I Second derivative of a curve defined by parametric equations

[Mr Davis 97]

Quick question. I know that if we have a curve defined by $x=f(t)$ and $y=g(t)$, then the slope of the tangent line is $\displaystyle \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$. I am trying to find the second derivative, which would be $\displaystyle \frac{d}{dx}\frac{dy}{dx} = \frac{\frac{d}{dx}\frac{dy}{dt}}{\frac{dx}{dt}}$. Now, the final correct formula is $\displaystyle \frac{d^2 y}{dx^2} = \frac{\frac{d}{dt}\frac{dy}{dx}}{\frac{dx}{dt}}$. My question is, why is it valid that $\displaystyle \frac{d}{dx} \frac{dy}{dt} = \frac{d}{dt} \frac{dy}{dx}$?

 

[andrewkirk]

The formula can be calculated by the chain rule, treating $x$ as the intermediate function (of $t$), as:
$$\frac{d}{dx}\left(\frac{dy}{dx}\right)=
\left(\frac d{dt}\frac{dy}{dx}\right)\frac{dt}{dx}$$
And that is equal to
$$\left(\frac d{dt}\frac{dy}{dx}\right)\ /\ \frac{dx}{dt}
$$
which is the same as what you describe as the final correct formula.

How did you get your own formula? It doesn't look correct to me. Doing it your way I get:
$$\frac{d}{dx}\left(\frac{dy}{dx}\right)=
\frac{d}{dx}\left(\frac{dy}{dt}\ /\ \frac{dx}{dt}\right)
=\left[\frac{dx}{dt} \frac{d}{dx}\frac{dy}{dt}-
\frac{dy}{dt} \frac{d}{dx}\frac{dx}{dt}
\right]\ /\ \left(\frac{dx}{dt}\right)^2
$$
(using the quotient rule for the second step), which is not like what you wrote.

 

[Mr Davis 97]

The formula can be calculated by the chain rule, treating $x$ as the intermediate function (of $t$), as:
$$\frac{d}{dx}\left(\frac{dy}{dx}\right)=
\left(\frac d{dt}\frac{dy}{dx}\right)\frac{dt}{dx}$$
And that is equal to
$$\left(\frac d{dt}\frac{dy}{dx}\right)\ /\ \frac{dx}{dt}
$$
which is the same as what you describe as the final correct formula.

How did you get your own formula? It doesn't look correct to me. Doing it your way I get:
$$\frac{d}{dx}\left(\frac{dy}{dx}\right)=
\frac{d}{dx}\left(\frac{dy}{dt}\ /\ \frac{dx}{dt}\right)
=\left[\frac{dx}{dt} \frac{d}{dx}\frac{dy}{dt}-
\frac{dy}{dt} \frac{d}{dx}\frac{dx}{dt}
\right]\ /\ \left(\frac{dx}{dt}\right)^2
$$
(using the quotient rule for the second step), which is not like what you wrote.

Okay, I see how you do it now, using the chain rule. However, why is it not correct reasoning that if $\displaystyle \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$, then $\displaystyle \frac{d}{dx}\frac{dy}{dx} = \frac{d}{dx}\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$?

 

[andrewkirk]

Okay, I see how you do it now, using the chain rule. However, why is it not correct reasoning that if $\displaystyle \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$, then $\displaystyle \frac{d}{dx}\frac{dy}{dx} = \frac{d}{dx}\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$?

It is correct. But what you wrote here is different from what you wrote in the OP. Perhaps your LaTeX got a bit muddled back there?