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A particle moves in the xy-plane so that its velocity vector at time t is v(t)=(t^2, sin(pi*t)) and the particle's position vector at time t=0 is (1, 0). Find the position vector of the particle when t=3?

Answer: (10, 2/pi)

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A particle (or any moving object) has an instantaneous velocity given by the derivative of the position vector with respect to time. This means that we can then get the position vector from the velocity vector by taking the integral with respect to time._{
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In this case, when we take the integral, x(t) = (1/3*t^3 + a, -cos(pi*t)/pi + b) where a and b are constants that are a result of doing an indefinite integral. We can find the values for a and b using the additional information given. we know that x(0) = (1,0). That means that 1/2*(0)^3 + a = 1 and so a = 1. That also means that -cos(pi*0)/pi + b = 0 and so b = cos(0)/pi = 1/pi.

Now we have all the information needed to find the position at t = 3. We know that x(t) = (1/3*t^3 +1, -cos(pi*t)/pi + 1/pi) so all we have to do is plug in t = 3. That would give us x(3) = (1/3*(3)^3 + 1, -cos(pi*3)/pi + 1/pi). Simplifying we get x(3) = (9+1,-(-1)/pi+1/pi). Thus you can see that the final answer will be x(3) = (10, 2/pi).

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