A parametric curve gives both coordinates as functions of a third variable: $x = x(t)$, $y = y(t)$, like the position of a moving point at time $t$. In eliminating the parameter you turned such curves into one equation in $x$ and $y$. Often that is messy or impossible, so here you do calculus directly in $t$, using the chain rule and the idea behind arc length.
Why it works
Near a point where $\frac{dx}{dt} \ne 0$, the curve is the graph of some function $y(x)$, and $y(t) = y(x(t))$. The chain rule says $\frac{dy}{dt} = \frac{dy}{dx}\cdot\frac{dx}{dt}$, so
$$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}.$$
Rise over run: in a short time $\Delta t$ the point moves about $\frac{dx}{dt}\Delta t$ across and $\frac{dy}{dt}\Delta t$ up, and the $\Delta t$ cancels. The distance moved is $$\begin{aligned} &\sqrt{(\Delta x)^{2} + (\Delta y)^{2}} \\ &\approx \sqrt{x'(t)^{2} + y'(t)^{2}}\,\Delta t: \end{aligned}$$ speed times time.
Key idea
$$\frac{dy}{dx} = \frac{dy/dt}{dx/dt},$$ $$L = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^{2} + \left(\frac{dy}{dt}\right)^{2}}\,dt.$$
Horizontal tangent: $\frac{dy}{dt} = 0$ with $\frac{dx}{dt} \ne 0$. Vertical tangent: $\frac{dx}{dt} = 0$ with $\frac{dy}{dt} \ne 0$.
Worked example: A slope
For $x = t^{2}$, $y = t^{3} - 3t$, find $\frac{dy}{dx}$ at $t = 2$.
- $\frac{dx}{dt} = 2t$ and $\frac{dy}{dt} = 3t^{2} - 3$.
- $\frac{dy}{dx} = \frac{3t^{2} - 3}{2t}$. At $t = 2$: $\frac{9}{4}$.
- Horizontal tangents where $3t^{2} - 3 = 0$: $t = \pm 1$. A vertical tangent at $t = 0$, where $\frac{dx}{dt} = 0$ but $\frac{dy}{dt} = -3$.
Answer: $\frac{9}{4}$
Common mistake
Dividing the other way, $\frac{dx/dt}{dy/dt}$. That gives $\frac{dx}{dy}$, the reciprocal of the slope. Test with a curve you know: $x = t$, $y = 3t$ is the line $y = 3x$, slope 3. The correct formula gives $\frac{3}{1} = 3$; the flipped one gives $\frac{1}{3}$. Slope is always rise ($y$) over run ($x$).
Worked example: A tangent line
Find the tangent line to $x = t^{2}$, $y = t^{3} - 3t$ at $t = 2$.
- The point: $x = 4$, $y = 8 - 6 = 2$.
- The slope from before: $\frac{9}{4}$.
- $y - 2 = \frac{9}{4}(x - 4)$, that is $y = \frac{9}{4}x - 7$.
Answer: $y = \frac{9}{4}x - 7$
Worked example: Arc length
Find the length of $x = t^{2}$, $y = \frac{2}{3}t^{3}$ for $0 \le t \le \sqrt{3}$.
- $\frac{dx}{dt} = 2t$, $\frac{dy}{dt} = 2t^{2}$, so the speed is $\sqrt{4t^{2} + 4t^{4}} = 2t\sqrt{1 + t^{2}}$ for $t \ge 0$.
- $$\begin{aligned} L &= \int_0^{\sqrt{3}} 2t\sqrt{1 + t^{2}}\,dt \\ &= \frac{2}{3}\left[(1 + t^{2})^{3/2}\right]_0^{\sqrt{3}}. \end{aligned}$$
- $= \frac{2}{3}(8 - 1) = \frac{14}{3}$.
Answer: $\frac{14}{3}$
Second derivative: $\frac{d^{2}y}{dx^{2}}$ is the derivative of $\frac{dy}{dx}$ with respect to $x$, so differentiate $\frac{dy}{dx}$ (a function of $t$) with respect to $t$ and divide by $\frac{dx}{dt}$ again. For the first curve, $\frac{d}{dt}\left(\frac{3t^{2} - 3}{2t}\right) = \frac{3}{2} + \frac{3}{2t^{2}}$, and dividing by $2t$ gives $\frac{3(t^{2} + 1)}{4t^{3}}$.