One bracket, two independent components BC only

Write $\mathbf{r}(t) = \langle x(t),\, y(t)\rangle$. Then

$$\mathbf{r}'(t) = \langle x'(t),\, y'(t)\rangle, \qquad \mathbf{r}''(t) = \langle x''(t),\, y''(t)\rangle.$$

That is the whole rule. The components do not interact, so the product rule, the quotient rule and the chain rule apply inside each slot exactly as they always have, and nothing new needs learning.

The errors CA5 names all come from letting the components touch each other:

1. Differentiating only the first. Both slots move.
2. Combining before differentiating. Taking $\sqrt{x^{2} + y^{2}}$ and differentiating that gives the rate at which the distance from the origin changes, which is a different quantity from the velocity and is not even a vector.
3. Swapping the slots on reassembly. Mechanical, common under time pressure, and undetectable in the algebra.

For $\mathbf{r}(t) = \langle 3t^{2},\, 2t^{3}\rangle$: $\mathbf{r}'(t) = \langle 6t,\, 6t^{2}\rangle$ and $\mathbf{r}''(t) = \langle 6,\, 12t\rangle$. At $t = 1$ those are $\langle 6, 6\rangle$ and $\langle 6, 12\rangle$.

Three quantities, three different types:

1. Position $\mathbf{r}(t)$: a vector. Where the particle is.
2. Velocity $\mathbf{v}(t) = \mathbf{r}'(t)$: a vector. How fast and in what direction.
3. Speed $|\mathbf{v}(t)|$: a scalar. How fast, with the direction discarded.

The magnitude is

$$|\mathbf{v}| = \sqrt{\left(\frac{dx}{dt}\right)^{2} + \left(\frac{dy}{dt}\right)^{2}},$$

and that expression should look familiar: it is 9.3's arc length integrand exactly. That is not a coincidence, and it is worth stating out loud, because it explains why distance travelled is the integral of speed, which is the whole content of 9.6.

At $t = 1$ the running example has $\mathbf{v} = \langle 6, 6\rangle$ and speed $\sqrt{36 + 36} = 6\sqrt{2} \approx 8.485$. Three wrong answers circulate, all CA6: reporting the ordered pair $\langle 6, 6\rangle$ when a number was asked for, adding the components to get $12$, and taking $|\mathbf{r}(1)| = |\langle 3, 2\rangle| = \sqrt{13}$, which measures distance from the origin rather than speed.

A type check catches all three. If the question says "how fast", the answer is one number. If it says "in what direction" or asks for the velocity, the answer has two components. An ordered pair offered as a speed is wrong before its entries are examined.

Acceleration is $\mathbf{a}(t) = \mathbf{v}'(t) = \mathbf{r}''(t)$, again componentwise. It is a vector, and it points in the direction the velocity is changing, which is generally not the direction of motion.

The cleanest illustration is uniform circular motion. For $\mathbf{r}(t) = \langle \cos t,\, \sin t\rangle$: $\mathbf{v} = \langle -\sin t,\, \cos t\rangle$ with speed $\sqrt{\sin^{2}t + \cos^{2}t} = 1$, constant, and $\mathbf{a} = \langle -\cos t,\, -\sin t\rangle = -\mathbf{r}$, pointing straight at the origin.

So here is a particle with constant speed and nonzero acceleration, which sounds contradictory only if speed and velocity have been conflated. The speed never changes; the direction does, and acceleration is what changes it.

One further caution. $|\mathbf{a}|$ is not the derivative of $|\mathbf{v}|$. The first is the size of the acceleration vector; the second is the rate at which the speed changes. In the circular example $|\mathbf{a}| = 1$ while $\frac{d}{dt}|\mathbf{v}| = 0$. Taking a magnitude and differentiating are operations that do not commute, and swapping their order is the same class of error as CA5's "combine before differentiating".

Nothing about the bracket changes how the pieces inside it are differentiated. Each of these is an ordinary Unit 2 or Unit 3 problem happening in one slot:

1. Chain rule: $\frac{d}{dt}\langle \sin(3t),\, e^{t^{2}}\rangle = \langle 3\cos(3t),\, 2te^{t^{2}}\rangle$.
2. Product rule: $\frac{d}{dt}\langle t^{2}\ln t,\, t\cos t\rangle = \langle 2t\ln t + t,\, \cos t - t\sin t\rangle$.
3. A constant component: $\frac{d}{dt}\langle 5,\, t^{3}\rangle = \langle 0,\, 3t^{2}\rangle$, and the particle moves vertically along the line $x = 5$.

Two facts about the vector as a whole are worth having, since both appear on the exam:

1. $\mathbf{v} = \mathbf{0}$ means the particle has stopped, and that requires both components to vanish at the same $t$. One zero component means motion parallel to an axis, not rest.
2. The velocity vector is tangent to the path. So $\frac{dy}{dx} = \frac{y'(t)}{x'(t)}$ is the slope of that tangent, which is 9.1's quotient recovered from the vector.