consider the curve C formed from the intersection of the plane y = 2x − 1 and the
paraboloid z = x^2 + y^2 − 4.
(a) Find a parametric equation of the curve C 
(b) Find the tangent vector to this curve at the point (2,3, 9).

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.

Explore the parametric equation and tangent vector of curve C formed by the intersection of the plane y = 2x - 1 and the paraboloid z = x^2 + y^2 - 4. Includes step-by-step solutions and calculations for finding the parametric equation and tangent vector at point (2, 3, 9).

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