Question

A particle is moving with the given data. Find the position of the particle. v(t)=sint-cost, s(0)=0

Answer 1

Integrate it:
\[\frac{dx}{dt}=\sin(t)-\cos(t) \implies \int\limits_0^x dx' = \int\limits_0^t \left[ \sin(t')-\cos(t')\right]dt'\]
\[\implies s(t)=\left[-\cos(t')-\sin(t') \right]_0^t=-\cos(t)-\sin(t)+1\]
When you integrate from 0 to x you are taking into account the initial condition, namely s(0)=0. That is also why you integrate from 0 to t. If you had a condition like s(0)=A then you would have:
\[\int\limits_A^x dx'\]
Or if you had something even more general like s(t_0)=s_0 we'd have:
\[\int\limits_{s_0}^s ds'=\int\limits_{t_0}^t f(t)dt'\]
This way we don't get a "C_1" and then have to resolve.

Answer 2

Verification:
http://www.wolframalpha.com/input/?i=solve+dx%2Fdt%3D-cos%28t%29%2Bsin%28t%29%2Cx%280%29%3D0

Answer 3

To find position, you find the anti-derivative of the velocity function (integrate it). Antiderivative of sint is -cost, and antiderivative of (-cost) is (-sint), so:

s(t) = -cost-sint + C

and since s(0) = 0, we can find C:

0 = -cos0 - sin0 + C
0 = -1 + C
C = 1

so s(t) = -cost-sint + 1

malevolence's method avoids needing to find C, but it's not difficult to in this case (you don't really even have to do the steps, if you know that -cos0 = -1 and sin0 = 0)

Answer 4

Thanks guys, helped a lot.