# I Does derivative formula work for all parametric equations

1. Jun 21, 2016

### Happiness

The derivative for the parametric equations $x=f(t)$ and $y=g(t)$ is given by

$\frac{dy}{dx}=\frac{\Big(\frac{dy}{dt}\Big)}{\Big(\frac{dx}{dt}\Big)}$

The proof of the above formula requires that $y$ be a function of $x$, as seen in http://tutorial.math.lamar.edu/Classes/CalcII/ParaTangent.aspx

But somehow the formula works even if $y$ is not a function of $x$, for example, it works for the unit circle $x^2+y^2=1$, whose parametric equations are $x=\cos t$ and $y=\sin t$.

Does the formula always work as long as the relation between $x$ and $y$ can be parametrised even if $y$ is not a function of $x$? If not, what is a counterexample? For what parametric equations does the formula fail?

Note: A function is a relation that passes the vertical-line test.

Edit: We exclude cases where the curve intersects itself because the derivative doesn't exist at the intersecting points.

Last edited: Jun 21, 2016
2. Jun 21, 2016

### Staff: Mentor

If both, $x=x(t)$ and $y=y(t)$ depend on a parameter $t$, then they depend on each other. In the simplest case it's $y(t) = y(x^{-1}(t))$. In general one might need some local inversion properties.
Your example is even without a parametrization no counterexample, since $x^2+y^2=1$ already establishes a dependency $y(x)$.

It won't be possible, if you say $x=x(t)$ and $y=y(s)$ and there is no (formal) relation between $s$ and $t.$ In this case $\frac{\partial{y}}{\partial{x}}$ wouldn't make sense.

3. Jun 21, 2016

### Happiness

I thought $y(x)$ means $y$ is a function of $x$? But for $x^2+y^2=1$, $y$ is not a function of $x$. The unit circle fails the vertical-line test.

4. Jun 21, 2016

### Staff: Mentor

You could resolve it by separately consider the two branches $y = ± \sqrt{1-x^2}$.
Whether you can explicitly write $y=y(x)$ or not doesn't change the dependency. It makes the difference between $\frac{dy}{dx} = 0$ and $\frac{dy}{dx} ≠ 0.$
$$0 = \frac{d}{dx} 1 = \frac{d{x^2}}{d{x}} + \frac{d{y^2}}{d{x}} = 2x + 2y \frac{d{y}}{d{x}} \text{ and therefore } x+ y \cdot \frac{d{y}}{d{x}} = 0$$
If $y$ and $x$ were independent then $x$ would be a constant function and not a coordinate of a circle.

5. Jun 22, 2016

### chiro

If one variable has a zero derivative with respect to another variable then no inverse function in that neighborhood exists - and this is part of a theorem known as the inverse function theorem which looks at determinants of matrices in higher dimensions.

If two things are related then there is a functional mapping between them of some sort (you may have to "unwind" the mapping so that you don't have branch cuts and you maintain an actual one to one functional correspondence).

If a derivative is zero everywhere then no relation between the two variables exists at all.