Vector Calculus: Velocity, Acceleration, and Fields
Example 1: Ferris Wheel Motion
A child is sitting on a Ferris wheel with a diameter of 10 meters, making one revolution every 2 minutes. Find the speed of the child and draw velocity vectors at two different times.
Solution: The child moves at a constant speed around a circle of radius 5 meters, completing one revolution every 2 minutes. One revolution around a circle of radius 5 is a distance of 10π, so the child’s speed is 10π/2 = 5π ≈ 15.7 m/min. Hence, the magnitude of the velocity vector is 15.7 m/min. The direction of motion is tangent to the circle and, therefore, perpendicular to the radius at that point.
Example 3: Tangent Line to a Parametric Curve
Find the tangent line at the point (1, 1, 2) to the curve defined by the parametric equation
r(t) = t2i + t3j + 2tk.
Solution: At time t = 1, the particle is at the point (1, 1, 2) with position vector r0 = i + j + 2k. The velocity vector at time t is r’(t) = 2ti + 3t2j + 2k, so at time t = 1, the velocity is v = r’(1) = 2i + 3j + 2k. The tangent line passes through (1, 1, 2) in the direction of v, so it has the parametric equation
r(t) = r0 + tv = (i + j + 2k) + t(2i + 3j + 2k).
Example 5: Straight-Line Motion
Consider the motion given by the vector equation
r(t) = 2i + 6j + (t3 + t)(4i + 3j + k). Show that this is straight-line motion in the direction of the vector 4i + 3j + k and relate the acceleration vector to the velocity vector.
Solution: The velocity vector is
v = (3t2 + 1)(4i + 3j + k).
Since (3t2 + 1) is a positive scalar, the velocity vector v always points in the direction of the vector 4i + 3j + k. In addition,
Speed = ||v|| = (3t2 + 1) * √(42 + 32 + 12) = √(26)(3t2 + 1).
Notice that the speed is decreasing until t = 0, then starts increasing. The acceleration vector is a = 6t(4i + 3j + k). For t > 0, the acceleration vector points in the same direction as 4i + 3j + k, which is the same direction as v. This makes sense because the object is speeding up. For t < 0, the acceleration vector 6t(4i + 3j + k) points in the opposite direction to v because the object is slowing down.
Vector Fields
Example 1: Sketch the vector field in 2-space given by F(x, y) = -yi + xj.
Solution: Table 17.1 shows the value of the vector field at a few points. Notice that each value is a vector. To plot the vector field, we plot F(x, y) with its tail at (x, y). (See Figure 17.20.)
Now we look at the formula. The magnitude of the vector at (x, y) is the distance from (x, y) to the origin since
||F(x, y)|| = ||-yi + xj|| = √(x2 + y2)
Therefore, all the vectors at a fixed distance from the origin have the same magnitude. The magnitude gets larger as we move farther from the origin. What about the direction? Figure 17.20 suggests that at each point (x, y), the vector F(x, y) is perpendicular to the position vector r = xi + yj. We confirm this using the dot product:
r · F(x, y) = (xi + yj) · (-yi + xj) = 0.
Thus, the vectors are tangent to circles centered at the origin and get longer as we go out.