# Thread: Finding a Vector Function

1. ## Finding a Vector Function

Hey guys,
I am having a bit of a problem with vector functions. This problem says: A circular cylinder of radius 1 and the parabolic cylinder z=2x^2 intersect. What is the vector function if the curve of intersection is x(0)=0?

I first set x = cos(t) and y = sin(t); then solved for z = 2 cos^2(t). Would this be correct? Or am I approaching this problem incorrectly?

Any feed and help is appreciated.

Thanks

2. ## Re: Finding a Vector Function

That problem doesn't make much sense to me as stated. Have you included the full statement of the problem?

Originally Posted by Beevo
A circular cylinder of radius 1 and the parabolic cylinder z=2x^2 intersect.
There are numerous "circular cylinder of radius 1" in 3-space. Is that freedom of positioning such a circlar cylinder what this problem is about?

Originally Posted by Beevo
What is the vector function if the curve of intersection is x(0)=0?
That reads that the "vector function" you're asked to find is a vector form for a curve.

$\text{It's asking you to find } F : \mathbb{R} \rightarrow \mathbb{R}^3.$

But the meaning of "the curve of intersection is x(0)=0" is very unclear. Is it the constant curve at the origin? Is it *some* curve satisfying x = 0 at time 0? That isn't even a point in 3 space. If it's *some* curve, then why the word "the". Saying "x(0)=0" is simply not describing "THE curve of intersection" for those two objects in 3 space.

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What is this circular cylinder of radius 1? It seems to me that both of the below satisfy "the curve of intersection is x(0)=0". I really just don't get it.

$x^2 + (z+1)^2 = 1 \text{ implies } F(t) = (0, t, 0) = t\hat{j} \in \mathbb{R}^3 \ \forall \ t \in \mathbb{R}.$

$y^2 + (z+1)^2 = 1 \text{ implies } F(t) = (0, 0, 0) = \vec{0} \in \mathbb{R}^3 \ \forall \ t \in \mathbb{R}.$

3. ## Re: Finding a Vector Function

Yes, that looks good to me. The parabolic cylinder and the circular cylinder intersect in a curve given by x= cos(t), y= sin(t), $z= 2cos^2(t)$

4. ## Re: Finding a Vector Function

Well the problem provided a graph. First, the graph of the cylinder has the z-axis as the central axis, so I am assuming the equation would be x^2 + y^2 = 1. Then the other equation is z=2x^2, which intersects with the cylinder. Then the main question states, which vector function has the curve of intersection as its graph if x(0)=0.

Also, does the x(0)=0 matter? What if it was x(0)=1.

5. ## Re: Finding a Vector Function

It is x(0) = 1 if x(t) = cos(t).
Although it's not how to usually do it, in this case, if that's basically the form you're seeking, make it x(t) = sin(t), y(t) = cos(t) in order to have x(0) = 0.
Though again, who knows what this problem is asking, so who knows if that's a reasonable answer. Surely not me.

Edit:
Originally Posted by Beevo
Well the problem provided a graph. First, the graph of the cylinder has the z-axis as the central axis, so I am assuming the equation would be x^2 + y^2 = 1.
Yes - that's right. That makes the problem much more clear.

$F(t) = (\sin(t), \cos(t), 2\sin^2(t)), \text{ or } \vec{F}(t) = \sin(t)\hat{i} + \cos(t)\hat{j} + 2\sin^2(t)\hat{k}, \text{ is what I bet they're looking for.}$

6. ## Re: Finding a Vector Function

Originally Posted by johnsomeone
It is x(0) = 1 if x(t) = cos(t).
Although it's not how to usually do it, in this case, if that's basically the form you're seeking, make it x(t) = sin(t), y(t) = cos(t) in order to have x(0) = 0.
Though again, who knows what this problem is asking, so who knows if that's a reasonable answer. Surely not me.

Edit:

Yes - that's right. That makes the problem much more clear.

$F(t) = (\sin(t), \cos(t), 2\sin^2(t)), \text{ or } \vec{F}(t) = \sin(t)\hat{i} + \cos(t)\hat{j} + 2\sin^2(t)\hat{k}, \text{ is what I bet they're looking for.}$
Yeah that is it. Thanks for your help.