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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?
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?

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shadowfiend
 5 years ago
Best ResponseYou've already chosen the best response.0Hmm... 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.

shadowfiend
 5 years ago
Best ResponseYou've already chosen the best response.0I 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You'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.

shadowfiend
 5 years ago
Best ResponseYou've already chosen the best response.0Ah. Hmm... Still a toughie for sure.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Indeed, got a whole bunch of similar ones, luckily only 1 showed up on 2 preparation exams. So hopefully there wont be one.

shadowfiend
 5 years ago
Best ResponseYou've already chosen the best response.0Ok, so: \[\begin{align} x(\theta) &= r \cos \theta \\ y(\theta) &= r \sin \theta \end{align}\]

shadowfiend
 5 years ago
Best ResponseYou've already chosen the best response.0So, if we derive twice, we get: \[\begin{align} x'(\theta) &= r \sin \theta & x''(\theta) &= r \cos \theta\\ y'(\theta) &= r \cos \theta & y''(\theta) &= r \sin \theta \end{align}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Your r should just be a constant 5 here, I don't think you need to have 2 variables

shadowfiend
 5 years ago
Best ResponseYou've already chosen the best response.0Right, that's irrelevant  we can plug that in.

shadowfiend
 5 years ago
Best ResponseYou've already chosen the best response.0Theta, however, is a function of time. We go pi of theta in one second.

shadowfiend
 5 years ago
Best ResponseYou've already chosen the best response.0But the time it takes for one revolution is irrelevant  it's a distractor.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You'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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, it shouldn't really matter because your angular velocity is just the natural period of sin and cos

shadowfiend
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think you could have left this in x and y and just done implicit differentiation since you've already been given coordinates

shadowfiend
 5 years ago
Best ResponseYou've already chosen the best response.0Or rather once you have the two derivatives as r cos theta and r sin theta.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0solved it for y and just used specifically the top part of the circle

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sounds 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.
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