Parametric differentiation: both coordinates ride on t

Usually a curve links $y$ directly to $x$. A parametric curve gives $x$ and $y$ separately, each in terms of a third letter called the parameter, usually $t$:

$x = f(t), \qquad y = g(t)$

Think of $t$ as time. At each moment $t$, the formulas tell you where a moving dot is. The path the dot traces out is the curve.

Example: $x = 4\cos t$, $y = 3\sin t$. At $t = 0$ the dot is at $(4, 0)$. At $t = \tfrac{\pi}{2}$ it's at $(0, 3)$. As $t$ keeps going, it traces an oval (an ellipse).

1. Differentiate $x$ with respect to $t$ to get $\dfrac{dx}{dt}$.
2. Differentiate $y$ with respect to $t$ to get $\dfrac{dy}{dt}$.
3. Divide: $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$. This is the gradient in terms of $t$.
4. Put in the given $t$ to get a number.
5. Find the point by putting the same $t$ into both $x$ and $y$.
6. Write the line with $y - y_1 = m(x - x_1)$. For a normal, use gradient $-\dfrac{1}{m}$ instead.

Students often find $x$ in step 5 and forget $y$. You need both.