Parametric Equations: Lines, Circles, and Motion in 3D Space

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. The circle has a radius of 3 in the yz-plane, centered at the 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, the parametric equations are:

• x = 0
• y = 3cos(t)
• z = -3sin(t)

Example 3: Describe in words the motion given parametrically by x = cos(t), y = sin(t), z = t.

Solution: The particle's 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 = cos(t), 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: 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 = 7 and 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 ≤ t ≤ 1.

Example 9: Two particles move through space, with equations r₁(t) = ti + (1 + 2t)j + (3 - 2t)k and r₂(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 r₁(t) = r₂(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 don't 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.