I have written a parametric function called DrawCircle that renders a circle at a given x and y point, with radius;
DrawCircle(t,position,radius)=(radiuscos(t)+position.x,radiussin(t)+position.y)
circle=DrawCircle(t,(0,0),1)
Is there any way to access the x position of the position variable?
Here’s my graph:
I’m not certain I understand your question. You’ve defined d_{ragMe} and I see you defined your function using dot notation (i.e. d_{ragMe}.x ) which seems to be what you’re asking for.
If you mean a point on the circle, you just need an angle value instead of t. I added a folder in the link below to demonstrate.
Side note: You don’t need to define the point in your evaluated function using dot notation, you can just enter the point, since you’re parsing the coordinates with the function anyway!
I was hoping to get the x-coordinate from the position variable. When calling DrawCircle, it takes in t, a coordinate-pair, and a radius. Is there any way to get the x-coordinate from position? For Example, when defining
circle=DrawCircle(t,(12,3),1)
circle.x should be equal to 12. Is this possible, or will I have to use a point to get it’s position?
You can’t extract it from circle, but if you use a variable for the position (as you did in your sample), then you can use the dot notation from that.
So, for
C_{enter} = (12,3)
c_{ircle} = D_{rawCircle}(t, C_{enter}, 1)
x_{circle} = C_{enter}.x
The circle is created is the collection of points equidistant from your position, so trying to extract an x-coordinate would really be a list of x-coordinates, not the center. You can see this if you use a list of values between 0 and 2pi instead of t.