3.4 Motion in Space

Big idea

This section spends the machinery of the chapter. If $\mathbf{r}(t)$ is the position of a moving object at time $t$, then the derivative is its velocity and the second derivative is its acceleration, exactly as in one dimension. What changes is that velocity now has a direction as well as a size, so the single word “rate” splits into two ideas: speed, which is the magnitude of the velocity, and direction of travel, which is the unit tangent vector.

Reversing the process is just as important. Physical laws usually prescribe acceleration, not position, and integrating twice recovers the trajectory once you supply an initial velocity and an initial position. Projectile motion is the standard case: near the surface of the earth, gravity gives a constant downward acceleration and nothing else, so two integrations produce a position function whose horizontal part is linear and whose vertical part is quadratic. Every fact about range, flight time and maximum height falls out of that formula rather than being memorized separately.

The deeper idea in the section is the decomposition of acceleration. Acceleration can do two different jobs: change how fast you are going, or change the direction you are going. The first job is done by the component along the direction of travel, and the second by the component perpendicular to it. Writing acceleration in terms of the tangent and normal vectors from the previous section separates the two jobs cleanly, and the separation is physically real: a car speeding up in a straight line feels only the first, a car rounding a curve at steady speed feels only the second.

That second component is where curvature returns. The normal component equals the curvature times the square of the speed, which is why a tight turn and a fast turn both press you against the door, and why the effect grows with the square of the speed rather than in proportion to it.

Decoder

The acceleration of a moving object always lies in the plane spanned by the unit tangent and the principal unit normal, and its tangential component is the rate of change of speed while its normal component is the curvature times the speed squared.

The first clause is a constraint worth trusting. Acceleration has no binormal component, ever. If you compute an acceleration and decompose it, whatever is left after removing the tangential and normal parts must be zero, and if it is not, something is wrong.

The second clause explains what each part does. The tangential part is a signed number, positive when speeding up and negative when slowing down, and it is literally the derivative of the speed function. The normal part is never negative and vanishes only where the curvature is zero or the object is at rest; it always points toward the concave side of the path. So an object moving at constant speed has zero tangential acceleration even though its acceleration vector is not zero, which is the standard case of uniform circular motion.

The classic mistake is treating speed as though it were a component of velocity. Speed is the length $\|\mathbf{v}(t)\|$, a nonnegative scalar, and the derivative of speed is the tangential component of acceleration, not the magnitude of acceleration. Those two numbers agree only when the path is straight. A useful arithmetic check exists because the tangential and normal directions are perpendicular: the squares of the two components must add up to $\|\mathbf{a}\|^2$.

Definitions and results

Position, velocity, speed. For a twice-differentiable position function $\mathbf{r}(t)$,

$$ \mathbf{v}(t) = \mathbf{r}'(t), \qquad \text{speed} = \|\mathbf{v}(t)\|, \qquad \mathbf{a}(t) = \mathbf{v}'(t) = \mathbf{r}''(t) $$

The velocity is tangent to the path; the speed is the rate at which arc length accumulates, so $\|\mathbf{v}(t)\| = ds/dt$.

Distance travelled. Integrating the speed over a time interval gives the length of the path covered, which is generally larger than the straight-line distance between the endpoints.

Recovering motion by integration. Given acceleration and the two initial conditions,

$$ \mathbf{v}(t) = \mathbf{v}(t_0) + \int_{t_0}^{t} \mathbf{a}(u)\,du, \qquad \mathbf{r}(t) = \mathbf{r}(t_0) + \int_{t_0}^{t} \mathbf{v}(u)\,du $$

Each integration introduces a constant vector, and each initial condition removes one.

Projectile motion. With gravity the only force, $\mathbf{a}(t) = \langle 0, -g \rangle$. Launching from the origin with speed $v_0$ at angle $\alpha$ above the horizontal gives

$$ \mathbf{r}(t) = \left\langle (v_0\cos\alpha)\,t, \; (v_0\sin\alpha)\,t - \tfrac{1}{2}gt^2 \right\rangle $$

Eliminating $t$ shows the path is a parabola.

Flight time, range and maximum height. From that model, the object returns to launch height at $t = 2v_0\sin\alpha / g$, giving range $v_0^2 \sin(2\alpha)/g$ and maximum height $v_0^2\sin^2\alpha/(2g)$ reached at half the flight time. The range is largest at $\alpha = 45$ degrees, since that is where $\sin(2\alpha)$ equals $1$.

Tangential component of acceleration.

$$ a_{\mathbf{T}} = \frac{d}{dt}\|\mathbf{v}(t)\| = \frac{\mathbf{v}(t) \cdot \mathbf{a}(t)}{\|\mathbf{v}(t)\|} $$

It is the signed rate of change of speed, and its sign is the sign of $\mathbf{v} \cdot \mathbf{a}$.

Normal component of acceleration.

$$ a_{\mathbf{N}} = \frac{\|\mathbf{v}(t) \times \mathbf{a}(t)\|}{\|\mathbf{v}(t)\|} = \kappa \, \|\mathbf{v}(t)\|^2 $$

It is never negative, and the second expression shows it grows with the square of the speed for a fixed path.

The decomposition. Combining the two,

$$ \mathbf{a}(t) = a_{\mathbf{T}}\,\mathbf{T}(t) + a_{\mathbf{N}}\,\mathbf{N}(t), \qquad a_{\mathbf{T}}^{\,2} + a_{\mathbf{N}}^{\,2} = \|\mathbf{a}(t)\|^2 $$

The last identity holds because $\mathbf{T}$ and $\mathbf{N}$ are perpendicular unit vectors, and it is the fastest check on a computed decomposition. It also gives an alternative route to $a_{\mathbf{N}}$ as $\sqrt{\|\mathbf{a}\|^2 - a_{\mathbf{T}}^{\,2}}$.

Worked examples

Velocity, speed and acceleration

Let $\mathbf{r}(t) = \left\langle 2t, \; t^2, \; \tfrac{1}{3}t^3 \right\rangle$. Differentiating twice,

$$ \mathbf{v}(t) = \langle 2, \; 2t, \; t^2 \rangle, \qquad \mathbf{a}(t) = \langle 0, \; 2, \; 2t \rangle $$

The speed simplifies unusually well:

$$ \|\mathbf{v}(t)\| = \sqrt{4 + 4t^2 + t^4} = \sqrt{(t^2+2)^2} = t^2 + 2 $$

which is positive for every $t$, so the curve is smooth. At $t = 1$ the velocity is $\langle 2,2,1 \rangle$ with speed $3$, matching $1 + 2$, and the acceleration is $\langle 0,2,2 \rangle$. Because the speed is a clean formula, the distance travelled from $t = 0$ to $t = 3$ is also clean:

$$ \int_0^3 (t^2 + 2)\,dt = \left[ \tfrac{1}{3}t^3 + 2t \right]_0^3 = 9 + 6 = 15 $$

Splitting the acceleration

Decompose the acceleration of the same motion at $t = 1$.

The tangential component is

$$ a_{\mathbf{T}} = \frac{\mathbf{v} \cdot \mathbf{a}}{\|\mathbf{v}\|} = \frac{0 + 4 + 2}{3} = 2 $$

Check it against the other formula: the speed is $t^2 + 2$, whose derivative is $2t$, and at $t = 1$ that is $2$. The two agree.

For the normal component,

$$ \mathbf{v} \times \mathbf{a} = \langle 2,2,1 \rangle \times \langle 0,2,2 \rangle = \langle 4 - 2, \; 0 - 4, \; 4 - 0 \rangle = \langle 2, -4, 4 \rangle $$

with magnitude $\sqrt{4 + 16 + 16} = 6$, so $a_{\mathbf{N}} = 6/3 = 2$.

Now the check: $a_{\mathbf{T}}^{\,2} + a_{\mathbf{N}}^{\,2} = 4 + 4 = 8$, and $\|\mathbf{a}\|^2 = 0 + 4 + 4 = 8$. The decomposition is consistent, and the curvature at that instant follows from $a_{\mathbf{N}} = \kappa \|\mathbf{v}\|^2$, giving $\kappa = 2/9$.

A projectile

A ball is launched from ground level at $49$ meters per second at $30$ degrees above the horizontal. Take $g = 9.8$ meters per second squared and ignore air resistance.

The initial velocity components are $v_0\cos 30^\circ = \tfrac{49\sqrt{3}}{2}$ and $v_0 \sin 30^\circ = 24.5$, so

$$ \mathbf{r}(t) = \left\langle \tfrac{49\sqrt{3}}{2}\,t, \; 24.5\,t - 4.9\,t^2 \right\rangle $$

The ball returns to the ground when $24.5t - 4.9t^2 = 0$, that is at $t = 24.5/4.9 = 5$ seconds. The range is the horizontal distance at that moment:

$$ \tfrac{49\sqrt{3}}{2}(5) = 122.5\sqrt{3} \approx 212.2 \text{ meters} $$

The vertical velocity $24.5 - 9.8t$ vanishes at $t = 2.5$ seconds, half the flight time as expected, and the height there is $24.5(2.5) - 4.9(6.25) = 61.25 - 30.625 = 30.625$ meters. The general formula agrees: $v_0^2\sin^2\alpha/(2g) = 600.25/19.6 = 30.625$. Launching at $45$ degrees instead would give a range of $49^2/9.8 = 245$ meters, the largest possible at this speed.

From acceleration back to position

An object has acceleration $\mathbf{a}(t) = \langle 0, 2, 6t \rangle$, initial velocity $\mathbf{v}(0) = \langle 1, 0, -1 \rangle$ and initial position $\mathbf{r}(0) = \langle 0,0,0 \rangle$.

Antidifferentiate once and fit the velocity condition:

$$ \mathbf{v}(t) = \langle 1, \; 2t, \; 3t^2 - 1 \rangle $$

Antidifferentiate again and fit the position condition:

$$ \mathbf{r}(t) = \langle t, \; t^2, \; t^3 - t \rangle $$

Check by differentiating back: $\mathbf{r}'(t) = \langle 1, 2t, 3t^2 - 1 \rangle$ and $\mathbf{r}''(t) = \langle 0, 2, 6t \rangle$, and both initial conditions hold at $t = 0$. Notice that the third component of velocity is negative for $t < 1/\sqrt{3}$, so the object initially descends before turning around, which the position function confirms: $t^3 - t$ is negative on $(0,1)$ and returns to zero at $t = 1$.

Practice

These problems cover the whole section: differentiating a position function to get velocity, speed and acceleration, integrating acceleration with initial conditions, computing distance travelled, working with the projectile model, and splitting acceleration into its tangential and normal components.

Quiz

Five items on velocity, speed and acceleration, recovering position by integration, projectile motion, and the tangential and normal components of acceleration.