5. Where does the curve r(t) = (t, t2 − 1, −1) intersect the plane given by x + 2y − z/2 = 24?

We are tasked with finding the point of intersection between the curve \( \mathbf{r}(t) = (t, t^2 - 1, -1) \) and the plane \( x + 2y - \frac{z}{2} = 24 \).

### Step 1: Parametrize the curve
The vector form of the curve is:
\[
\mathbf{r}(t) = (x(t), y(t), z(t)) = (t, t^2 - 1, -1)
\]
From this, we extract the parametric equations:
\[
x(t) = t, \quad y(t) = t^2 - 1, \quad z(t) = -1
\]

### Step 2: Substitute into the plane equation
The equation of the plane is:
\[
x + 2y - \frac{z}{2} = 24
\]
Substitute the parametric expressions for \( x(t), y(t), z(t) \) into the plane equation:
\[
t + 2(t^2 - 1) - \frac{-1}{2} = 24
\]
Simplify:
\[
t + 2(t^2 - 1) + \frac{1}{2} = 24
\]
\[
t + 2t^2 - 2 + \frac{1}{2} = 24
\]
\[
t + 2t^2 - \frac{3}{2} = 24
\]
Multiply through by 2 to eliminate the fraction:
\[
2t + 4t^2 - 3 = 48
\]
Now solve for \( t \):
\[
4t^2 + 2t - 3 = 48
\]
\[
4t^2 + 2t - 51 = 0
\]

### Step 3: Solve the quadratic equation
The equation is:
\[
4t^2 + 2t - 51 = 0
\]
We can solve this using the quadratic formula:
\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 4 \), \( b = 2 \), and \( c = -51 \). Substituting into the quadratic formula:
\[
t = \frac{-2 \pm \sqrt{2^2 - 4(4)(-51)}}{2(4)}
\]
\[
t = \frac{-2 \pm \sqrt{4 + 816}}{8}
\]
\[
t = \frac{-2 \pm \sqrt{820}}{8}
\]
\[
t = \frac{-2 \pm 28.64}{8}
\]
Thus, the two solutions for \( t \) are:
\[
t_1 = \frac{-2 + 28.64}{8} = 3.33, \quad t_2 = \frac{-2 - 28.64}{8} = -3.81
\]

### Step 4: Find the intersection points
For \( t_1 = 3.33 \):
\[
x(3.33) = 3.33, \quad y(3.33) = (3.33)^2 - 1 = 10.09, \quad z(3.33) = -1
\]
Thus, one intersection point is approximately \( (3.33, 10.09, -1) \).

For \( t_2 = -3.81 \):
\[
x(-3.81) = -3.81, \quad y(-3.81) = (-3.81)^2 - 1 = 13.52, \quad z(-3.81) = -1
\]
Thus, the other intersection point is approximately \( (-3.81, 13.52, -1) \).

### Conclusion:
The curve intersects the plane at two points: approximately \( (3.33, 10.09, -1) \) and \( (-3.81, 13.52, -1) \).

Would you like further details on any step?

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Here are 5 related questions to explore:
1. How do you derive parametric equations from a given curve?
2. What is the geometric interpretation of solving a system of a curve and a plane?
3. How does the quadratic formula work for solving second-degree equations?
4. How can we verify if both solutions of \( t \) are valid in this context?
5. How does the method of substitution work in general for intersections?

**Tip**: Always verify your intersection points by plugging them back into both the curve's parametric equations and the plane equation to ensure they satisfy both.

Where does the curve r(t) = (t, t^2 − 1, −1) intersect the plane given by x + 2y − z/2 = 24?

Find the intersection points of the curve r(t) = (t, t^2 - 1, -1) and the plane x + 2y - z/2 = 24. The solution involves substituting the parametric equations into the plane equation and solving a quadratic equation. This problem covers key concepts in calculus and analytic geometry, ideal for undergraduate students.

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