A slope built from two derivatives BC only

A parametric curve gives $x$ and $y$ each as a function of $t$. Along the curve, $y$ still depends on $x$, so the chain rule applies:

$$\frac{dy}{dt} = \frac{dy}{dx}\cdot\frac{dx}{dt}.$$

Solve for the quantity you want and the formula is forced rather than remembered:

$$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}, \qquad \frac{dx}{dt} \neq 0.$$

Deriving it this way is worth the thirty seconds, because the inverted version $\frac{dx/dt}{dy/dt}$ is exactly as memorable and completely wrong. The chain rule puts $\frac{dy}{dt}$ on top because $\frac{dy}{dt}$ is what you started from.

For $x = t^{2} + 1$ and $y = t^{3} - 3t$, the pieces are $\frac{dx}{dt} = 2t$ and $\frac{dy}{dt} = 3t^{2} - 3$, so

$$\frac{dy}{dx} = \frac{3t^{2} - 3}{2t}.$$

Note what this is a function of: $t$, not $x$. To find the slope at a point you need the parameter value there, which is why parametric problems almost always hand you a $t$ rather than an $x$.

The quotient is zero when its numerator is and undefined when its denominator is, and each says something geometric.

1. Horizontal tangent: $\frac{dy}{dt} = 0$ and $\frac{dx}{dt} \neq 0$. The particle is moving, but not vertically.
2. Vertical tangent: $\frac{dx}{dt} = 0$ and $\frac{dy}{dt} \neq 0$. The slope is undefined, which is a feature of the curve rather than a failure of the arithmetic.
3. Both zero: the formula says nothing. The particle has momentarily stopped, and the curve may have a cusp there. This needs a limit, not a substitution.

In the running example, $3t^{2} - 3 = 0$ at $t = \pm 1$, and $\frac{dx}{dt}$ is $\mp 2$ there, so both give horizontal tangents, at $(2, 2)$ and $(2, -2)$. And $2t = 0$ at $t = 0$, where $\frac{dy}{dt} = -3 \neq 0$, so there is a vertical tangent at $(1, 0)$.

The habit that prevents the most common slip: set the numerator and the denominator to zero as two separate equations. Solving the whole quotient at once loses the distinction, and an undefined slope reported as "no tangent" throws away a real feature of the curve.

A parametrisation carries more information than the set of points it traces. It carries where the motion starts, which direction it goes, and how many times it covers each point.

Follow the running example from $t = -2$ to $t = 2$. The particle starts at $(5, -2)$, sweeps up and left through $(4, 0)$, reaches the top of the loop at $(2, 2)$, turns back through the leftmost point $(1, 0)$, drops to $(2, -2)$, returns through $(4, 0)$ a second time, and finishes at $(5, 2)$. The loop is traced counterclockwise, and the point $(4, 0)$ is visited twice, at $t = -\sqrt{3}$ and $t = \sqrt{3}$.

Three consequences that show up on the exam:

1. A self-intersection has two parameter values. The curve crosses itself; the particle does not collide with itself, because it is at the two visits at different times.
2. The slope at a crossing is not single-valued. Each visit has its own $\frac{dy}{dx}$, so "the slope at $(4, 0)$" is an incomplete question. Here the two are $\frac{6}{-2\sqrt{3}}$ and $\frac{6}{2\sqrt{3}}$, which differ in sign.
3. The interval may not cover the curve once. Always check the endpoints and whether the motion doubles back.

It is often possible to solve for $t$ and produce a rectangular equation. Doing so is legitimate and it throws information away.

Take $x = \cos t$, $y = \sin t$ on $[0, 2\pi]$ and $x = \cos t$, $y = -\sin t$ on the same interval. Both eliminate to $x^{2} + y^{2} = 1$. They are the same curve and opposite motions: the first counterclockwise, the second clockwise. No rectangular equation can tell them apart, because a rectangular equation describes a set of points and says nothing about time.

Two more losses worth naming:

1. The domain. $x = t^{2}$, $y = t$ eliminates to $x = y^{2}$, but the parametrisation on $t \in [0, 2]$ traces only the upper half from $(0,0)$ to $(4,2)$. The rectangular equation describes the whole sideways parabola.
2. The speed. $x = t$, $y = t$ and $x = t^{3}$, $y = t^{3}$ trace the same line, one at constant speed and one slowing to a stop at the origin and accelerating away.

So a question about the curve may be answered after elimination, and a question about the motion may not. Deciding which one is being asked is the first step, and it is the step CA4 names.