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1142004    Consider the parametric equations $x=f(t)$ and $y=g(t)$. To "find" $\frac{d^2y}{dx^2}$, there are three ways to go: (1) the correct one, that is, $\frac{\frac{d^2y}{{dt}{dx}}}{\frac{dx}{dt}}$, and two wrong ones that are (2) $\frac{\frac{d^2y}{dt^2}}{\frac{d^2x}{dt^2}}$, and (3) taking the derivative of the result of $\frac{dy}{dx}$ in terms of $t$, that is, $\frac{ \frac{d^2y}{dt^2}·\frac{dx}{dt}-\frac{d^2x}{dt^2}·\frac{dy}{dt} }{\left(\frac{dx}{dt}\right)^2}$.

Most textbooks warn students to avoid the second one, and to illustrate, work with a concrete example and show the result gained from the second one differs from the first! That is obviously not a constructive approach, to say the least. But, interesting, it is the third approach that is most common, though it looks messy at the first glance. Consider that when working with a concrete example, it is very natural to take the derivative of the result of $\frac{dy}{dx}$ (usually written at the right of the equal sign in terms of $t$) and consider it as $\frac{d^2y}{dx^2}$.

The question: Is there any interesting constructive way to convince students that the second and the third is wrong without referring to the first?

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  • $\begingroup$ There is a math educators beta SE that this would probably be better for: matheducators.stackexchange.com $\endgroup$ Feb 10, 2015 at 12:47
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    $\begingroup$ I'm surprised you feel that a counterexample disproving a general statement is not “constructive.” $\endgroup$ Feb 10, 2015 at 12:49
  • $\begingroup$ @MatthewLeingang I am aware of that. But, for the mathematics involved, I thought it is better to try this one first $\endgroup$ Feb 10, 2015 at 12:50
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    $\begingroup$ First thing: I think you made a typo in the correct formula; it should be (d^2 y/dtdx)/(dx/dt). Second thing: if you want to eliminate these specific mistakes, you may want to introduce units (say, y are kilograms, x are meters, t are seconds) and assert that the final formula must have the good units (here, kilograms per square meters). $\endgroup$
    – D. Thomine
    Feb 10, 2015 at 12:51
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    $\begingroup$ For the purposes of killing (3) I would use an example with $dx/dt$ negative. If a student divides by $(dx/dt)^2$ instead of the correct $(dx/dt)^3$ (which is what you get from the correct formula), then you get the wrong sign for the second derivative $d^2y/dx^2$. $\endgroup$ Feb 10, 2015 at 13:00

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Write $F(t)$ for the derivative. Where the curve is suitably 'nice', we have that $F(t)$ is in fact given by $F(x(t))$ as the curve locally looks like a function. Now differentiate using the Chain Rule to get

$$\frac{dF}{dt}=\frac{dF}{dx}\cdot \frac{dx}{dt}\Rightarrow \frac{dF}{dx}=\frac{\frac{dF}{dt}}{\frac{dx}{dt}}.$$

Where of course $\displaystyle F(t)=\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ so we have $$\frac{d^2y}{dx^2}=\frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}.$$

That is how I would show the result. For the others I would revert to Matthew's comment.

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  • $\begingroup$ It is the exact way that I derive the correct formula. But, one of the current tasks that I gave to students showed me that the temptation of using the third approach is too high when working with a concrete example. $\endgroup$ Feb 10, 2015 at 15:17
  • $\begingroup$ How do you stop students using the second derivative test incorrectly? e.g. "$f''<0$ at a stationary point implies a min"? $\endgroup$ Feb 10, 2015 at 16:10
  • $\begingroup$ Pursuant to the Wikipedia article, "symmetry of second derivatives," $$ $$ Can the order of the double differentiation in the top of the correct method (1) be reversed, I wonder? $\endgroup$ Feb 15, 2015 at 1:05
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I'd suggest that the best way to convince students the second and third are wrong is to make certain that they have a clear idea of what they're trying to compute.

While I'm the world's greatest fan of Leibniz notation, I find that in problems like these, it's a rare student who can actually say what $\frac{d^2y}{dx^2}$ might possibly mean; given that problem, it's no surprise that they can't distinguish among different proposed ways to compute it.

I should say that in rare cases, there's a student who can compute it without knowing what the heck it might me. In some ways, this worries me even more. :(

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I'm ready to fill out my second comment to an answer: Consider the parametric curve with equations $x=t^3$, $y=t^6$. This curve has equation $y=x^2$. Therefore we know $\frac{dy}{dx} = 2x$ and $\frac{d^2y}{dx^2} = 2$.

Let's find the derivative $\frac{dy}{dx}$ using the parametrization. We have $\frac{dy}{dt} = 6t^5$ and $\frac{dx}{dt} = 3t^2$. Therefore $$ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{6t^5}{3t^2} = 2t^3 $$ And this makes sense because $2t^3 = 2x$.

Now apply the proposed second derivative formulas: $$ \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}} = \frac{6t^2}{3t^2} = 2 \tag{1} $$ $$ \frac{\frac{d^2y}{dt^2}}{\frac{d^2x}{dt^2}} = \frac{30t^4}{6t} = 5t^3 \tag{2} $$ $$ \frac{d}{dt}\left(\frac{dy}{dx}\right) = 6t^2 \tag{3} $$ Only equation (1) provides a second derivative agreeing with our non-parametric result. So (2) and (3) can't be the right formula.

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  • $\begingroup$ I'm going to share this page with my students. Hope it helps. $\endgroup$ Feb 10, 2015 at 21:54
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There is a simple mnemonic allowing to avoid such wrong approaches. There is a difference of $d^2 x$ and $dx^2$. The first one is the differential linear operator applied twice $d^2 x=d(dx)$; The second is the SQUARE of the result of the differential operator applied once. $dx^2 =(dx)^2$. So $d^2 y / dx^2$ cannot be $d^2 y/d^2 x$.

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