## anonymous 5 years ago Word problem: A particle moves around the circle x^2+y^2=25 at constant speed, making one revolution in 2 s. Find it's acceleration when it is at (3,4). How is my thinking process supposed to be here?

Hmm... So you're going to need the tangential acceleration, which would be the second derivative of the function. But, the function is in terms of both x and y.

I think you'll have to find the second partial derivatives with respect to x and y, but I'm not 100% sure about that. Been a while since I've done a problem like this.

3. anonymous

You'll find the solution by converting to polar coordinates and using the given speed. It's a bit tricky, and I really suck at these kind of assignments. You'll have to treat the vector as r= x(t) + y(t) and make use of the chain rule when finding the derivatives. How would you approach it? I just find it really hard to realise all the steps without seeing them first.

Ah. Hmm... Still a toughie for sure.

Hm... Wait...

6. anonymous

Indeed, got a whole bunch of similar ones, luckily only 1 showed up on 2 preparation exams. So hopefully there wont be one.

Ok, so: \begin{align} x(\theta) &= r \cos \theta \\ y(\theta) &= r \sin \theta \end{align}

So, if we derive twice, we get: \begin{align} x'(\theta) &= -r \sin \theta & x''(\theta) &= -r \cos \theta\\ y'(\theta) &= r \cos \theta & y''(\theta) &= -r \sin \theta \end{align}

9. anonymous

Your r should just be a constant 5 here, I don't think you need to have 2 variables

10. anonymous

Correct Flying.

Right, that's irrelevant -- we can plug that in.

Theta, however, is a function of time. We go pi of theta in one second.

But the time it takes for one revolution is irrelevant -- it's a distractor.

(I think.)

15. anonymous

You're restricted to the first quadrant by the coordinates you're given, I feel like this problem just sounds difficult. It really isn't too bad though

16. anonymous

Yeah, it shouldn't really matter because your angular velocity is just the natural period of sin and cos

Yeah, I think once you have the second derivative as r cos theta, you can find theta easily using the point you're given and the tangent, and then plug in r and theta into the two above. Then you just need to normalize it to a magnitude + direction for the acceleration vector.

18. anonymous

I think you could have left this in x and y and just done implicit differentiation since you've already been given coordinates

Or rather once you have the two derivatives as -r cos theta and -r sin theta.

20. anonymous

solved it for y and just used specifically the top part of the circle

21. anonymous

Sounds good, the trick my teacher did was to use the points given 3 = 5cos(theta) and 4 = 5sin(theta) when you get the vector readily differentiated for acceleration. Quite neat.