Calculus/Parametric Differentiation

Taking Derivatives of Parametric Systems

Just as we are able to differentiate functions of $x$ , we are able to differentiate $x$ and $y$ , which are functions of $t$ . Consider:

We would find the derivative of $x$ with respect to $t$ , and the derivative of $y$ with respect to $t$ :

In general, we say that if

then:

It's that simple.

This process works for any amount of variables.

Slope of Parametric Equations

In the above process, $x^{\prime }$ has told us only the rate at which $x$ is changing, not the rate for $y$ , and vice versa. Neither is the slope.

In order to find the slope, we need something of the form $\frac{dy}{dx}$ .

We can discover a way to do this by simple algebraic manipulation:

$\frac{y^{\prime }}{x^{\prime }}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{dy}{dx}$

So, for the example in section 1, the slope at any time $t$ :

$\frac{1}{\cos (t)}=\sec (t)$

In order to find a vertical tangent line, set the horizontal change, or $x^{\prime }$ , equal to 0 and solve.

In order to find a horizontal tangent line, set the vertical change, or $y^{\prime }$ , equal to 0 and solve.

If there is a time when both $x^{\prime },y^{\prime }$ are 0, that point is called a singular point.

Concavity of Parametric Equations

Solving for the second derivative of a parametric equation can be more complex than it may seem at first glance.

When you have take the derivative of $\frac{dy}{dx}$ in terms of $t$ , you are left with $\frac{\frac{d^{2}y}{dx}}{dt}$ :

$\frac{d}{dt}\left[\frac{dy}{dx}\right]=\frac{\frac{d^{2}y}{dx}}{dt}$ .

By multiplying this expression by $\frac{dt}{dx}$ , we are able to solve for the second derivative of the parametric equation:

$\frac{\frac{d^{2}y}{dx}}{dt}\times \frac{dt}{dx}=\frac{d^{2}y}{d{x}^2}$ .

Thus, the concavity of a parametric equation can be described as:

$\frac{d}{dt}\left[\frac{dy}{dx}\right]\times \frac{dt}{dx}$

So for the example in sections 1 and 2, the concavity at any time $t$ :

$\frac{d}{dt}[\csc (t)]\times \cos (t)=-{\csc }^2(t)\times \cos (t)$