Unit Normal Vector: (2,4,-1) at (2,1,4)

[string_656]

Homework Statement

Find unit normal vector to the surface z = 2xy at (2,1,4)

Homework Equations

-

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?

 

[jtyler05si]

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

 

[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 I am a bit stuck because it will just end up being zero?

 

[jtyler05si]

why would it be zero?

 

[jtyler05si]

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

 

[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.

 

[HallsofIvy]

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.

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 2\vec{i}+ 4\vec{j}- \vec{k}[/math]. That is not a "unit" vector because it has length \sqrt{2^2+ 4^2+ 1^2}= \sqrt{21}[/math].<br /> <br /> The unit normal vector is <br /> (2/sqrt{21})<b>i</b>+ (4/sqrt{21})<b>j</b>- (1/sqrt{21})<b>k</b>.<br /> (For some reason I simply cannot get the LaTex to show this correctly.)<br /> <br /> I have no idea why you are now trying to take the <b>derivative</b> of the normal vector.

 

[string_656]

well that's what I am 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 textbook 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)?

 

[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.