Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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

Answer 5

Hm... Wait...

Answer 6

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

Answer 7

Ok, so:

\[\begin{align}
x(\theta) &= r \cos \theta \\
y(\theta) &= r \sin \theta
\end{align}\]

Answer 8

So, if we derive twice, we get:

\[\begin{align}
x'(\theta) &= -r \sin \theta & x''(\theta) &= -r \cos \theta\\
y'(\theta) &= r \cos \theta & y''(\theta) &= -r \sin \theta
\end{align}\]

Answer 9

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

Answer 10

Correct Flying.

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

(I think.)

Answer 15

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

Answer 16

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

Answer 17

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.

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

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.