Example 2 Find parametric equations for a particle that starts at (0, 3, 0) and moves around a circle as shown
in Figure 17.1.
: Circle of radius 3 in the yz-plane, centered at origin
Solution Since the motion is in the yz-plane, we have x = 0 at all times t. Looking at the yz-plane from the
positive x-direction we see motion around a circle of radius 3 in the clockwise direction. Thus,
x = 0, y= 3cost, z = -3 sint.
Example 3 Describe in words the motion given parametrically by
x = cost, y = sint, z = t.
Solution The particles x- and y-coordinates give circular motion in the xy-plane, while the z-coordinate
increases steadily. Thus, the particle traces out a rising spiral, like a coiled spring. (See Figure 17.2.)
This curve is called a helix.
Figure 17.2: The helix x =cost, y = sin t, z = t
Example 4 Find parametric equations for the line parallel to the vector 2i + 3j + 4k and through the point
(1, 5, 7).
Solution Lets imagine a particle at the point (1, 5, 7) at time t = 0 and moving through a displacement of
2i + 3j +4k for each unit of time, t. When t = 0, x = 1 and x increases by 2 units for every unit
of time. Thus, at time t, the x-coordinate of the particle is given by
x = 1 + 2t.
Similarly, the y-coordinate starts at y = 5 and increases at a rate of 3 units for every unit of time.
The z-coordinate starts at y = 7and increases by 4 units for every unit of time. Thus, the parametric
equations of the line are
x = 1 + 2t, y = 5 + 3t, z = 7 + 4t.
We can generalize the previous example as follows:
Parametric Equations of a Line through the point (x0, y0, z0) and parallel to the vector
ai + bj + ck are
x = x0 + at, y = y0 + bt, z = z0 + ct.Example 5 (a) Describe in words the curve given by the parametric equations x = 3+t, y = 2t, z = 1- t.
(b) Find parametric equations for the line through the points (1, 2,-1) and (3, 3, 4).
Solution (a) The curve is a line through the point (3, 0, 1) and parallel to the vector i + 2j -k.
(b) The line is parallel to the vector between the points P = (1, 2,-1) and Q = (3, 3, 4).
PQ = (3 - 1)i + (3 - 2)j + (4 -(-1))k = 2i +j + 5k .
Thus, using the point P, the parametric equations are
x = 1 + 2t, y = 2+t, z = -1+ 5t.
Using the point Q gives the equations x = 3 + 2t, y = 3+t, z = 4 + 5t, which represent the
same line.Example 7 (a) Find parametric equations for the line passing through the points (2,-1, 3) and (-1, 5, 4).
(b) Represent the line segment from (2,-1, 3) to (-1, 5, 4) parametrically.
Solution (a) The line passes through (2,-1, 3) and is parallel to the displacement vector v = -3i +6j +k
from (2,-1, 3) to (-1, 5, 4). Thus, the parametric equation is
r (t) = 2i -j + 3k + t(-3i + 6j +k ).(b) In the parameterization in part (a), t = 0 corresponds to the point (2,-1, 3) and t = 1 corresponds
to the point (-1, 5, 4). So the parameterization of the segment is
r (t) = 2i -j + 3k + t(-3i + 6j +k ), 0 <=>=><=>=>
Example 9 Two particles move through space, with equations r 1(t) = ti + (1 + 2t)j + (3 - 2t)k and
r 2(t) = (-2 - 2t)i + (1 - 2t)j + (1+t)k . Do the particles ever collide? Do their paths cross?Solution To see if the particles collide, we must find out if they pass through the same point at the same time
t. So we must find a solution to the vector equation r 1(t) = r 2(t), which is the same as finding a
common solution to the three scalar equations
t = -2 -2t, 1+ 2t = 1? 2t, 3 ? 2t = 1+t.
Separately, the solutions are t = -2/3, t = 0, and t = 2/3, so there is no common solution, and the
particles dont collide. To see if their paths cross, we find out if they pass through the same point at
two possibly different times, t1 and t2. So we solve the equations
t1 = -2 -2t2, 1 + 2t1 = 1-2t2, 3 - 2t1 = 1+t2.
We solve the first two equations simultaneously and get t1 = 2, t2 = -2. Since these values also
satisfy the third equation, the paths cross. The position of the first particle at time t = 2 is the same
as the position of the second particle at time t = -2, namely the point (2, 5,-1).
The exact area enclosed by a 4 petaled rose.