Unit Vector of f(x) at point

In summary, the normal vector of a vector <x,y> is either <-y,x> or <y,-x>. The unit vector of <x,y> is always <cos(x),sin(x)>.
  • #1
haxtor21
46
0

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
 
Last edited by a moderator:
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  • #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
haxtor21 said:

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]
 
Last edited by a moderator:
  • #4
transphenomen said:
<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
haxtor21 said:
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
transphenomen said:
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
haxtor21 said:
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
transphenomen said:
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?
 

What is the definition of a unit vector of f(x) at a point?

A unit vector of f(x) at a point is a vector with a magnitude of 1 that points in the same direction as the slope of the function at that specific point. It represents the direction in which the function is increasing at that point.

How is the unit vector of f(x) at a point calculated?

The unit vector of f(x) at a point can be calculated by taking the gradient of the function at that point and then dividing it by its magnitude. This will result in a vector with a magnitude of 1 and the same direction as the gradient.

What does the unit vector of f(x) at a point represent?

The unit vector of f(x) at a point represents the direction of the steepest ascent or descent of the function at that specific point. It can also be interpreted as the direction of the tangent line to the function at that point.

Why is the unit vector of f(x) at a point important?

The unit vector of f(x) at a point is important because it provides valuable information about the behavior of the function at that specific point. It can be used to determine the direction of change in the function and can aid in understanding the overall behavior of the function.

How can the unit vector of f(x) at a point be used in applications?

The unit vector of f(x) at a point can be used in various applications, such as optimization problems, where it can help determine the direction of maximum or minimum change in a function. It can also be used in physics to represent the direction of a force acting on an object along a curved path.

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