Question

Sketch the space curve represented by the intersection of the surfaces. Then represent the curve by a vector-valued function using the given parameter. $$\text { Surfaces } \quad \text { Parameter }$$ $$ z=x^{2}+y^{2}, \quad z=4 \quad x=2 \cos t $$

Short answer

The space curve, represented by a vector-valued function, which is an intersection of the given surfaces, is given by \(r(t) = <2cos(t), 2sin(t), 4>\)

Step-by-step solution

Understanding the surfaces

Start by acknowledging the two given surfaces: \(z=x^{2}+y^{2}\) and \(z=4\). The intersection between these two surfaces will help create a 3D curve, basically a circle in this case.

Finding the intersection

Replace \(z\) from the equation \(z=4\) into the equation \(z=x^{2}+y^{2}\), so you get \(4 = x^{2} + y^{2}\). This equation represents a circle in the xy-plane with radius \(r=2\). Because z-value is 4 for every point on the curve, this represents a circle of radius 2 in the plane z = 4.

Implementing the parameter

From the equation \(x = 2\cos t\), we can express \(y\) in terms of \(t\). Since \(4 = x^{2} + y^{2}\), it can be simplified to \(y = \sqrt{4 - x^{2}}\). Replacing \(x = 2\cos t\) in y's equation yields \(y = \sqrt{4 - (2\cos t)^{2}} = 2\sin t\).

Formulating a vector-valued function

The vector-valued function \(r(t)\) representing the space curve can be obtained by combining the parameterized x, y and z values: \(r(t) = = <2cos(t), 2sin(t), 4>\)