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Hello everyone.

Imagine an object moving around a certain point on a circular orbit. Magnitude of the velocity is constant during the motion ($|v|$). The orbit radius is $r$. (I'd better notice that we're just talking about kinematic view of this motion.)

According to the image I've uploaded, we'll have:

$\large v_x(\theta)=|v|\cdot \cos\theta$

$\large v_y(\theta)=|v|\cdot \sin\theta$

Since perimeter of the circular path is $2\pi r$, and magnitude of the velocity is constant, we'll have:

$\large\theta (t)=\frac{|v|\cdot t}{2\pi r} \times 2\pi =\frac{|v|\cdot t}{ r}$

Now we can combine these equations:

$\large v_x(\theta)=|v|\cdot \cos(\frac{|v|\cdot t}{ r})$

$\large v_y(\theta)=|v|\cdot \sin(\frac{|v|\cdot t}{ r})$

By this point, everything is okay. But the problem occurs here, where we try to get derivative of $v_x(t)$ and $v_y(t)$ in order to find $a_x(t)$ and $a_y(t)$. As we know by differentiation we have:

$\cos^{\prime}(x)=-\sin(x)$

$\sin^{\prime}(x)=\cos(x)$

And we know that acceleration(time) function is derivative of velocity(time). So:

$\large a_x(t)=(v_x(\theta))'=|v|\cdot -\sin(\frac{|v|\cdot t}{ r})$

$\large a_y(t)=(v_y(\theta))'=|v|\cdot \cos(\frac{|v|\cdot t}{ r})$

Well, now something is wrong: These two equations result a $m/s$ unit (or something like that) for acceleration, but that's wrong. Acceleration unit must be $m/s^2$, (or something like that).

The question is that: Where does this problem come from? I couldn't figure it out at all. I don't know, maybe some kind of misunderstanding about derivative concepts cause that. So please try to answer simple, clean as much as possible.

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You have to apply the chain rule, because of the $\frac{|v|}{ r}$ factor:

$$\frac{d\cos(u(x))}{dx}=-\frac{du}{dx}\sin(u)$$

I think you can do the rest (it will multiply the rest by $1/s$) .

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  • $\begingroup$ Great hint. Heretofore, if you'd ask me "What's derivative of $sin(2x)$", I'd instantly answer "It's $cos(2x)$" and that was really a silly mistake! Now I know that derivative of $sin(2x)$ is $2cos(x)$ regarding to your answer and chain rule. (Of course I had to notice that, because it's clear that derivative of $sin(2x)$ sometimes reaches to 2 but maximum of $cos(2x)$ is 1). Thanks a galaxy $\endgroup$ – Moctava Farzán Jan 28 '14 at 14:08
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    $\begingroup$ The derivative of $\sin (2x)$ is not $2\cos x$! It's $2\cos (2x)$. $\endgroup$ – mikhailcazi Jan 29 '14 at 15:02
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    $\begingroup$ This common mistake was made because you didn't write explicitly with respect to what variable you are taking the derivative. It would be better to always use notations like $f(x)'_x$ than just $f(x)'$. Then you can deal with situations like $F(x(y))'_y$ without any problems. $\endgroup$ – Edvard Jan 29 '14 at 15:07
  • $\begingroup$ Yeah, @mikhailcazi, you're right. That was a mistake. Thanks $\endgroup$ – Moctava Farzán Jan 30 '14 at 9:40

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