# how do you figure this out??

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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.

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

$x(t) = \int \cos(t^2)\, dt$

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)?

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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?

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oh yea the class is AP Calculus BC, i havent learned how ot integrate that type of problem yet

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And yea, as long as the method approximate the position at t = 3, yea i think its ok

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oh and dave one more thing, do you think you could help me figure out another problem? Its on the Homework forum, and youll see that it was my name as the thread starter. Here's the link:

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Just my two cents but can you do integration by parts?

$\int udv = uv - \int vdu$

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on what? how will it help?

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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.

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yes ive learned power series but i dunno how that would help in this problem 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?

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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.

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:-( :-( too muich work oh well might as well try it out, i gotta review power series anyway. Thanx guys!

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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)??

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OH WAIT nvr mind stupid me, i just realized you need both x and y coordinates for a position sry,

dave can you help me figure out this problem?

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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?

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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.

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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...

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What did you get as the power series?

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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.

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