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I'm currently a student in an online AP BC calc class, and I'm having a little trouble with a question about derivatives and parametric equations.

Anyways, the problem is written as follows: "The position of an object is described by the parametric equations $x=\ln t$ and $y=5t^2$. What is the acceleration of the object in $m/sec^2$ when $t=2$?"

I've been wondering the whole time what I should be calculating - my sister says I should be calculating $\frac{d^2y}{dx^2}$, but personally I think I should be calculating $\sqrt{\frac{d^2y}{dt^2}+\frac{d^2x}{dt^2}}$ for t=2 - I'm a little confused, because I thought acceleration had direction as well, meaning that a single scalar would not accurately describe the acceleration of a particle with motion described in x and y axis?


Correction:$\sqrt{\frac{d^2y}{dt^2}+\frac{d^2x}{dt^2}}$ should be $\sqrt{(\frac{d^2y}{dt^2})^2+(\frac{d^2x}{dt^2})^2}$

EDIT: After submission and feedback it appears that I have gotten the problem wrong. The answer and explanation given are as follows: "We know that dy/dx=(dy/dt)/(dx/dt). x=lnt, therefore dx/dt=1/t*y=5t^2, therefore dy/dt=10t. So, dy/dx=(dy/dt)/(dx/dt)=(10t)/(1/t)=10t^2 and a=d^2y/dx^2=(d/dx(dy/dx))/(dx/dt)=(20/t)/1/t)=20t^2. So, when t=2, a=20(2)^2=80m/sec^2."

I do disagree with this answer, which assumes that a=d^2y/dx^2. At the time we were learning about parametric differentiation - I guess I should have known better, but something bothered me about the intuition of what d^2y/dx^2 represents. I have talked a bit with my teacher about the problem, but I'm guessing that I might want to call her to understand what she's trying to say better.

In short, does a=d^2y/dx^2, or does a=sqrt((d^2y/dt^2)^2+(d^2x/dt^2))? And if a is not equal to d^2y/dx^2, how should I explain that to my teacher? Thanks.

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up vote 4 down vote accepted

You are right on all counts!

Acceleration is indeed a vector: $\vec a = \left(\frac{d^2x}{dt^2},\frac{d^2y}{dt^2}\right)$. But if we only care about the magnitude of the acceleration, then we take the magnitude of the vector, which gives your expression with the square root.

Edit: As Matt pointed out below, your expression for $||\vec a||$ as it stands is actually not quite right, but I assume this is a typo!

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So I guess I should just assume the problem is asking for magnitude - and not direction? In which case I should simply evaluate t=2 for this expression and I would arrive at my result, correct? – inkyvoyd Nov 20 '12 at 14:03
It's hard to tell without more context. And yes, the general formula (which is still wrong in you question, by the way; see Matt's answer below) will give you the acceleration as a function of time, so you need to set $t=2$ to get the final answer. – Rhys Nov 20 '12 at 14:05
why does a=$\sqrt{(\frac{d^2y}{dt^2})^2+(\frac{d^2x}{dt^2})^2}$ and not d^2y/dx^2? – inkyvoyd Dec 13 '12 at 13:06

Almost right :) As Rhys said, $\vec{a} = (\frac{d^2x}{dt^2}, \frac{d^2y}{dt^2})$, but to get the magnitude of that you want $\| \vec{a} \| = \sqrt{(\frac{d^2x}{dt^2})^2 + (\frac{d^2y}{dt^2})^2}$, which you want to evaluate for $t=2$.

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I had actually meant to use $\sqrt{(\frac{d^2y}{dt^2})^2+(\frac{d^2x}{dt^2})^2}$, but I guess I must've forgotten... In any case, thanks for pointing that out! – inkyvoyd Nov 20 '12 at 13:57
Whoops, good point! I didn't read carefully enough. :-) – Rhys Nov 20 '12 at 13:57

As Rhys explained, your answer is correct. Now, why the alternative $\frac{d^2y}{dx^2}$ isn't ? First thing to look at, the dimension of the formulas : $d^2y$ is in $\mathrm{meter}$. $dx^2$ is in $\mathrm{meter}^2$ So $\frac{d^2y}{dx^2}$ is in $\frac{1}{\mathrm{meter}}$ which is obviously not the dimension of an acceleration.

In fact the formulas $\frac{d^2y}{dx^2}$ comes from a mix-up about parametric equations and classic functions. This formula would give you an acceleration if the path of your object would be a single line and the function $y = f(x)$ represent the position of your object on this line. But in that case, $x$ would be a time (for example in second).

If you want to give an interpretation to the formula $\frac{d^2y}{dx^2}$, first interpret the formula $\frac{dy}{dx}$: which is equal to $\tan(\alpha)$ where $\alpha$ is the angle of the speed vector with the abscissa.

But: $$\frac{d^2y}{dx^2} = \frac{d}{dx}.\frac{dy}{dx}$$

So we have: $$\frac{d^2y}{dx^2} = \frac{d}{dx}\cdot\tan(\alpha)$$

$\frac{d^2y}{dx^2}$ can be interpreted as the variation of the direction of the speed vector when $x$ changes. However in a context of parametric equations, you should avoid to try to interpret variation on a single parameter because your likely to go into complicated cases (for example, $tan(\alpha)$ can have two different values for a same $x$ value. This is because $\tan(\alpha)$ is not only a function of $x$ but also of $y$.)

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$\frac{dx}{dt}=\frac1t$ and $\frac{dy}{dt}=10t$.

$\therefore \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$.

Now $\frac{d^2y}{dx^2}=\frac d{dx}\big(\frac{dy}{dx}\big)=\frac d{dt} \big(\frac{dy}{dx}\big)*\frac{dt}{dx}$ because $\frac{dy}{dx}$ is a function of $t$.

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