To find the parametric equations for the tangent line to the given curve at the specified point, we need to follow these steps:

1. Find the derivatives of $x(t)$, $y(t)$, and $z(t)$ with respect to $t$.
2. Evaluate these derivatives at the given $t$-value corresponding to the point $(0, 6, 1)$.
3. Use the point and the derivatives to write the parametric equations of the tangent line.

Step 1: Derivatives of $x(t)$, $y(t)$, and $z(t)$

Given:

• $x(t) = 3 \ln(t)$
• $y(t) = 6t$
• $z(t) = t^5$

Find the derivatives:

1. $x'(t) = \frac{d}{dt} [3 \ln(t)] = \frac{3}{t}$.
2. $y'(t) = \frac{d}{dt} [6t] = 6$.
3. $z'(t) = \frac{d}{dt} [t^5] = 5t^4$.

Step 2: Determine $t$-value for the specified point $(0, 6, 1)$

• For $x(t) = 0$:
  $0 = 3 \ln(t)$
  $\ln(t) = 0$
  $t = e^0 = 1$.

• For $y(t) = 6$:
  $6 = 6t$
  $t = 1$.

• For $z(t) = 1$:
  $1 = t^5$
  $t = 1$.

The parameter $t = 1$ corresponds to the given point $(0, 6, 1)$.

Step 3: Evaluate derivatives at $t = 1$

1. $x'(1) = \frac{3}{1} = 3$.
2. $y'(1) = 6$.
3. $z'(1) = 5 \times 1^4 = 5$.

Step 4: Parametric Equations of the Tangent Line

The parametric equations of the tangent line can be expressed as:

$x = x_0 + x'(1) \cdot t = 0 + 3t = 3t$

$y = y_0 + y'(1) \cdot t = 6 + 6t = 6 + 6t$

$z = z_0 + z'(1) \cdot t = 1 + 5t = 1 + 5t$

So, the parametric equations for the tangent line at the given point are:

$\boxed{x = 3t}$

$\boxed{y = 6 + 6t}$

$\boxed{z = 1 + 5t}$

Do you want further details or have any questions?

Related Questions:

1. How do you find the normal line to a parametric curve at a given point?
2. What is the geometric interpretation of parametric equations for tangent lines?
3. How do you find the curvature of a parametric curve at a given point?
4. How do you determine if a parametric curve is smooth at a point?
5. How do you find the arc length of a parametric curve?

Tip: For finding tangent lines, always ensure the point corresponds to a specific parameter value.