How do I determine the y values for the tail of a vector function plot?

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Discussion Overview

The discussion revolves around how to plot the vector function F(x,y) = i + cos(x) j, focusing on determining the y-values for the tail of the vectors at various x-values. Participants explore the implications of the function's definition in Cartesian coordinates and the resulting vector components.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Homework-related

Main Points Raised

  • One participant describes the components of the vector function at specific x-values, noting that the i-component remains constant while the j-component varies.
  • Another participant clarifies that x represents the x-coordinate, not an angle, which is essential for understanding the function's behavior.
  • A participant expresses confusion about how to obtain a sinusoidal shape from the varying j-component while the i-component remains constant.
  • Further clarification is provided regarding the tail of the vector, indicating that for a fixed x-value, the y-value can be any real number, as the output does not depend on y.
  • Participants discuss the need to locate the tail of the vector and draw the arrow based on the i and j components, questioning how to determine the y-values for the tails in their examples.

Areas of Agreement / Disagreement

Participants express confusion and seek clarification on the plotting process, particularly regarding the y-values for the vector tails. There is no consensus on a definitive method for determining these values, and multiple viewpoints on the interpretation of the function's behavior are present.

Contextual Notes

The discussion highlights the dependence of the vector function on the x-coordinate while indicating that y can take any value, which may lead to ambiguity in plotting the vectors.

bksree
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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 doesn't look corect

TIA
 
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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.
 

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  • {1,Cos[x]}.jpg
    {1,Cos[x]}.jpg
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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
 
I'm not sure I get what your doubt is. Could you say what you expect it to look like?
 
Thank you for the reply.
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
 
bksree said:
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 ?

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.
 

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