It looks like you've uploaded a problem involving the intersection of curves and surfaces. There are two separate problems here, so let's go through them one by one.

First Problem: Curve Intersection with a Paraboloid

The first question asks for the points where the curve r(t) = t i + (6t - t²) k intersects the paraboloid z = x² + y². The goal is to find the t-values where these curves intersect.

1. Parameterizing the curve: The vector function r(t) gives us the parametric equations for the curve:

   • x = t
   • y = 0 (since there is no j component)
   • z = 6t - t²

2. Substitute into the paraboloid equation: The paraboloid equation is: $z = x^2 + y^2$ Substituting x = t and y = 0 into the equation gives: $6t - t^2 = t^2$ Simplifying the equation: $6t = 2t^2$ Dividing both sides by 2t (assuming t ≠ 0): $3 = t$

Thus, the value of t = 3 is a potential solution.

Second Problem: Curve Intersection with a Cylinder and a Surface

The second problem involves finding the vector function r(t) representing the intersection of the cylinder x² + y² = 64 and the surface z = xy.

1. Equation of the cylinder: The cylinder is given by x² + y² = 64, which represents a circular cylinder with radius 8 (since √64 = 8).

2. Equation of the surface: The surface is given by z = xy.

To parametrize this curve, we can use polar coordinates:

• Let x = 8cos(t), y = 8sin(t), as this satisfies x² + y² = 64.
• Now, substitute these into the equation for z: $z = xy = (8cos(t))(8sin(t)) = 64sin(t)cos(t)$ Using the identity sin(2t) = 2sin(t)cos(t), we can simplify the expression for z: $z = 32sin(2t)$

Thus, the parametric vector function for the curve is: $r(t) = 8cos(t)i + 8sin(t)j + 32sin(2t)k$

Third Problem: Curve Intersection with a Cone and a Plane

The third problem involves finding the vector function r(t) for the curve of intersection between the cone z = √(x² + y²) and the plane z = 4 + y.

1. Substitute the equation of the plane into the cone equation: Since z = 4 + y, substitute this into the equation of the cone: $4 + y = √(x² + y²)$ Squaring both sides: $(4 + y)^2 = x² + y²$ Expanding the left side: $16 + 8y + y² = x² + y²$ Simplifying: $16 + 8y = x²$ Thus, the equation for x becomes: $x = √(16 + 8y)$

2. Parametrize the curve: We can parametrize y as y = t, then substitute this into the equation for x: $x = √(16 + 8t)$ Now, using the plane equation, the z-value is: $z = 4 + t$

Thus, the parametric vector function is: $r(t) = √(16 + 8t)i + tj + (4 + t)k$

Would you like any further details or explanations on any of these steps?

Additional Related Questions:

1. How would you graph these curves and surfaces to visualize their intersections?
2. What happens if we change the parameters of the curve or the surfaces involved?
3. How do you solve for t-values when given more complex surface equations?
4. Can you apply these methods to other types of curves, like ellipsoids or hyperboloids?
5. How would you compute the tangent vectors or velocity vectors for these curves?

Tip: When working with vector functions, always remember to first parameterize the surfaces or curves using convenient coordinate systems like polar or cylindrical coordinates for easier manipulation.