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Unit Vector of f(x) at point

  • Thread starter haxtor21
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  • #1
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Homework Statement


Find a unit vector (a) parallel to and (b) normal to the graph of f(x) at the given point. Then sketch a graph of the vectors and the functions.

Function:
f(x)=xx
Point:
(3,9)


Homework Equations





The Attempt at a Solution



http://img543.imageshack.us/img543/6964/gedc0203.th.jpg [Broken]

Uploaded with ImageShack.us
 
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Answers and Replies

  • #2
<cos(x),sin(x)> is always a unit vector.

For the normal vector, just find a vector orthogonal to the one you have. Both <-x,y> and <x,-y> are orthogonal to <x,y>, or if you want to cheat, use <0,0,1>.
 
  • #3
SammyS
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Homework Statement


Find a unit vector (a) parallel to and (b) normal to the graph of f(x) at the given point. Then sketch a graph of the vectors and the functions.

Function:
f(x)=xx
Point:
(3,9)


Homework Equations





The Attempt at a Solution



http://img543.imageshack.us/img543/6964/gedc0203.th.jpg [Broken]

Uploaded with ImageShack.us
Your vector, [tex]\vec{v}\,,[/tex] is a unit vector.

[tex]\left\|\vec{v}\right\|=\sqrt{\cos^2(\,\arctan(6))+\sin^2(\arctan(6))\,\,}=\sqrt{1}=1[/tex]

BTW,

[tex]\sin(\arctan(6)) =\frac{6}{\sqrt{1+6^2\,}}=\frac{6}{\sqrt{37}}[/tex]

and

[tex]\cos(\arctan(6)) =\frac{1}{\sqrt{1+6^2\,}}=\frac{1}{\sqrt{37}}[/tex]
 
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  • #4
46
0
<cos(x),sin(x)> is always a unit vector.

For the normal vector, just find a vector orthogonal to the one you have. Both <-x,y> and <x,-y> are orthogonal to <x,y>, or if you want to cheat, use <0,0,1>.
Ok, so far I have a) done. How exactly am I supposed to do this? The back of the book says the answer is +-(1/sqrt(37))<6,-1>. Where does the +- come from and how did they get that answer (looks iverse. According to what you said, I would just stick a negative in either component (?) ).
 
  • #5
Ok, so far I have a) done. How exactly am I supposed to do this? The back of the book says the answer is +-(1/sqrt(37))<6,-1>. Where does the +- come from and how did they get that answer (looks iverse. According to what you said, I would just stick a negative in either component (?) ).
Correct. The +- means that either answer is acceptable, though to be thorough you can have both answers.
 
  • #6
46
0
Correct. The +- means that either answer is acceptable, though to be thorough you can have both answers.
So you are saying that I can have (1/sqrt(37))<-1,6> or (1/sqrt(37))<1,-6> for b) even though the book says the answer is (1/sqrt(37))<6,-1>?

So then I would guess the +- reffers to the direction of the vector compared to what it was previously, but it has no relationship with the signs inside the < , > components correct?
 
  • #7
So you are saying that I can have (1/sqrt(37))<-1,6> or (1/sqrt(37))<1,-6> for b) even though the book says the answer is (1/sqrt(37))<6,-1>?

So then I would guess the +- reffers to the direction of the vector compared to what it was previously, but it has no relationship with the signs inside the < , > components correct?
I'm sorry, I was wrong. The normal vector of <x,y> is <-y,x> or <y,-x>.
 
  • #8
46
0
I'm sorry, I was wrong. The normal vector of <x,y> is <-y,x> or <y,-x>.
How do you know that is the normal vector? is there a proof sort of thing somewhere?
 

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