# Why do these equations result an incorrect unit for acceleration?

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.

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 $1/s$) .
• 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 Jan 28 '14 at 14:08
• The derivative of $\sin (2x)$ is not $2\cos x$! It's $2\cos (2x)$. Jan 29 '14 at 15:02
• 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. Jan 29 '14 at 15:07