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 for the tangent line at \( t = 4 \) for the given vector function, we need to:

1. Calculate the point on the curve at \( t = 4 \).
2. Find the derivative of each component (to determine the tangent vector) and evaluate it at \( t = 4 \).
3. Use the point and tangent vector to write the parametric equations for the tangent line.

The given functions are:
\[
x(t) = (t - 1)^2, \quad y(t) = 3, \quad z(t) = 2t^3 - 3t^2
\]

### Step 1: Compute the Point on the Curve at \( t = 4 \)
Evaluate each component at \( t = 4 \):

- \( x(4) = (4 - 1)^2 = 3^2 = 9 \)
- \( y(4) = 3 \) (since \( y \) is constant)
- \( z(4) = 2(4)^3 - 3(4)^2 = 2 \cdot 64 - 3 \cdot 16 = 128 - 48 = 80 \)

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

### Step 2: Compute the Tangent Vector by Differentiating Each Component

Now we differentiate each component with respect to \( t \) to get the tangent vector.

1. **For \( x(t) \):**
   \[
   \frac{dx}{dt} = \frac{d}{dt} (t - 1)^2 = 2(t - 1)
   \]
   At \( t = 4 \), \( \frac{dx}{dt} = 2(4 - 1) = 2 \cdot 3 = 6 \).

2. **For \( y(t) \):**
   Since \( y(t) = 3 \), which is constant, \( \frac{dy}{dt} = 0 \).

3. **For \( z(t) \):**
   \[
   \frac{dz}{dt} = \frac{d}{dt} (2t^3 - 3t^2) = 6t^2 - 6t
   \]
   At \( t = 4 \), \( \frac{dz}{dt} = 6(4)^2 - 6(4) = 6 \cdot 16 - 24 = 96 - 24 = 72 \).

So, the tangent vector at \( t = 4 \) is \( (6, 0, 72) \).

### Step 3: Write the Parametric Equations for the Tangent Line
Using the point \( (9, 3, 80) \) and the tangent vector \( (6, 0, 72) \), we can write the parametric equations for the tangent line as:
\[
x = 9 + 6t, \quad y = 3, \quad z = 80 + 72t
\]

Thus, the parametric equations for the tangent line are:
\[
x = 9 + 6t, \quad y = 3, \quad z = 80 + 72t
\]

Would you like more details or have any questions?

---

1. How do you find tangent lines for curves given in vector form?
2. How do you compute derivatives for functions in parametric form?
3. Why is the \( y \)-component constant in this problem?
4. How would the tangent vector change if \( y(t) \) were variable?
5. What happens to the tangent line if the derivative at a certain point is zero?

**Tip:** When dealing with parametric equations, always differentiate each component separately to find the tangent vector accurately.

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

Learn how to find parametric equations for the tangent line at t = 4 for the functions x = (t − 1)^2, y = 3, z = 2t^3 − 3t^2. This problem involves key calculus concepts, including derivatives and tangent lines. Suitable for students in Grades 11-12.

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