# Does derivative formula work for all parametric equations

• I
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:

fresh_42
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2021 Award
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.

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.

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.

fresh_42
Mentor
2021 Award
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.
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.

chiro