Let's solve the problem step by step.

(a) Find a Parametric Equation of the Curve $C$

The curve $C$ is the intersection of the plane $y = 2x - 1$ and the paraboloid $z = x^2 + y^2 - 4$.

1. Substitute the plane equation into the paraboloid equation:

   The plane equation is $y = 2x - 1$. Substituting this into the equation of the paraboloid:

   $z = x^2 + (2x - 1)^2 - 4$

2. Simplify the equation:

   Expand the square:

   $z = x^2 + (4x^2 - 4x + 1) - 4$

   Simplify further:

   $z = 5x^2 - 4x - 3$

3. Express the curve parametrically:

   The curve $C$ can be parametrized by $x$. Set $x = t$, then:

   $y = 2t - 1$ $z = 5t^2 - 4t - 3$

   Therefore, the parametric equation of the curve $C$ is:

   $\mathbf{r}(t) = \langle t, 2t - 1, 5t^2 - 4t - 3 \rangle$

(b) Find the Tangent Vector to the Curve at the Point $(2, 3, 9)$

To find the tangent vector, we need to differentiate the parametric equation $\mathbf{r}(t)$ with respect to $t$ and evaluate it at the specific value of $t$ that corresponds to the point $(2, 3, 9)$.

1. Differentiate the parametric equation:

   The parametric equation is $\mathbf{r}(t) = \langle t, 2t - 1, 5t^2 - 4t - 3 \rangle$.

   Differentiate each component with respect to $t$:

   $\mathbf{r}'(t) = \langle 1, 2, 10t - 4 \rangle$

2. Find the value of $t$ corresponding to the point $(2, 3, 9)$:

   For $x = 2$, $y = 2(2) - 1 = 3$, and $z = 5(2)^2 - 4(2) - 3 = 9$.

   So, the corresponding $t$ value is $t = 2$.

3. Evaluate the derivative at $t = 2$:

   Substitute $t = 2$ into $\mathbf{r}'(t)$:

   $\mathbf{r}'(2) = \langle 1, 2, 10(2) - 4 \rangle = \langle 1, 2, 20 - 4 \rangle = \langle 1, 2, 16 \rangle$

   Thus, the tangent vector at the point $(2, 3, 9)$ is:

   $\mathbf{r}'(2) = \langle 1, 2, 16 \rangle$

Summary

• The parametric equation of the curve $C$ is $\mathbf{r}(t) = \langle t, 2t - 1, 5t^2 - 4t - 3 \rangle$.
• The tangent vector to the curve at the point $(2, 3, 9)$ is $\mathbf{r}'(2) = \langle 1, 2, 16 \rangle$.

Would you like more details or have any questions?

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Here are 5 related questions to consider:

1. How do you find the normal vector to a curve at a given point?
2. Can the parametric equation be represented using other variables instead of $t$?
3. How would the curve change if the plane equation was $y = 3x + 2$?
4. What is the geometric interpretation of the tangent vector at a point on a curve?
5. How can you find the second derivative of a parametric curve?

Tip: When dealing with parametric curves, always check the consistency of the parametrization by substituting it back into the original equations.