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Find a Vector Function That Represents the Curve of Intersection x2+y2=4x ^ { 2 } + y ^ { 2 } = 4

Question 3

Multiple Choice

Find a vector function that represents the curve of intersection of the two surfaces:
The circular cylinder x2+y2=4x ^ { 2 } + y ^ { 2 } = 4 and the parabolic cylinder z=xyz = x y .


A) r(t) =4costi+4tj+4cos2tk\mathbf { r } ( t ) = 4 \cos t \mathbf { i } + 4 t \mathbf { j } + 4 \cos ^ { 2 } t \mathbf { k }
B) r(t) =2cos(t) i+2sin(t) jsin(t) cos(t) k\mathbf { r } ( t ) = 2 \cos ( t ) \mathbf { i } + 2 \sin ( t ) \mathbf { j } - \sin ( t ) \cos ( t ) \mathbf { k }
C) r(t) =costi+sintj4cos2tk\mathbf { r } ( t ) = \cos t \mathbf { i } + \sin t \mathbf { j } - 4 \cos 2 t \mathbf { k }
D) r(t) =cos(t) i+sin(t) j+4sin(t) cos(t) k\mathbf { r } ( t ) = \cos ( t ) \mathbf { i } + \sin ( t ) \mathbf { j } + 4 \sin ( t ) \cos ( t ) \mathbf { k }
E) r(t) =2cos(t) i+2sin(t) j+4sin(t) cos(t) k\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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