# Unit normal vector

1. Apr 29, 2010

### string_656

1. The problem statement, all variables and given/known data
Find unit normal vector to the surface z = 2xy at (2,1,4)

2. Relevant equations

-

3. The attempt at a solution
fx = 2y
fx(2,1,4)=2
fy = 2x
fy = 4

= (2,4,-1)

is this correct? and in the right format?

2. Apr 29, 2010

### jtyler05si

That looks good to me so far. I don't believe that is a "unit" normal vector yet.

3. Apr 29, 2010

### string_656

unit mormal vector is given by N(t) = T'(t)/|T'(t)| where T(t) = r'(t)/|r'(t)|, and i think in this case r(t) = 2i +4j - 1k. so im a bit stuck because it will just end up being zero?

4. Apr 30, 2010

### jtyler05si

why would it be zero?

5. Apr 30, 2010

### jtyler05si

the i component of your normal unit vector that I got is (2/sqrt(21))i

6. May 1, 2010

### string_656

umm ok?.. so r(t) = 2(t)i +4(t)j - 1(t)k
so you went, r'(t) = 2 + 4 - 1
and.... |r'(t)| = (sqrt(21)
so T(t) = 2/(sqrt(21)i + 4/(sqrt(21)j - 1/(sqrt(21)k....?

...
T'(t) = 0
|T'(t)| = 0
N(t) = 0??

Thanks for the help by the way.

7. May 1, 2010

### HallsofIvy

What are you calculating here? You originally asked for the unit normal vector to z= 2xy at (2, 1, 4). Okay if you write $F(x,y,z)= 2xy- z$ then $\nabla F= 2y\vec{i}+ 2x\vec{j}- \vec{k}$ is normal to the surface. At (2, 1, 4), that is [itex]2\vec{i}+ 4\vec{j}- \vec{k}[/math]. That is not a "unit" vector because it has length [itex]\sqrt{2^2+ 4^2+ 1^2}= \sqrt{21}[/math].

The unit normal vector is
(2/sqrt{21})i+ (4/sqrt{21})j- (1/sqrt{21})k.
(For some reason I simply cannot get the LaTex to show this correctly.)

I have no idea why you are now trying to take the derivative of the normal vector.

Last edited by a moderator: May 1, 2010
8. May 1, 2010

### string_656

well thats what im not sure about. The book states that the "normal unit vector" is defined as N(t) = T'(t)/|T'(t)|
where T(t) = r'(t)/|r'(t)|

the text book uses 1 example (its very brief about this topic), but it uses cos, and sin. so constantly differentiating them dosn't reduce it to 0.

they used, r(t) = cos(t)i+ sin(t)j + (t)k,
r'(t) = -sin(t)i+ cos(t)j + k
|r'(t)| = sqrt(2) ------->(No idea how its the root of 2)
... and so on. for T'(t)...
untill they get N(t) = -cos(t)i - sin(t)j - 0k
= (-cos(t), -sin(t), 0)

am i making a mistake when defining my r(t)?

9. May 1, 2010

### jtyler05si

It looks as though they found some general formula for the normal unit vector for that particular surface.

If that is true, then you should be able to the same and then plug in the appropriate values at some point.

I have never seen that before so I am no help, I'm sorry.

but, I have seen problems exactly like your question before and working them the way you did in post #1 works great (with the addition of finding the unit vector)! If this is for a class, I would ask the prof/etc. about what is expected. When I took the class pertaining to this, this was how I was taught.