Jump to content

Recommended Posts

Posted

guys i need help please ive been stuck on this problem for days:

 

An object moving along a curve in the xy-plane has position (x(t), y(t)) at time t with

 

dx/dt = cos(t^3) and dy/dt = 3sin(t^2)

 

for 0<= t <= 3. At time t = 2, the object is at position (4,5).

 

Find the position of the object at time t = 3.

 

Please help! im stuck!

Posted

I had originally posted something thinking that the problem is relatively trivial. Normally, one would integrate both sides of the differential equation and get:

 

[math]x(t) = \int \cos(t^2)\, dt[/math]

 

However, as far as I know, there's no easy way of evaluating that right hand side integral. Is this problem just for fun, or is it an assignment of some sort? If the latter, which class are you taking it in, and are you allowed to use some sort of numerical method (e.g. forward Euler)?

Posted

Hey dave. Yes I know it's very hard to integrate that equation. No, its not for fun, its a take home test for easter break :(. I think there's a way you can at east approximate it...do you know how?

Posted

Have you learned about power serries (I think they teach this in Calc BC)? You can express cos(t3) and sin(t2) as power serries then integrate those.

Posted

yes ive learned power series but i dunno how that would help in this problem :confused: If power series do work, can sum1 help me? i dunt really know how ot apply them to this situation.

 

Oh yea Matt grime yur a maths expert can you help?

Posted

Well, you should know the power series expansion for cosine and sine. Work out the power series expansion for cos(t3) and sin(t2), and then integrate term-by-term. You can work out the answer by plugging the first few terms into a calculator.

Posted

wait hold on a sec...do i need to do power series of both the x and y functions? Dont i need to do it only for the x function? why the sin(t^2)??

Posted

OK now im CONFUSED. I dunt know how ot get the ans to this problem. I tried power series but i dunt think its a viable method cuz the power series of a cos function or sin function will be an alternating series. How could you get the position at t = 3?

Posted

Well, assuming you don't have to give a precise answer, you could just evaluate the first few terms of the power series to get a decent approximation.

Posted

yea i did. But i got more disparate ans every time i evaluated a term. For example, since it was an alternating series, the first term would be 3, then the next term like -50, then 200, then -700, 10000, -23462456....sumthin like that. The ans get farther and farther away with each term. Maybe im doin it wrong...

Posted

I used the formula for powerseries for cos x --> 1 - (x^2)/2! + (x^4)/4! - (x^6)/6! + (x^8)/8! ... in which i plugged in x^3 as the function for x (the function was cos (x^3). That was how i got my power series.

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.