Find parametric equations for the tangent line at 
t = 4
 for 
x = (t − 1)2,
 
y = 3,
 
z = 2t3 − 3t2
.
(Enter your answers as a comma-separated list of equations.)


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.

Find parametric equations for the tangent line at t = 4 for x = (t − 1)^2, y = 3, z = 2t^3 − 3t^2.

Discover how to find parametric equations for the tangent line at t = 4 for the curve defined by x = (t - 1)^2, y = 3, and z = 2t^3 - 3t^2. The solution includes differentiation steps and evaluates derivatives to find the tangent line direction and point on the curve. This solution is ideal for students in upper-level high school calculus or Calculus I.

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