Question

Find a vector function that represents the curve of intersection of the two surfaces

28. The cylinder \({x^2} + {y^2} = 4\) and the surface \(z = xy\)

Short answer

\(r(t) = \left\langle {2\sin t,2\cos t,2\sin 2t} \right\rangle \)

Step-by-step solution

Rationalization

We know that one way of the parameterization of a circle of radius r in \({\mathbb{R}^2}\)centered at the origin O(0,0) on the xy plane is

\(x = r\sin t\;\;\;\;\;y = r\cos t\)

We can use these values to substitute to the plane \(z = xy\) since the circle and the plane are intersecting.

Simplification

Using the aforementioned values, we can find the parameterization for the intersection of circular cylinder\({x^2} + {y^2} = 4\) in \({\mathbb{R}^3}\)and the plane\(z = xy\) by substituting \(x = r\sin t\;\& \;y = r\,\cos t\) with \(r = \sqrt 4 = 2\) to get:

\(\begin{aligned}{c}z &= xy\\ &= (2\sin t)(2\cos t)\\ &= 4\sin t\cos t\\ &= 2\sin 2t\end{aligned}\)

Thus we can write the vector function as:

\(r(t) = \left\langle {2\sin t,2\cos t,2\sin 2t} \right\rangle \)

There are many possible answers to this.