Does derivative formula work for all parametric equations

In summary, the formula for the derivative of 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)}##. This formula is valid even when ##y## is not a function of ##x##, as long as the relation between ##x## and ##y## can be parametrized. However, if there is no formal relation between two variables, the derivative does not exist. The inverse function theorem states that if a derivative is zero everywhere, there is no functional mapping between the variables.
  • #1
Happiness
679
30
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.
 
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  • #2
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
fresh_42 said:
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.
 
  • #4
Happiness said:
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.##
Your example becomes
$$0 = \frac{d}{dx} 1 = \frac{d{x^2}}{d{x}} + \frac{d{y^2}}{d{x}} = 2x + 2y \frac{d{y}}{d{x}} \text{ and therefore } x+ y \cdot \frac{d{y}}{d{x}} = 0$$
If ##y## and ##x## were independent then ##x## would be a constant function and not a coordinate of a circle.
 
  • #5
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.
 

1. What is a derivative formula?

A derivative formula is a mathematical expression that calculates the rate of change of a function at a specific point. It is used to find the slope of a curve at a given point.

2. How is the derivative formula used for parametric equations?

The derivative formula can be applied to parametric equations by finding the first derivative of both the x and y components of the equation with respect to the independent variable. This will give the slope of the curve at any given point.

3. Does the derivative formula work for all parametric equations?

Yes, the derivative formula can be used for all parametric equations as long as the equations are differentiable (have a well-defined slope) and the independent variable is continuous.

4. What is the significance of using the derivative formula for parametric equations?

Using the derivative formula for parametric equations allows us to find the instantaneous rate of change or slope of a curve at any given point. This is useful in various fields such as physics, engineering, and economics.

5. Are there any limitations to using the derivative formula for parametric equations?

One limitation is that the derivative formula only gives the slope of the curve at a specific point, so it cannot provide information about the overall shape of the curve. Additionally, the derivative may not exist at certain points or may be undefined for certain parametric equations.

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