Is equation of line or curve in 3 space have to be a parametric equation?

yungman
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I have been confused about this. When come to equation of lines and curves in even 2 space, they automatically go parametric equation x(t), y(t) etc. I just want to verify my understanding:

For 3 space, equation using x, y and z will automatically produce a plane or a surface. In order to produce a line or curve, they have not choice but to represent x, y and z in form of x(t), y(t) and z(t) so it become line or curve.

Therefore a linear line or a curve HAS to be represented by a parametric equation.


Am I correct?
 
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The set of solutions of these two equation defines a curve in R3:
x2 + y2 = 1
z = 2
 
Mark44 said:
The set of solutions of these two equation defines a curve in R3:
x2 + y2 = 1
z = 2

In the calculus books I studied include Sherman Stein and Howard Anton. They only represent lines and curves in 3 space with parametric equation. Also in the chapter of vector value function. All the books I have immediately go to parametric equations.
 
My point is that curves in space don't have to be represented by parametric equations.
 
Thanks.
Alan
 
Curves are one dimensional. That means that every point can, theoretically, be determined by using a single number. If you write a curve in terms of parametric equations, then that single number, the parameter, is apparent.

But any time you have n equations in n+1 variables, you can, theoretically, solve for n of the variables in terms of the other one. For example, if you have 3x+ 2y+ z= 4 and x- y- z= 5, adding the two equations eliminates z giving 4x+ y= 9. You can then solve for y in terms of x: y= 9- 4x. Putting that into the second equation, x- (9- 4x)- z= -3x- 9- z= 5 or -z= 14+ 3x so that z= -14- 3x. We could then use x as parameter:
x= t, y= 9- 4t, z= -14- 3t. But the two equations 3x+ 2y+ z= 4 and x- y- z= 5 represent that line (geometrically you can think of the line as the intersection of the two planes given by those equations.

Another way of representing a curve is in what is called the "symmetric form": f(x,y,z)= g(x, y, z)= h(x, y, z). That is just a way of representing the two equations f(x,y,z)= g(x,y,z), g(x,y,z)= h(x,y,z). Again, since there are two equations in three variables, you could, theoretically, solve for two of the variables in terms of the third- its graph is one-dimensional, a curve.
 
HallsofIvy said:
Curves are one dimensional. That means that every point can, theoretically, be determined by using a single number. If you write a curve in terms of parametric equations, then that single number, the parameter, is apparent.

But any time you have n equations in n+1 variables, you can, theoretically, solve for n of the variables in terms of the other one. For example, if you have 3x+ 2y+ z= 4 and x- y- z= 5, adding the two equations eliminates z giving 4x+ y= 9. You can then solve for y in terms of x: y= 9- 4x. Putting that into the second equation, x- (9- 4x)- z= -3x- 9- z= 5 or -z= 14+ 3x so that z= -14- 3x. We could then use x as parameter:
x= t, y= 9- 4t, z= -14- 3t. But the two equations 3x+ 2y+ z= 4 and x- y- z= 5 represent that line (geometrically you can think of the line as the intersection of the two planes given by those equations.

Another way of representing a curve is in what is called the "symmetric form": f(x,y,z)= g(x, y, z)= h(x, y, z). That is just a way of representing the two equations f(x,y,z)= g(x,y,z), g(x,y,z)= h(x,y,z). Again, since there are two equations in three variables, you could, theoretically, solve for two of the variables in terms of the third- its graph is one-dimensional, a curve.

Thanks a million, this is exactly what I need, a comparison.

Alan
 
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