About the slope of a parametric curve at a point. (James Stewart Calculus 8th Edition.)

Question

I am reading "Calculus 8th Edition" by James Stewart.

Suppose $f$ and $g$ are differentiable functions and we want to find the tangent line at a point on the parametric curve $x=f(t),y=g(t)$, where $y$ is also a differentiable function of $x$. Then the Chain Rule gives $$\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}$$ If $\frac{dx}{dt} \neq 0$, we can solve for $\frac{dy}{dx}$: $$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$

What is the differentiable function $y$ of $x$?
We name the function $\phi$.

Consider $x = t^2$ and $y = t^3-3t$.

If $x_0 = 0$, then $x_0 = t^2$ has a unique solution $t_0 = 0$.
And $y_0 = t_0^3 - 3t_0 = 0$.
So, $\phi(0) = 0$.

If $x_0 \in (0, +\infty)$, then, $x_0 = t^2$ has two solutions $t_0 = \pm\sqrt{x_0}$. And $y_0 = t_0^3 - 3t_0 = \pm{x_0^{\frac{3}{2}} \mp 3 x_0^{\frac{1}{2}}}$.
Is $\phi(x_0) = {x_0^{\frac{3}{2}} - 3 x_0^{\frac{1}{2}}}$?
Is $\phi(x_0) = -{x_0^{\frac{3}{2}} + 3 x_0^{\frac{1}{2}}}$?

Can $\phi(x)$ be a differentiable function if we choose the values of $\phi$ appropriately?

Answer 1

If you draw a curve from left to right (so that it passes the vertical line test) in the $xy$-plane, then you have defined a function. The curve of your parametric equations doesn't satisfy the vertical line test if you let $t$ range over all the real numbers, but if you restrict $t$ in various ways, you can define a function. In particular, if you insist that $t>0$ the graph of $\phi$ is the graph of a function. This choice forces $t=\sqrt{x}$. You can also choose $t<0$, which gives a different $\phi.$

This is an interesting curve because at the point where $t=\sqrt{3}$, there are two different tangent lines.

Answer 2

A curve does not have to be the graph of a function in order to have a tangent line at some given point. One possible interpretation of $\frac{dy}{dx}$ is that it is simply a notation for the slope of that line. (Provided that $\frac{dx}{dt}\neq 0,$ the line is non-vertical and actually is the graph of a function that can be differentiated.)

Another way to interpret $\frac{dy}{dx}$ is to erase all of the curve except a small part of it containing the point in question. If you erase enough of the curve you can make it so that what is left is the graph of a function over some interval on the real line, and then $\frac{dy}{dx}$ is the derivative of that function.