Thread: Intersection of functions --> vector function

I'm a bit confused as to how to approach this problem, the only piece of information I can decipher is the statement at the end of this post, if anyone can shed some light, I'd greatly appreciate it. Here's the question...

A circular cylinder of radius 1 and the parabolic cylinder z = 2x^2 - 1 intersect as shown in the attached image.

Which vector function has the curve of intersection as its graph if x(0) = 1.

The answer choices are...

(cost)i + (sint)j + (cos2t - 1)k
(cost)i + (sint)j + (cos2t)k
(cost)i + (sint)j + (1 - cos2t)k
(sint)i + (cost)j + (cos2t - 1)k
(sint)i + (cost)j + (cos2t)k
(sint)i + (cost)j + (1 - cos2t)k


I'm under the impression that since x(0) has to equal 1, then i must be cost, since cos(0) = 1. If that is correct that eliminates three of the six choices.

Originally Posted by ocryan

I'm a bit confused as to how to approach this problem, the only piece of information I can decipher is the statement at the end of this post, if anyone can shed some light, I'd greatly appreciate it. Here's the question...

A circular cylinder of radius 1 and the parabolic cylinder z = 2x^2 - 1 intersect as shown in the attached image.

Which vector function has the curve of intersection as its graph if x(0) = 1.

The answer choices are...

(cost)i + (sint)j + (cos2t - 1)k
(cost)i + (sint)j + (cos2t)k
(cost)i + (sint)j + (1 - cos2t)k
(sint)i + (cost)j + (cos2t - 1)k
(sint)i + (cost)j + (cos2t)k
(sint)i + (cost)j + (1 - cos2t)k


I'm under the impression that since x(0) has to equal 1, then i must be cost, since cos(0) = 1. If that is correct that eliminates three of the six choices.

Yes and which of the remaining three choices satisfies ?