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.