
We discuss derivatives of parametrically defined curves.

### Derivatives

Recall from differential calculus that the tangent line provides the best linear approximation to a curve at a given point. Thus, we are often interested in calculating the tangent line. If the curve can be expressed as a function of either $x$ or $y$ then the slope of the tangent line is obtained by taking the derivative at the given point. If the curve has a Cartesian description $F(x,y)=0$, then even if we can’t express it as a function, we can still use implicit differentiation to find the slope of the tangent line.

Suppose we have a curve given parametrically. Remember that for some parametric curves would be difficult or impossible to find Cartesian forms. We would like to be able to find the slope of the tangent line directly from the parametric description without having to convert to a Cartesian form.

Suppose we have a curve $C$ that is traced out by the parametric equations

In order to find the slope of the tangent line we need to compute the rate of change $\dd [y]{x}$ of $y$ with respect to $x$,

We use the chain rule to write

Recall from the Inverse Function Theorem, we have $\dd [t]{x}=\frac {1}{\dd [x]{t}}$. This means we can rewrite $\dd [y]{x}$ as

Notice that this formula allows us to calculate $\dd [y]{x}$ directly from our parametric description of $C$.

The definition leaves two special cases to consider. When the tangent line is horizontal, the normal line is undefined by the above definition as $y'(t_0)=0$. Likewise, when the normal line is horizontal, the tangent line is undefined. It seems reasonable that these lines be defined (one can draw a line tangent to the “right side” of a circle, for instance), so we add the following to the above definition.

• If the tangent line at $t=t_0$ has a slope of $0$, the normal line to $C$ at $t=t_0$ is the vertical line $x=x(t_0)$.
• If the normal line at $t=t_0$ has a slope of $0$, the tangent line to $C$ at $t=t_0$ is the line $x=x(t_0)$.

Let’s look at some examples.