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The other day I saw what was, seemingly, a fairly simple question.

A curve in the $xy$-plane is given parametrically by the equations: $$\begin{align} x &=\phantom{3}t^2+2t \\ y &=3t^4+4t^3 \end{align}$$ for all $t>0$. Find the value of $\frac{d^2y}{dx^2}$ at $(8,80)$.

My method to obtain the correct answer is as follows:

$$\frac{dx}{dt}=2t+2 \tag1$$

$$(\frac{dx}{dt})^2=\frac{dx^2}{dt^2}=4t^2+8t+4 \tag2$$

$$\frac{dy}{dt}=12t^3+12t^2\implies\frac{d^2y}{dt^2}=36t^2+24t \tag3$$

$$\frac{\frac{d^2y}{dt^2}}{\frac{dx^2}{dt^2}}=\frac{d^2y}{dt^2}\frac{dt^2}{dx^2}=\frac{d^2y}{dx^2}=\frac{36t^2+24t}{4t^2+8t+4}\tag4$$

$(8,80)$ occurs when $t=2$. Plugging this value of $t$ into the equation above yields an answer of $\frac{16}{3}$. The answer, however, should be $4$.

Where did I go wrong and why, exactly, did this particular method fail to give the correct answer?

Blue
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Lkryat
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    The division in the last step is the suspect. You cannot do that with second derivatives differentials as one always could with the first derivative differentials. – Jack LeGrüß Oct 26 '20 at 07:54
  • Nowhere in this solution do I assume that $\frac{d^2f}{dt^2}=(\frac{df}{dt})^2$. – Lkryat Oct 26 '20 at 07:56
  • This seems to be a case where the derivative symbol looking like a fraction causes problems. Your approach is actually somewhat clever for what you think should happen; unfortunately, the second derivative for a parametric curve is a bit trickier than that. See Wikipedia's "Parametric derivative" entry. – Blue Oct 26 '20 at 08:01
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    I see. I suppose, then, that the prevailing question is why is it that $\frac{df}{dt}\frac{dt}{dx}=\frac{df}{dx}$,but $\frac{d^2f}{dt^2}\frac{dt^2}{dx^2}\neq\frac{d^2f}{dx^2}$. – Lkryat Oct 26 '20 at 08:10
  • Try doing the same with $(x,y)=(t,t^2)$ and you’ll end up dividing by zero. – Thomas Andrews Oct 26 '20 at 08:43
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    @Lkryat: "the prevailing question is why is it that ..." ... The fact that $\frac{df}{dt}\frac{dt}{dx}=\frac{df}{dx}$ is something of a "happy accident" with the derivative symbol looking like a fraction. Nevertheless, the derivative symbol ISN'T a fraction and cannot legitimately be manipulated as if it is one. (This has been a source of confusion ever since the notation was developed. The symbol is a bit of a blessing and a curse.) Appreciate such accidents when they happen, but guard against letting them mislead you about what's really going on. – Blue Oct 26 '20 at 08:47
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    @Thomas Andrews Notwithstanding the fact that I think Blue, Jack LeGrüß, and a couple others have already sufficiently answered why this method will not work, you would not end up dividing by zero. Reread the post...carefully. I'm not dividing by the individual second derivatives. – Lkryat Oct 26 '20 at 09:22

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As mentioned in the comments, you cannot divide the individual second derivatives to get the answer. The correct way of doing this is as follows: $$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dt}\left(\frac{dy}{dx}\right)\cdot\frac{dt}{dx}$$ we have $$\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{12t^2(t+1)}{2(t+1)} = 6t^2$$ and $$\frac{dt}{dx} = \frac{1}{\frac{dx}{dt}} = \frac{1}{2(t+1)}$$ which means that the final answer would be $$\frac{d^2y}{dx^2} = \frac{3t^2}{t+1}$$ substituting $t=2$ yields $4$, which is the correct answer.

Aniruddha Deb
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  • I would like to point out that I am not dividing individual second derivatives. I am dividing the second deriviative of $y(t)$ by the square of the first derivative of $x(t)$. – Lkryat Oct 26 '20 at 08:04
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    @Lkryat: I thought I should make this comment here, which I have done above. What you are doing in the last step of your attempt is presuming that $dt^2=(dt)^2$ because you expect $\frac{d^2y}{dt^2}\left( \frac{dt}{dx}\right)^2$ to reduce to what you are seeking when the $dt^2$ and $(dt)^2$ cancel out, which is incorrect . – Jack LeGrüß Oct 26 '20 at 08:06
  • Thank you for the answer Jack LeGrüß. That makes sense, and explains why I keep getting the incorrect answer. – Lkryat Oct 26 '20 at 08:15