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) $\mathbf { r } ( t ) = 2 \cos ( t ) \mathbf { i } + 2 \sin ( t ) \mathbf { j } + 4 \sin ( t ) \cos ( t ) \mathbf { k }$

Question

Quizplus · Accepted Answer

The vector function must satisfy both the circular cylinder equation $x^2 + y^2 = 4$ and the parabolic cylinder equation $z = xy$. Choice E uses $x = 2\cos(t)$ and $y = 2\sin(t)$, which satisfy the circular cylinder equation. Substituting these into the parabolic cylinder equation gives $z = (2\cos(t))(2\sin(t)) = 4\sin(t)\cos(t)$, which matches the $z$-component of choice E.

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

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