# Thread: Stuck here--not sure how to differentiate

1. ## Stuck here--not sure how to differentiate

Here's my problem. I have a rotating wheel, radius 40cm, and a connecting rod AP with length 1.2m. The pin P slides back and forth along the x-axis as the wheel rotates counterclockwise at a rate of 360 revolutions per minute. Find the angular velocity of the connecting rod, $\displaystyle \frac{d\alpha}{dt}$, in radians per second, when $\displaystyle \theta=\frac{\pi}{3}$.

I have attached a diagram, and my work on the problem so far. (NOTE: later in my work, for the sake of simplicity, i replaced $\displaystyle \alpha$ with "a" and $\displaystyle \theta$ with "O".)

Now, this is my problem. At the bottom, where i differentiated it, the derivative doesnt have an alpha in the equation, but i need it in the problem to solve it. I had to use the calculator to differentiate it, because i havent worked with differentiating $\displaystyle \sin^{-1}$ before. I just became completely lost at this point...am i going about the problem the wrong way?....................(The first picture is the top half, the second is the bottom half)

2. You didn't differentiate with respect to the correct variable, and you didn't differentiate correctly. You have

$\displaystyle \displaystyle\alpha(t)=\sin^{-1}\left(\frac{1}{3}\sin(\theta(t))\right).$

What you want to compute is

$\displaystyle \displaystyle\dot{\alpha}(t)=\dots$ using the chain rule. What do you get?

3. I get $\displaystyle \frac{d}{dt}a(t)=\frac{\cos(\theta(t))\theta(1)}{\ sqrt{9-\sin(\theta(t))^2}}$

I'm not sure about this part, either...If i replace $\displaystyle \theta(t)$ with $\displaystyle \frac{\pi}{3}$, because $\displaystyle \theta=\frac{\pi}{3}$ is the point at which Im trying to find the speed, I get $\displaystyle .174*\theta(1)$. The correct answer is 6.563. This means that, for the equation above to be correct, $\displaystyle \theta(1)$ has to equal 37.718....Right? But this makes no sense :P

4. I get

$\displaystyle \dot{\alpha}(t)=\dfrac{\cos(\theta(t))\,\dot{\thet a}(t)}{\sqrt{9-\sin^{2}(\theta(t))}}.$ I think you're on the right track.

So you've got, correctly, that $\displaystyle \theta=\pi/3.$ What is $\displaystyle \dot{\theta}(t)?$ Are there any unit conversion issues you might have?

5. Originally Posted by Chokfull
I get $\displaystyle \frac{d}{dt}a(t)=\frac{\cos(\theta(t))\theta(1)}{\ sqrt{9-\sin(\theta(t))^2}}$

I'm not sure about this part, either...If i replace $\displaystyle \theta(t)$ with $\displaystyle \frac{\pi}{3}$, because $\displaystyle \theta=\frac{\pi}{3}$ is the point at which Im trying to find the speed, I get $\displaystyle .174*\theta(1)$. The correct answer is 6.563. This means that, for the equation above to be correct, $\displaystyle \theta(1)$ has to equal 37.718....Right? But this makes no sense :P

Instead of: $\displaystyle \displaystyle \frac{d}{dt}\alpha(t)=\frac{\cos(\theta(t))\theta( 1)}{\sqrt{9-\sin(\theta(t))^2}},$

you should have: $\displaystyle \displaystyle \frac{d}{dt}\alpha(t)=\frac{\cos(\theta(t))}{\sqrt {9-\sin^2(\theta(t))}}{{d\theta}\over{dt}}\,.$

Alternatively you could have differentiated a preceding equation with respect to time:

$\displaystyle \displaystyle {{d}\over{dt}} \left( \sin(\alpha(t))\right) = {{d}\over{dt}} \left({{1}\over{3}}\sin(\theta(t))\right)$

to get:

$\displaystyle \displaystyle \left( \cos(\alpha(t))\right){{d\alpha}\over{dt}} = \left({{1}\over{3}}\cos(\theta(t))\right){{d\theta }\over{dt}}\,.$

But then you would need to figure out what $\displaystyle \displaystyle cos\alpha$ is when $\displaystyle \displaystyle \theta = {{\pi}\over{3}}\,.$

6. Okay....If my $\displaystyle \theta(1)$ is supposed to be $\displaystyle \frac{d\theta}{dt}$, then all i have to do is multiply by $\displaystyle \frac{d\theta}{dt}$. But to get the right answer, this needs to be 37.718. This is really just the rate at which the wheel turns (360 revolutions per minute, or 6 per second). Since we're working in radians and seconds, and $\displaystyle 2\pi$ radians equals one full turn, it's rotating at a rate of....$\displaystyle 12\pi$ radians per second! This is exactly he answer I'm looking for. Thank you both for your help

7. 360 revolutions per minute is how many radians per second?