Problem Breakdown:

We are tasked with finding where the normal line to the paraboloid $z = x^2 + y^2$ at the point $(3, 3, 18)$ intersects the paraboloid a second time. If no second intersection exists, the answer should be $\text{DNE}$.

Step-by-Step Solution:

1. Find the Gradient (Normal Vector):

The gradient of $z = x^2 + y^2$ is: $\nabla f = \left( \frac{\partial z}{\partial x}, \frac{\partial z}{\partial y}, \frac{\partial z}{\partial z} \right) = (2x, 2y, -1).$ At the point $(3, 3, 18)$, the gradient becomes: $\nabla f = (2(3), 2(3), -1) = (6, 6, -1).$ This is the direction vector of the normal line.

2. Parametrize the Normal Line:

Using the point $(3, 3, 18)$ and the direction vector $(6, 6, -1)$, the parametric equations for the line are: $x = 3 + 6t, \quad y = 3 + 6t, \quad z = 18 - t.$

3. Find the Intersection with the Paraboloid:

The normal line intersects the paraboloid $z = x^2 + y^2$. Substituting the parametric equations into the paraboloid: $z = x^2 + y^2.$ Substitute $x = 3 + 6t$, $y = 3 + 6t$, and $z = 18 - t$: $18 - t = (3 + 6t)^2 + (3 + 6t)^2.$ Simplify: $18 - t = 2(3 + 6t)^2.$ Expand $(3 + 6t)^2$: $(3 + 6t)^2 = 9 + 36t + 36t^2 = 9 + 72t + 36t^2.$ Substitute back: $18 - t = 2(9 + 72t + 36t^2).$ Simplify: $18 - t = 18 + 144t + 72t^2.$ Rearrange: $72t^2 + 145t = 0.$ Factor: $t(72t + 145) = 0.$

4. Solve for $t$:

From $t(72t + 145) = 0$, we get: $t = 0 \quad \text{or} \quad t = -\frac{145}{72}.$

• $t = 0$ corresponds to the original point $(3, 3, 18)$.
• $t = -\frac{145}{72}$ gives the second intersection.

5. Find the Coordinates of the Second Intersection:

Substitute $t = -\frac{145}{72}$ into the parametric equations: $x = 3 + 6\left(-\frac{145}{72}\right) = 3 - \frac{870}{72} = 3 - \frac{145}{12} = \frac{-109}{12},$ $y = 3 + 6\left(-\frac{145}{72}\right) = \frac{-109}{12},$ $z = 18 - \left(-\frac{145}{72}\right) = 18 + \frac{145}{72} = \frac{1296}{72} + \frac{145}{72} = \frac{1441}{72}.$

Final Answer:

The second intersection point is: $\boxed{\left( \frac{-109}{12}, \frac{-109}{12}, \frac{1441}{72} \right)}.$

Let me know if you have questions or want any clarification!

Follow-Up Questions:

1. What is the geometric meaning of a normal line to a surface?
2. How do you calculate the gradient vector for other surfaces?
3. What are other ways to find intersections of lines and surfaces?
4. What happens if a line is tangent to a paraboloid?
5. How can this problem be generalized to other quadratic surfaces?

Tip:

The gradient vector is always perpendicular to the surface. For intersections, substitute the parametric equations into the surface equation!