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| This has the side effect of associating a direction with movement along the line. At ''t=0'', the solution is (0,0). At ''t=1'', the solution is (-1,1). | This has the side effect of associating a direction with movement along the line. At ''t=0'', the solution is (0,0). At ''t=1'', the solution is (-1,1). Applications of parametric equations then have three parts: * ''does'' a line pass through some point, line, or plane? * ''when'' does it pass through? * ''where'' does it pass through? Given a system described by ''x=2t-1'', ''y=t+2'', and ''z=-3t+2'' and a plane ''x+2y+4z=7'', the first and second questions are answered by substitution. ''(2t-1)+2(t+2)+4(-3t+2)≟7'' ''2t-1+2t+4-12t+8≟7'' ''-8t+11≟7'' ''-8t≟-4'' ''t=.5'' If there was no solution to satisfy equality, then the line must not pass through. (The third question is then answered by substituting this solution into the original system.) Given a vector a⃗ and a starting point A,,0,, (e.g., the origin), a parametric equation can be expressed as ''A,,t,, = A,,0,, + a⃗t''. The same can be said when given two points in ''R^3^'' as ''A,,0,,'' and ''A,,1,,''. These trivially yield a vector a⃗ as ''[A,,1x,,-A,,0x,, A,,1y,,-A,,0y,, A,,1z,,-A,,0z,,]''. |
Parametric Equation
A parametric equation is a reformulation of an equations in terms of time, such that there is now a direction associated with movement along the equation.
Contents
Description
A line can often be described with a single equation. This below line, however, requires two equation to be expressed: y=√(x+1)+1 and y=-√(x+1)+1.
This can be addressed by reformulating the equation in terms of time t: y=f(t) and x=g(t). In this specific example, the parametric equations are y=t and x=(t-1)2-1.
This has the side effect of associating a direction with movement along the line. At t=0, the solution is (0,0). At t=1, the solution is (-1,1). Applications of parametric equations then have three parts:
does a line pass through some point, line, or plane?
when does it pass through?
where does it pass through?
Given a system described by x=2t-1, y=t+2, and z=-3t+2 and a plane x+2y+4z=7, the first and second questions are answered by substitution.
(2t-1)+2(t+2)+4(-3t+2)≟7
2t-1+2t+4-12t+8≟7
-8t+11≟7
-8t≟-4
t=.5
If there was no solution to satisfy equality, then the line must not pass through. (The third question is then answered by substituting this solution into the original system.)
Given a vector a⃗ and a starting point A0 (e.g., the origin), a parametric equation can be expressed as At = A0 + a⃗t. The same can be said when given two points in R3 as A0 and A1. These trivially yield a vector a⃗ as [A1x-A0x A1y-A0y A1z-A0z].