projection of a 2d curve in 3d plane onto each of the planes (1 Viewer)

[stickyricebuns]

sorry if title doesn't make sense, im talking about parametric equations where one variable x y or z is 0 so the curve is technically only a 2d curve but it still exists in the 3d plane. would the projections onto 2 of the planes of either xy, xz or yz (in cartesian form) be x/y/z=0 or something else? because if you do it algebraically (subbing into 0 just gives 0) two of the three projections will be 0. but if you look at it from the side of those two planes there is still a line segment.

 

[username_2]

umm... some visuals maybe - i cannot really understand what your saying but from what I can make of your statement, I can imagine the curve y=x^2 in the 3-d space. Now its parametric equation would be y = a^2 and x = a and z = 0. But if I, per se, want this parabola on the y-z plane, then it would be x=0 y=a and z=x^2 and so on... is that what you're asking?

 

[stickyricebuns]

umm... some visuals maybe - i cannot really understand what your saying but from what I can make of your statement, I can imagine the curve y=x^2 in the 3-d space. Now its parametric equation would be y = a^2 and x = a and z = 0. But if I, per se, want this parabola on the y-z plane, then it would be x=0 y=a and z=x^2 and so on... is that what you're asking?

ok so if i used a parabola as an example then
r(t)=(t, t^2, 0)
so x=t, y=t^2, z=0 so the cartesian equation of the projection of the curve on the xy plane is y=x^2. but the projections onto the yz plane is z=0 since u can't sub anything into z=0 and the projection onto the xz plane is also z=0 for the same reason. but if u graph it in 3d in geogebra, when u move the view perspective into a certain angle so u are perpendicular to the xz plane, the projection of the graph is a line segment and isn't just 0. hopefully this made sense?

 

[username_2]

yes. i think i understand a bit. So what you're basically doing is this:

Imagine you had a piece of paper on a table and a pencil that is perpendicular to the table (this is the z-axis). Now if I look from the top (which is the x-y plane) i see a parabola (doesn't really matter what the eqn is). Aight. Now if I want to project this on to day the y-z plane then I will be looking from the side of the table (which is directly facing the y-z plane). Hence when you see this paper from the side, what you see is the edge of the paper - which is a straight line! The length of the line is the difference in the range extremes and hence you have "projected" the so-called curve onto the y-z plane but you haven't really affected the equation of the graph. hope this makes sense?

 

[stickyricebuns]

yes. i think i understand a bit. So what you're basically doing is this:

Imagine you had a piece of paper on a table and a pencil that is perpendicular to the table (this is the z-axis). Now if I look from the top (which is the x-y plane) i see a parabola (doesn't really matter what the eqn is). Aight. Now if I want to project this on to day the y-z plane then I will be looking from the side of the table (which is directly facing the y-z plane). Hence when you see this paper from the side, what you see is the edge of the paper - which is a straight line! The length of the line is the difference in the range extremes and hence you have "projected" the so-called curve onto the y-z plane but you haven't really affected the equation of the graph. hope this makes sense?

yess but if i were asked to find the equation of the projection, would it be 0 or a line?

 

[username_2]

line - z = 0 to be precise from the initial value of y's range to the final value of final value of y's range

 

[stickyricebuns]

line - z = 0 to be precise from the initial value of y's range to the final value of final value of y's range

i’m not sure what u mean by z=0 is a line? shouldnt it be z=(any variable) and u restrict the domain of that variable according to the range of the line accordingly

 

[username_2]

i’m not sure what u mean by z=0 is a line? shouldnt it be z=(any variable) and u restrict the domain of that variable according to the range of the line accordingly

no - I mean the projection's parametric equation would be z=0 and y=a where a is restricted within the original function's range (basically your x-component becomes 0). It is not what you are imagining - this are 3-d geometry and you cannot assume 2-d rules... atleast that is where I think the error is in your thinking.

 

[stickyricebuns]

no - I mean the projection's parametric equation would be z=0 and y=a where a is restricted within the original function's range (basically your x-component becomes 0). It is not what you are imagining - this are 3-d geometry and you cannot assume 2-d rules... atleast that is where I think the error is in your thinking.

ohhh.....ok thank you )