How to make a vector function from a f(x) function

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To convert the function f(x) = x² into a vector function, one can represent it as the position vector <x, f(x)> or <x, x²>. This representation captures the x and y coordinates relative to the origin. The length of the vector can be calculated using the formula √(x² + (x²)²), which provides the magnitude. The discussion clarifies that the vector function primarily represents the position rather than explicitly including direction and length as separate components. Ultimately, the position vector for f(x) = x² is expressed as xi + f(x)j.
DrummingAtom
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Let's say I want to turn f(x) = x2 into a vector function. How would I do that?

I know I can take plots of f(x) = x2 then plug them into the Pythagorean theorem to get the distance from the origin and then I would also know the direction. But is that doesn't seem the same as a vector valued function.
 
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DrummingAtom said:
Let's say I want to turn f(x) = x2 into a vector function. How would I do that?

I'm not sure what you mean by this. Do you just want a vector that represents the x and y values wrt the origin? If so, all you need is:

<x,f(x)> = < x , x2 >
 
Wouldn't the length be needed for a vector function? The whole "magnitude and direction" thing for vectors. Something like V=(direction, length).
 
DrummingAtom said:
Wouldn't the length be needed for a vector function? The whole "magnitude and direction" thing for vectors. Something like V=(direction, length).

For any vector <x,y>, you can find the length by \sqrt{x^2 + y^2}.

In this case, if you want the length from the origin to a point (x,f(x)), all you need is \sqrt{(x)^2+(x^2)^2}

----

Just out of curiousity, what are you trying to do with this?
 
Ok, I came up with this too: \sqrt{(x)^2+(x^2)^2}.

I guess I just wanted to figure out if the vector function would be in terms of (direction, length) or only the position. Thanks for help.
 
It sounds to me like you have some function f(x) and want to construct a vector that represents the postion vector of (x, f(x)).

The position vector of any point (x, y) is xi+ yj so the position vector of (x, f(x)) is xi+ f(x)j.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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