Derivative of parametric equations

Question

Consider the parametric equations $x=x(t), y=y(t)$.

I'm told that the derivative can be expressed as $$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$ provided $dx/dt\neq 0$.

However, in the derivation of this, it was assumed that we can express $y$ as a function of $x$, i.e. $y=f(x),$ for some $f$.

In my experience with parametric equations, it is not always possible to write $y=f(x)$, so how can we make this assumption?

Answer 1

Consider the case of the cycloid

$$\begin{cases}x=t-\sin t,\\y=1-\cos t\end{cases}.$$

You cannot solve the transcendental equation $x=t-\sin t$ analytically for $t$, so that you cannot express $y=f(x)$ with a closed-form expression. Anyway,

$$\frac{dy}{dx}=\frac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}=\frac{1-\cos t}{\sin t}$$ allows you to compute the slope at any point of the curve.

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Addendum:

Notice that we can eliminate $t$ by

$$x=\arccos(1-y)-\sin(\arccos(1-y))$$

which gives the inverse of $f$ (but one has to consider several branches of the arc cosine).

Answer 2

The definition of $\frac{dy}{dx}$ is more general; it's just saying that if you change $x$ by a small amount, then you change $\frac{dy}{dx}$ times this small amount in the $y$-coordinate. If you have some strange curve which isn't a function in $x$, you can still define a "tangent" in this way near every point, and this gives you $\frac{dy}{dx}$.