Find a Vector Function That Represents the Curve of Intersection $x ^ { 2 } + y ^ { 2 } = 4$

Find a vector function that represents the curve of intersection of the two surfaces:
The circular cylinder $x ^ { 2 } + y ^ { 2 } = 4$ and the parabolic cylinder $z = x y$ .

A) $\mathbf { r } ( t ) = 4 \cos t \mathbf { i } + 4 t \mathbf { j } + 4 \cos ^ { 2 } t \mathbf { k }$
B) $\mathbf { r } ( t ) = 2 \cos ( t ) \mathbf { i } + 2 \sin ( t ) \mathbf { j } - \sin ( t ) \cos ( t ) \mathbf { k }$
C) $\mathbf { r } ( t ) = \cos t \mathbf { i } + \sin t \mathbf { j } - 4 \cos 2 t \mathbf { k }$
D) $\mathbf { r } ( t ) = \cos ( t ) \mathbf { i } + \sin ( t ) \mathbf { j } + 4 \sin ( t ) \cos ( t ) \mathbf { k }$
E) $\mathbf { r } ( t ) = 2 \cos ( t ) \mathbf { i } + 2 \sin ( t ) \mathbf { j } + 4 \sin ( t ) \cos ( t ) \mathbf { k }$

Correct Answer:

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