To find the parametric equations of the tangent line to the curve at $t = 4$, we follow these steps:

Step 1: Calculate the Derivative of Each Component

Given: $x = (t - 1)^2, \quad y = 3, \quad z = 2t^3 - 3t^2$

1. Differentiate $x$: $\frac{dx}{dt} = 2(t - 1)$

2. Differentiate $y$: Since $y = 3$, which is constant, its derivative is: $\frac{dy}{dt} = 0$

3. Differentiate $z$: $\frac{dz}{dt} = 6t^2 - 6t$

Step 2: Evaluate Each Derivative at $t = 4$

1. For $x$: $\frac{dx}{dt} \Big|_{t=4} = 2(4 - 1) = 2 \cdot 3 = 6$

2. For $y$: $\frac{dy}{dt} \Big|_{t=4} = 0$

3. For $z$: $\frac{dz}{dt} \Big|_{t=4} = 6(4)^2 - 6(4) = 6 \cdot 16 - 24 = 96 - 24 = 72$

The tangent vector at $t = 4$ is: $\left( \frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt} \right) = (6, 0, 72)$

Step 3: Find the Point on the Curve at $t = 4$

1. For $x$: $x(4) = (4 - 1)^2 = 3^2 = 9$

2. For $y$: $y(4) = 3$

3. For $z$: $z(4) = 2(4)^3 - 3(4)^2 = 2 \cdot 64 - 3 \cdot 16 = 128 - 48 = 80$

So, the point on the curve at $t = 4$ is $(9, 3, 80)$.

Step 4: Write the Parametric Equations for the Tangent Line

Using the point $(9, 3, 80)$ and the direction vector $(6, 0, 72)$, the parametric equations for the tangent line are: $x = 9 + 6t, \quad y = 3, \quad z = 80 + 72t$

Final Answer

$x = 9 + 6t, \quad y = 3, \quad z = 80 + 72t$

Would you like further explanation on any part of this solution?

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Here are some related questions:

1. How do you find parametric equations for tangent lines to a curve in general?
2. What are the conditions for a curve to have a tangent line at a given point?
3. How does the parametric form of a line differ from its Cartesian form?
4. How would these calculations change if one of the functions was a trigonometric function?
5. Why is the derivative of a constant function (like $y = 3$) zero?

Tip: When finding a tangent line, the derivative vector at a point gives the direction of the tangent line at that specific point.