrt{3}}{2} = 3\sqrt{3}$

3. Compute $\frac{dz}{dt}$: $\frac{dz}{dt} = -16 \sin \left(\frac{\pi}{3}\right) = -16 \times \frac{\sqrt{3}}{2} = -8\sqrt{3}$

Thus, the tangent vector at $t = \pi/6$ is:

$(-3, 3\sqrt{3}, -8\sqrt{3})$

Step 3: Write the Parametric Equations for the Tangent Line

Using the point $(3,3,4)$ and direction vector $(-3, 3\sqrt{3}, -8\sqrt{3})$, the parametric equations of the tangent line are:

$x = 3 - 3t$ $y = 3 + 3\sqrt{3} t$ $z = 4 - 8\sqrt{3} t$

Would you like a step-by-step explanation for each calculation?

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Related Questions:

1. How do you find the normal vector to a curve at a given point?
2. What is the geometric interpretation of a tangent vector in three dimensions?
3. How do you determine if a point lies on a given parametric curve?
4. What are the differences between tangent and normal lines for parametric curves?
5. How do you find the osculating plane for a space curve at a given point?

Math Tip:

When verifying the parameter value $t$, always check all coordinates to avoid miscalculations!