Chapter 3, Problem 2RWDT

To determine

To show : Using the chain rule, to explain one way to find locations where the tangent line to the curve is horizontal.

Explanation of Solution

Given information :

The given equations:

  $\begin{array}{l}x=x\left(t\right),\\ y=y\left(t\right)\end{array}$

Where $x\text{and}y$ are functions of $t$ .

Formula used:

The plane curve defined by the parametric equations $x=x\left(t\right),y=y\left(t\right)$

Also suppose that $x\text{'}\left(t\right)\text{and}y\text{'}\left(t\right)$ exist and $x\text{'}\left(t\right)\ne 0$ then the derivative is:

  $\begin{array}{c}\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\\ =\frac{y\text{'}\left(t\right)}{x\text{'}\left(t\right)}\end{array}$

Proof:

This can be proven by using chain rule. Assume that the parameter $t$ can be eliminated, a differentiable function $y=F\left(x\right)$ .

Then, $y\left(t\right)=F\left(x\left(t\right)\right)$

Differentiating using Chain Rule, then rearranging:

  $\begin{array}{c}y\text{'}\left(t\right)=F\text{'}\left(x\left(t\right)\right)\cdot x\text{'}\left(t\right)\\ F\text{'}\left(x\left(t\right)\right)=\frac{y\text{'}\left(t\right)}{x\text{'}\left(t\right)}\end{array}$

Where, $F\text{'}\left(x\left(t\right)\right)=\frac{dy}{dx}$ which proves the theorem.

To find the points at which the tangent line is horizontal.

One way to find where the slope of the function is $0$ because a horizontal line's slope is $0$ .

This can be used to calculate derivatives of plane curves, as well as critical points.

A critical point of a differentiable function $y=f\left(x\right)$ is at any point $x=x_0$ such that either $f\text{'}\left(x_0\right)=0\text{or}f\text{'}\left(x_0\right)$ does not exist.

  $F\text{'}\left(x\left(t\right)\right)=\frac{dy}{dx}$ gives a formula for the slope of a tangent line to a curve defined parametrically regardless of whether the curve can be described by a function $y=f\left(x\right)$ or not.