How to take parametric equations (x, y) to create a derivative formula?

Question

I always thought that if I take the derivative of the y and x equation and divide y' by x', then that would be the derivative in formula form.

Is this correct?

Answer 1

Yes, in many cases you may compute $$y'(t) = \frac{dy}{dt}\;\;\text{ and}\;\;x'(t) = \frac{dx}{dt} \quad \text{when}\;\;x'(t)\neq 0; \text{ and not both} \; x' = y' = 0$$

Then $$ \dfrac{dy}{dx} = \dfrac{y'(t)}{x'(t)}= \dfrac{dy/dt}{dx/dt}$$

But note that you will need to do more work if $\dfrac{dy}{dx}$ at some given $t$(s) evaluates in the form $\dfrac 00$, i.e., when both $x'(t) = y'(t) = 0$, for given $t$.

Answer 2

From the chain rule, we have $$\dfrac{dy}{dt} = \dfrac{dy}{dx} \cdot \dfrac{dx}{dt} \implies \dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$$

Hence, what you have is correct from the chain rule, provided $dy/dt$ and $dx/dt$ both exist. Note that we also need $dx/dt$ to be non-zero.

Answer 3

Your intuition is essentially correct, although certain parametrizations "come to a stop," so $\frac{dy}{dx}$ is of the form $\frac{0}{0}$.

Consider, for example, the parametrized curve $$ \begin{cases} x(t) = (t - 1)^3 \\ y(t) = (t - 1)^6. \end{cases} $$ It's clear that $y = x^2$, and $$ \begin{cases} x'(t) = 3(t - 1)^2 \\ y'(t) = 6(t - 1)^5, \end{cases} $$ so $$ \frac{~\frac{dy}{dt}~}{\frac{dx}{dt}} = \frac{6(t - 1)^5}{3(t - 1)^2} = 2(t - 1)^3 \quad \text{only for } t \ne 1. $$ But if we imagine $t$ measuring time, then $(x(t), y(t))$ is the position of a point at time $t$. The point traces out the curve $y = x^2$, but in such a way that it slows to a complete stop at $t = 1$ (notice that $(x'(1), y'(1)) = (0, 0)$), only to resume moving afterwards.

What direction is the particle moving at $t = 1$? (It's a trick question: the particle's not moving.) But we want to say that it's moving horizontally, or equivalently, in a direction of slope $\frac{dy}{dx} = 0$.

In a sense, that's correct, since we can the limit. $$ \frac{dy}{dx}\Bigg|_{(x, y) = (1, 1)} = \lim_{t \to 1} \frac{~\frac{dy}{dt}~}{\frac{dx}{dt}} = 0. $$