Parametric equations of lines General parametric equations In this part of the unit we are going to look at parametric curves. This is simply the idea that a point moving in space traces out a path over time. Thus there are four variables to consider, the position of the point (x, y, z) and an independent variable t, which we can think of as time. (If the point is moving in plane there are only three variables, the position of the point (x, y) and the time t.) Since the position of the point depends on t we write x = x(t), y = y(t), z = z(t) to indicate that x, y and z are functions of t. We call t the parameter and the equations for x, y and z are called parametric equations. In physical examples the parameter often represents time. We will see other cases where the parameter has a diﬀerent interpretation, or even no interpretation. Parametric equations of lines Later we will look at general curves. Right now, let’s suppose our point moves on a line. The basic data we need in order to specify a line are a point on the line and a vector parallel to the line. That is, we need a point and a direction. P = (x, y, z) P0 = point v = direction Example 1: Write parametric equations for a line through the point P0 = (1, 2, 3) and parallel to the vector v = �1, 3, 5�. Answer: If P = (x, y, z) is on the line then the vector −−→ P0 P = �x − 1, y − 2, z − 3� −−→ is parallel to �1, 3, 5�. That is, P0 P is a scalar multiple of �1, 3, 5�. We call the scale t and write: �x, y, z� = �x − 1, y − 2, z − 3� = t�1, 3, 5� ⇔ x − 1 = t, y − 2 = 3t, z − 3 = 5t ⇔ x = 1 + t, y = 2 + 3t, z = 3 + 5t. Example 2: In example 1, if our direction vector was �2, 6, 10� = 2v we would get the same line with a diﬀerent parametrization. That is, the moving point’s trajectory would follow the same path as the trajectory in example 1 , but would arrive at each point on the line at a diﬀerent time. Example 3: In general, the line through P0 = (x0 , y0 , z0 ) in the direction of (i.e., parallel to) v = �v1 , v2 , v3 � has parametrization �x, y, z� = �x0 + tv1 , y0 + tv2 , z0 + tv3 � ⇔ x = x0 + tv1 , y = y0 + tv2 z = z0 + tv3 . Example 4: Find the line through the point P0 = (1, 2, 3) and P1 = (2, 5, 8). Answer: We use the data given to ﬁnd the basic data (a point and direction vector) for the line. −−−→ We’re given a point, P0 = (1, 2, 3). The direction vector v = P0 P1 = �1, 3, 5�. So, we get −−−→ �x, y, z� = OP0 + tv = �1 + t, 2 + 3t, 3 + 5t� ⇔ x = 1 + t, y = 2 + 3t, z = 3 + 5t.