Surface Given By Parametric Equation

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  • #1
wubie
Hello,

First I will post the question.

Let S be the surface given by the parametric equations

x = u + v, y = u - v, z = (u^2 + v^2)/2, u^2 + v^2 =< 16

a) Represent the suface by a vector equation function r(u,v) and find r sub u X r sub v.

b) Find the parametric equations of the normal line to the surface at (4,2,5).

c) Find an equation of the tangent plane to the suface at (4,2,5).

d) Find the area of S.

Now I see that my instructor is trying to progressively guide us through the steps to find the area of the surface S. I have done part a. And I think I know how to do part c and d. But I am confusing myself with part b. Which is frustrating since this should be pretty elementry.

I guess I can show what I have so far:

a)

The vector function is

r(u,v) = (u + v)i + (u - v)j + ((u^2 + v^2)/2)k

and r sub u X r sub v is

(v + u)i - (v - u)j - 2k

For part c) I think that the normal vector at point (4,2,5) is

4i + 2j - 2k where v = 1 and u = 3.

So the equation of the tangent plane at (4,2,5) would be

4(x - 4) + 2(y - 2) - 2(z - 5) = 0.

I haven't done part d) yet but I don't think it will be a problem. However I am not sure what part b is asking.

I know that the normal vector to the tangent plane is

(v + u)i - (v - u)j - 2k.

So what is part b asking then? Is it just a precusor to part c)? I know that the normal vector at point (4,2,5) is

4i - 2j - 2k.

So are the parametric equations

x=4, y=-2, z=-2?

I am assuming that normal vector and the normal line are the same. But then again then normal vector at point (4,2,5) isn't the same as the normal line at point (4,2,5) is it?

Any help is appreciated. Thankyou.
 
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Answers and Replies

  • #2
HallsofIvy
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"The vector function is

r(u,v) = (u + v)i + (u - v)j + ((u^2 + v^2)/2)k "

Yes, that's the easy part!

"and r sub u X r sub v is

(v + u)i - (v - u)j - 2k"

Yes, that's right too.

"For part c) I think that the normal vector at point (4,2,5) is

4i + 2j - 2k where v = 1 and u = 3."

Yes. The point is that ru and rv are tangent to the plane so their cross product is normal to it. Since u+v= 4, u-v= 2 give u= 3, v= 1, evaluating (v+u)i- (v-u)j- 2k there gives 4i+ 2j- 2k which is a normal vector.
HOWEVER, the problem said "Find the parametric equations" and you haven't done that yet!

"So the equation of the tangent plane at (4,2,5) would be

4(x - 4) + 2(y - 2) - 2(z - 5) = 0."

Exactly right.

In (a), you found the "fundamental vector product", a vector normal to the surface at each point. The length of that vector is the differential of area (with respect to the projection into the u-v plane.)

Since that vector is (v + u)i - (v - u)j - 2k, its length is
&radic;((v+u)2+(v-u)2+ 4)= &radic;(2v2+ 2v2+ 4). The area will be

[tex]\int \sqrt(2v^2+ 2u^2+4)dudv[/tex]

where the integration is taken over the circle centered on (0,0) with radius 4. You might want to convert to polar coordinates for that.
 
  • #3
HallsofIvy
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I just tried to post a reply to this but got a

"zero sized error" message!
 
  • #4
wubie
I still am confused about find the parametric equations of the normal line at (4,2,5) - part b. I think I am confusing the concept of normal vector with normal line. I need clarification on this point. I don't remember some of the material from my last calculus course over a year ago.

Am I supposed to find the vector equation of the normal line at (4,2,5)?

If so I this is what I did:

let r = r sub 0 + n (the normal vector).

where r sub 0 is the position vector of the point (4,2,0). So then

r = <4,2,5> + n.

n is the normal vector at point (4,2,5) which is

<4,2,-2>.

So then n is the product of a scalar and a vector parallel to n.

I will stop here since I am getting confused now. I think I am mixing things up.
 
  • #5
HallsofIvy
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Originally posted by wubie
I still am confused about find the parametric equations of the normal line at (4,2,5) - part b. I think I am confusing the concept of normal vector with normal line. I need clarification on this point. I don't remember some of the material from my last calculus course over a year ago.

Am I supposed to find the vector equation of the normal line at (4,2,5)?

If so I this is what I did:

let r = r sub 0 + n (the normal vector).

where r sub 0 is the position vector of the point (4,2,5). So then

r = <4,2,5> + n.

n is the normal vector at point (4,2,5) which is

<4,2,-2>.

So then n is the product of a scalar and a vector parallel to n.

I will stop here since I am getting confused now. I think I am mixing things up.

Yes, you are getting confused! Did you notice that your
"r= <4,2,5>+ n" doesn't have a variable in it?

I'm not sure what "n is the product of a scalar and a vector parallel to n" means. Every vector is the product of a scalar and a vector parallel to it! I think it is that "scalar" that you intend as the variable.

If r0 is the position vector of a point, and v is a vector then r(t)= r0+ vt is the vector equation of the line through r0 in the direction of v (t is the variable).

In this particular case you have already determined that the point is (4,2,5) and that the normal vector to the surface at that point is
4i+ 2j- 2k. The vector function describing the line normal to the surface at (4,2,5) is r(t)= (4i+ 2j+ 5k)+ (4i+ 2j- 2k)t.
That could also be written as r(t)= (4+4t)i+ (2+2t)j+ (5-2t)k.

The "parametric equations" corresponding to that vector function are
x= 4+ 4t, y= 2+ 2t, z= 5- 2t.
 
  • #6
wubie
Yes, you are getting confused! Did you notice that your
"r= <4,2,5>+ n" doesn't have a variable in it?

Yes. That's when I knew I was in trouble.

If r0 is the position vector of a point, and v is a vector then r(t)= r0+ vt is the vector equation of the line through r0 in the direction of v (t is the variable)

Yes I knew this. I was letting the r0 be the position vector of point (4,2,5). Then I was letting the normal vector, n , be equal to vt where v was the unit vector of n and, this case, I had let t be 1/|n|.

What I was supposing was that v was the unit vector of n, and could also be a vector parallel to n with the same direction and same magnitude originating at the origin (a positional vector of some point). When v was multiplied by 1/|n| I would get the normal vector, n.

So I was letting t = 1/|n| in this case. But I knew that t could be any scalar. So I could have a line that was normal to the surface S, passing through (4,2,5) and parallel to my derived unit vector v at the origin by simple multiplying the v at the origin by an scalar t. If t was equal to 1/|n| then I would get the normal vector (vt = n).

Anyway, that is what I was thinking just so you can see perhaps the error of my "logic". 8)


In this particular case you have already determined that the point is (4,2,5) and that the normal vector to the surface at that point is 4i+ 2j- 2k.

Right.

The vector function describing the line normal to the surface at (4,2,5) is r(t)= (4i+ 2j+ 5k)+ (4i+ 2j- 2k)t. That could also be written as r(t)= (4+4t)i+ (2+2t)j+ (5-2t)k.

I see what happened to me now. In my case r(t) was ALWAYS equal to the positional vector <4,2,5> plus the normal vector <4,2,-2>. I was only expressing part of the line r(t)= (4+4t)i+ (2+2t)j+ (5-2t)k for a particular point on the line with an origin of (4,2,5).

The "parametric equations" corresponding to that vector function are x= 4+ 4t, y= 2+ 2t, z= 5- 2t.

I easily saw that once the equation of the normal line was derived.

I was also thinking that there was a relationship between the parameter t and the parameters of (u,v). I see that t is completely independent of (u,v) and that r0 at point (4,2,5) can be expressed by the normal line when t = 0. Or can be expressed as the vector equation function when v =1 and u = 3.


Thanks HallsOfIvy. This helps alot. Please post any more comments if you think that my thinking needs adjustment.

Cheers.
 

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