Question

Find a vector-valued function whose graph is the indicated surface. The part of the plane \(z=4\) that lies inside the cylinder \(x^{2}+y^{2}=9\)

Short answer

The vector-valued function representing the part of the plane \(z=4\) that lies inside the cylinder \(x^{2}+y^{2}=9\) is \(r(4, v) = <4\cos(v), 4\sin(v), 4>\)

Step-by-step solution

Identifying the geometry

First, identify the geometric interpretation of the problem. The equation \(z=4\) is the plane that is situated 4 units above the x-y plane, as the z-coordinate is constant. The equation \(x^{2}+y^{2}=9\) is the cylinder with its axis along the z-axis and radius of 3 units.

Parameterizing the area

Next, to represent the area inside the cylinder and on the plane, parameterize it with two variables \(u\) and \(v\). For this scenario, polar coordinates (\(r, \theta\)) are more suitable but here \(r\), the radial distance, is a constant (3) and \(\theta=v\) is a free variable that can range from 0 to \(2\pi\). Also, \(u=4\) is a constant, which corresponds to z.

Finding the vector-valued function

The vector valued function that gives the graph of this surface can be written in terms of parameters \(u\) and \(v\) as: \(r(u, v) = \). But in this case, since \(u = 4\), the function simplifies to \(r(4, v) = <4\cos(v), 4\sin(v), 4>\).