# How to plot vector function

1. Jul 15, 2011

### bksree

Hi
How do I plot this vector function
F(x,y) = i + cos x j where i and j are unit vectors
Spose I take
x = 0, then the components in i and y direction are 1, 1
x = pi/4, then " " 1, 1/sq rt 2
x = pi/2 " " 1, 0

This doesnt look corect

TIA

2. Jul 15, 2011

### Rasalhague

That's how you plot it. Remember that, if the function F is defined in terms of Cartesian coordinates, x just means the x coordinate, not the angle of the point from the x-axis.

#### Attached Files:

• ###### {1,Cos[x]}.jpg
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3. Jul 15, 2011

### bksree

Thabk you
My doubt is as follows :
When x = 0, the comp ibn i direction is 1 and in j dirn is 1
When x= pi/4 = 0.78, the comp in i dirn is 1 and in j is 1/sq rt 2
When x= 3pi/4 = 2.36, the comp in i dirn is 1 and in j is -1/sq rt 2
When x = pi/2 = 1.57, the comp in i dirn is 1 and in j is 0, and so on

But here the comp in i dirn is always 1 while j is varying.

How then do you get the sinusoidal shape ?

TIA

4. Jul 15, 2011

### Rasalhague

I'm not sure I get what your doubt is. Could you say what you expect it to look like?

5. Jul 15, 2011

### bksree

My question is : How do I plot this vector function manually

F(x,y) = i + cos x j where i and j are unit vectors
Spose I take
(i) When x = 0, the comp in i direction is 1 and in j dirn is 1
(ii) When x= pi/4 = 0.78, the comp in i dirn is 1 and in j is 0.707
(iii) When x = pi/2 = 1.57, the comp in i dirn is 1 and in j is 0
(iv) When x= 3pi/4 = 2.36, the comp in i dirn is 1 and in j is -0.707
and so on
To plot, I have to locate the tail of vector and draw the arrow in the direction given by the i and j components.
eg. in (i) the x for the tail is 0 and dirn is given by 1 and 1
(ii) the x for the tail is 0.78, dirn is given by 1 and 0.707
But what are the y values of the tail in these cases ?

TIA

6. Jul 15, 2011

### Rasalhague

Okay, take the vector (1,1). When x = 0, we can calculate that F(x,y) = (1,1). But this is true for all values of y. If you look at the plot I posted, you'll see that all the vectors along the vertical line where x =0 are identical. That's because the output doesn't depend on the value of y. So if you're asking where is the tail of the vector (1,1) = i + j, such that x = 0, there's no one, unique answer: y can equal any real number. Likewise, when F(x,y) = (1,0,707), y can take any value.