Vector valued functions and motion in the plane

When you reach for it

An object moving in a flat plane, a puck on an air hockey table or a sailplane seen from above, has two coordinates that change with time. Package them as one position vector \mathbf{r}(t) = \langle x(t), y(t) \rangle and every question about the motion becomes a question about a derivative or an integral of that vector. Which way is it going, and how fast? Is it at rest, or moving straight up the page? Is it gaining speed? Given only its acceleration and a start, where is it later? How far has it flown, as against how far it ended up from where it began?

Position, velocity and acceleration did this on a line, where the sign of v is the direction; Arc length and speed on parametric and polar curves found the speed \displaystyle \frac{ds}{dt} = \sqrt{x'(t)^2 + y'(t)^2} and the length of a parametric curve. Here velocity becomes a vector, speed is its length, and direction is the vector’s direction. Vectors (component form, length, scalar multiples, the dot product) come from Component form of a vector, Magnitude and direction of a vector, Scaling a vector, Forces and velocities as vectors, The dot product and the angle between vectors and Orthogonal vectors; adding vectors and scaling them are done component by component (Vector addition and scalar multiplication).

The procedure

Day 1: derivatives, velocity, speed and direction

Vector valued functions. A vector valued function assigns to each t in its domain a vector \mathbf{r}(t) = \langle x(t), y(t) \rangle = x(t)\,\mathbf{i} + y(t)\,\mathbf{j}. Drawn from the origin, the tip of \mathbf{r}(t) traces the parametric curve x = x(t), y = y(t), with the orientation of Parametric equations. A limit of \mathbf{r} is taken component by component: \lim_{t \to c} \mathbf{r}(t) = \langle \lim_{t \to c} x(t), \lim_{t \to c} y(t) \rangle when both limits exist. This is right because \mathbf{r}(t) gets close to a vector \mathbf{L} exactly when each component gets close to the matching component of \mathbf{L}: the distance between two vectors is at least each component difference and at most their sum.

The derivative. As for a function of one variable,

The numerator is the vector from one point of the path to a later one, and dividing by h scales it. The quotient’s components are the difference quotients of x and y, so the limit is taken component by component:

So \mathbf{r} is differentiable at t exactly when x and y both are. Where \mathbf{r}'(t) \ne \mathbf{0}, it points along the curve in the direction of travel, and its slope \displaystyle \frac{y'(t)}{x'(t)} is the slope \displaystyle \frac{dy}{dx} of Derivatives of parametric curves. Because derivatives are taken componentwise, the familiar rules carry over: (\mathbf{u} + \mathbf{w})' = \mathbf{u}' + \mathbf{w}', (c\,\mathbf{u})' = c\,\mathbf{u}', (f\mathbf{u})' = f'\mathbf{u} + f\mathbf{u}' for a scalar function f, and the dot product rule

For the last, write \mathbf{u} \cdot \mathbf{w} = u_1 w_1 + u_2 w_2 and apply the ordinary product rule to each term.

Motion. When \mathbf{r}(t) is the position of a moving object, its velocity and acceleration are

The speed is the length of the velocity, \displaystyle \lVert \mathbf{v}(t) \rVert = \sqrt{x'(t)^2 + y'(t)^2}, which is exactly the \displaystyle \frac{ds}{dt} of Arc length and speed on parametric and polar curves. Where \mathbf{v}(t) \ne \mathbf{0}, the direction of motion is the unit vector \displaystyle \frac{\mathbf{v}(t)}{\lVert \mathbf{v}(t) \rVert}, or equivalently the angle the velocity makes with the positive x direction. The signs of the components say which way it goes across (x' > 0 right, x' < 0 left) and up or down (y' > 0 up).

Rest, vertical and horizontal motion. On a line, a turn needs v = 0. In the plane, stopping needs both components to vanish at the same instant:

• the object is at rest at t when \mathbf{v}(t) = \mathbf{0}, that is x'(t) = 0 and y'(t) = 0;
• it is moving vertically when x'(t) = 0 and y'(t) \ne 0;
• it is moving horizontally when y'(t) = 0 and x'(t) \ne 0.

So the work is to solve x'(t) = 0 and y'(t) = 0 separately and compare the two lists.

Day 2: integrals, displacement, distance and speeding up

Integrating a vector. An antiderivative and a definite integral are also componentwise:

and the two constants of integration form one constant vector \mathbf{C} = \langle C_1, C_2 \rangle. The The fundamental theorem of calculus, applied to each component, gives

From acceleration back to position. Integrating \mathbf{a} gives \mathbf{v} up to a constant vector, and one known velocity \mathbf{v}(t_0) fixes it. Integrating \mathbf{v} gives \mathbf{r} up to another constant vector, fixed by one known position. Two vector conditions are needed, one for each integration.

Displacement and distance. The displacement over [a, b] is the vector \mathbf{r}(b) - \mathbf{r}(a) = \int_a^b \mathbf{v}(t)\,dt, from start to finish. The distance traveled is the integral of the speed,

the arc length of Arc length and speed on parametric and polar curves, counted again on any stretch covered twice. The length of the displacement is never more than the distance (Exercise 17 proves it), with equality only when the object keeps one direction. On a line this was Position, displacement and distance from velocity; in the plane the two differ even with no stop at all, because a path that bends is longer than its chord.

Speeding up and slowing down. Apply the dot product rule to \lVert \mathbf{v} \rVert^2 = \mathbf{v} \cdot \mathbf{v}:

Where \mathbf{v} \cdot \mathbf{a} > 0, the square of the speed increases, so the speed does (square root is increasing): the object speeds up. Where \mathbf{v} \cdot \mathbf{a} < 0 it slows down, and where \mathbf{v} \cdot \mathbf{a} = 0 on a whole interval the speed is constant. Where \mathbf{v} \ne \mathbf{0}, the rate itself is \displaystyle \frac{d}{dt}\lVert \mathbf{v} \rVert = \frac{\mathbf{v} \cdot \mathbf{a}}{\lVert \mathbf{v} \rVert}. In words: the part of \mathbf{a} along \mathbf{v} changes the speed, and the part perpendicular to \mathbf{v} (Orthogonal vectors) only turns the object.

Worked examples

Boundaries

• Everything here works component by component only in fixed Cartesian components \mathbf{i} and \mathbf{j}. Polar r and \theta are not components of a fixed basis, and \langle r'(t), \theta'(t) \rangle is not a velocity.
• \mathbf{r} is differentiable at t only if both components are. One differentiable component is not enough.
• At an instant with \mathbf{v} = \mathbf{0} there is no direction of motion, and \displaystyle \frac{\mathbf{v}}{\lVert \mathbf{v} \rVert} is undefined. “Moving vertically” needs y' \ne 0 as well as x' = 0.
• An antiderivative of a vector carries a constant vector. Velocity alone, or acceleration with one condition, does not fix the position; it takes one condition per integration.
• \int_a^b \lVert \mathbf{v} \rVert\,dt and \left\lVert \int_a^b \mathbf{v}\,dt \right\rVert are different numbers unless the object keeps one direction throughout. Distance needs \mathbf{v} continuous, or continuous on pieces, with the integral split where the velocity jumps (Exercise 14).
• The test \mathbf{v} \cdot \mathbf{a} > 0 reads speeding up on an interval; at a single instant where \mathbf{v} = \mathbf{0} the formula \displaystyle \frac{\mathbf{v} \cdot \mathbf{a}}{\lVert \mathbf{v} \rVert} does not apply. The signs of the components of \mathbf{a} alone say nothing about speeding up (Exercise 18).

Where this goes next

The same componentwise calculus works for \mathbf{r}(t) = \langle x(t), y(t), z(t) \rangle in space, for paths that climb as well as turn, and splitting acceleration into a part along the velocity and a part across it leads to curvature. Integrating the speed from a fixed start gives the arc length function, which lets a curve be traced at unit speed.

Exercises

Day 1: derivatives, velocity and speed

1. Find \mathbf{r}'(t) and \mathbf{r}''(t) for \mathbf{r}(t) = \langle \ln(t^2 + 1), \arctan t \rangle.
2. A particle moves with \mathbf{r}(t) = \langle 2t^3 - 15t^2 + 36t, t^2 - 4t \rangle, t \ge 0. Find the times and positions at which it is at rest, moving vertically, or moving horizontally.
3. For \displaystyle \mathbf{r}(t) = \langle 4\sqrt{t}, t^2 - 3t \rangle, find the position, velocity, acceleration and speed at t = 4, and the direction of motion as a unit vector and as an angle.
4. Seen from the side, Sade’s sailplane follows \displaystyle \mathbf{r}(t) = \langle 30t, 600 + 40\sin\frac{t}{5} \rangle, in meters and seconds, with y the altitude, as it rides a series of rising and sinking air bands. Find its speed, the largest and smallest speeds, and the times t \ge 0 when it moves horizontally. Is it ever moving vertically?
5. For \mathbf{r}(t) = \langle t^2 - 4t, 2t + 1 \rangle, find the speed as a function of t and its smallest value. Where is the particle then, and which way is it moving?

Day 1: reasoning

6. Use the limit definition to show that \mathbf{r}'(t) = \langle x'(t), y'(t) \rangle whenever x and y are differentiable at t.
7. Show that if \lVert \mathbf{r}(t) \rVert is constant on an interval, then \mathbf{r}(t) \cdot \mathbf{r}'(t) = 0 there. What does this say about motion on a circle centered at the origin?

Day 2: integrals and initial values

8. Find \displaystyle \int \langle 3\sqrt{t}, \frac{1}{1 + t^2} \rangle\,dt.
9. Evaluate \int_0^{\pi/2} \langle \cos t, \sin^2 t \rangle\,dt.
10. A particle has \mathbf{a}(t) = \langle 6t, -2 \rangle, with \mathbf{v}(1) = \langle 0, 2 \rangle and \mathbf{r}(1) = \langle 4, 3 \rangle. Find \mathbf{v}(t) and \mathbf{r}(t).
11. Amira’s air hockey puck, slowed by drag, has velocity \mathbf{v}(t) = \langle 0.9e^{-t/3}, 1.2e^{-t/3} \rangle feet per second and starts at \mathbf{r}(0) = \langle 0.5, 1 \rangle. Find \mathbf{r}(t), the point the puck approaches as t \to \infty, and the distance it travels for 0 \le t \le 3. Why does the distance equal the length of the displacement here?

Day 2: displacement and distance

12. For \displaystyle \mathbf{r}(t) = \langle 2t^2, \frac{4}{3}t^3 - t \rangle, 0 \le t \le 1, find the distance traveled and the length of the displacement.
13. For the sailplane of Exercise 4, find the distance flown during one full wave, 0 \le t \le 10\pi, to two decimal places, and compare it with the horizontal distance covered.
14. Amira’s puck leaves \langle 1, 1 \rangle with velocity \langle 2, 1.5 \rangle feet per second on a table whose side rail is the line x = 4. It bounces off the rail, which reverses the x component of its velocity and leaves the y component alone. Find its position at t = 3, its displacement, and the distance it traveled.

Day 2: speeding up and slowing down

15. For the particle of Exercise 5, find where it speeds up and where it slows down, for t \ge 0.
16. For the sailplane of Exercise 4, find where it speeds up and where it slows down during 0 < t < 10\pi, and say where on each wave the changes happen.

Day 2: reasoning

17. Prove that \left\lVert \int_a^b \mathbf{v}(t)\,dt \right\rVert \le \int_a^b \lVert \mathbf{v}(t) \rVert\,dt for continuous \mathbf{v}. (Hint: call the left integral \mathbf{D} and look at \mathbf{D} \cdot \mathbf{D}.)
18. Someone claims: “if both components of the acceleration are positive, the object is speeding up.” Refute it with \displaystyle \mathbf{r}(t) = \langle \frac{t^2}{2} - 3t, \frac{t^2}{2} + t \rangle at t = 0.
19. Show that a particle moves with constant speed on an interval exactly when \mathbf{v} \cdot \mathbf{a} = 0 throughout it. Then explain why Sade’s acceleration in Example 3 has constant length but her speed is not constant.