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?

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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.