Prove that a given curve is planar

[Electrophy6]

Homework Statement

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Show that the given parametric curve decribed by the following notations:

x=cos(t), y=sin(t), z=2+2cos(t)

lies in a single plane ,find the normal vector to this plane

Homework Equations

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[/B]
r(t)=cos(t) i + sin(t) j + 2+2cos(t) k

The Attempt at a Solution

My attempt was finding the osculating plane to this parametric curve using the follwing equations

Tangent(t)=(r'(t)/|r'(t)|) ===> Normal(t)=(T'(t)/|T'(t)|)
Binomial=T X B
[/B]

I tried this method , yet it takes a much longer time than I guess required for solving this question---and actually I was told there is a simple way to solve this problem ...

Another attempt was trying to isolate t and substituing it in 'z' and 'y' ...Yet it doesn't help me any way

my general idea is actually proving the normal vector points the same direction in every point on the curve
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I would like to mention ,someone who solved a simillar problem on another forum used the formula

Normal vector to the plane = (r' x r'')/ |r' x r''|.
Yet I couldn't find this formula or any proof of it in any of Calculus textbooks(Stewart,Thomas,Anton...) -and in class we have never encountered this formula -so we are not allowed to use it...(BTW ,in which textbook can I find this formula and its proof?)
-------------------------------------------------------------------------------------------------------------Thank you!

 

[pasmith]

If you could find a $\mathbf{c}$ and a $\mathbf{d}$ such that $$(\mathbf{r}(t) - \mathbf{d}) \cdot \mathbf{c} = 0$$ for every $t$ then you would know that your curve lies in a plane normal to $\mathbf{c}$. Can you do that? Perhaps if you express $\mathbf{r}(t)$ in the form $\mathbf{r}_0 + \mathbf{a}\cos(t) + \mathbf{b}\sin(t)$ ...

 

[HallsofIvy]

You are given that x= cos(t), y= sin(t), z= 2+ 2 cos(t).

Isn't it obvious that z- x= 2?

 

[Electrophy6]

Yes it is(that's what I did) , yet ,the answer key tells me the normal vector is different from (-1,0,1)...

 

[Ray Vickson]

Yes it is(that's what I did) , yet ,the answer key tells me the normal vector is different from (-1,0,1)...

So, what did the answer key say was the normal vector?

 

[Electrophy6]

The complete solution is attached in this picture below...(without that ,you will not be able to understand the answer)
I do understnad the way it was solved yet , I still wanted to solve it by my own in the ways presented above...

 

[Electrophy6]

So again , is there any alternative way to solve this ?

 

[pasmith]

So again , is there any alternative way to solve this ?

Did you try my earlier suggestion?

 

[Electrophy6]

Oh sorry Pasmith, first of all thank you for your reply!
I have never seen this formula before...it makes sense of course
However, I need another way to solve it (as I mentioned, I guess I can't use it ,since I wasn't taught that...)

BTW:what is the source of this formua ,is it taken from a textbook?

 

[Ray Vickson]

The complete solution is attached in this picture below...(without that ,you will not be able to understand the answer)
I do understnad the way it was solved yet , I still wanted to solve it by my own in the ways presented above...

We don't need to see the picture; we just need to see the final result, which gives the normal $\vec{n}$ as
$$\vec{n} = \text{scalar} \cdot ( \vec{e}_z - 2 \vec{e}_x )$$
Isn't that what the last line says? Isn't that consistent with the fact that on the curve we have $z = 2 + 2x$?

 

[SammyS]

You are given that x= cos(t), y= sin(t), z= 2+ 2 cos(t).

Isn't it obvious that z- x= 2?

Yes it is(that's what I did) , yet ,the answer key tells me the normal vector is different from (-1,0,1)...

It's obvious in comparing what HallsofIvy said with the published solution that HallsofIvy has a typo in his post.

z = 2 + 2cos(t) gives z = 2 + 2x which results in -2x + z = 2 .

This gives the normal as (-2, 0, 1) as desired.

 

[Electrophy6]

It's obvious in comparing what HallsofIvy said with the published solution that HallsofIvy has a typo in his post.

z = 2 + 2cos(t) gives z = 2 + 2x which results in -2x + z = 2 .

This gives the normal as (-2, 0, 1) as desired.

yes you are totally right , but take a look at the solution my lecturer uploaded - why didn't he just use that method ?

 

[vela]

He may not have noticed there was a simpler solution, or he may have simply wanted to illustrate some concepts he wants you to learn.