Question

Calculus 3

Answer 1

Given an acceleration vector, initial velocity \[<u _{0}, v _{0}>\] and initial position \[<x _{0}, y _{0}>\], find the velocity and position vectors for \[t \ge 0.\]

a(t)=<cos t, 2 sin t>
\[<u _{0}, v _{0}> = <0, 1>\]
\[<x _{0}, y _{0}> = <1,0>\]

Answer 2

Hey Bambi :)
I don't really remember my vectors all that well.
Mmmm we can just integrate though, yes?\[\Large\rm \int\limits \vec a=\int\limits \left<\cos t, 2\sin t\right>~dt\]\[\Large\rm \vec v=\left<\sin t,-2 \cos t\right>+\left<c_1,c_2\right>\]And they told us that our initial velocity (velocity at time t=0) is <0,1>.
Plugging in gives us:\[\Large\rm \left<0,1\right>=\left<\sin0+c_1,c_2-2\cos0\right>\]\For the first component:\[\Large\rm 0=\sin 0+c_1\]For the second:\[\Large\rm 1=c_2-2\cos0\]
Allowing you to find your c1 and c2.

Answer 3

Mmmm yah? Make sense?
Or am I forgetting some fundamental thing perhaps? hehe

Answer 4

The back of my book says the \[v(t) = <\sin t, -2\cos t + 3>\] and \[r(t) = <-\cos t +2, -2\sin t + 3t>\]

Answer 5

Yah that's where we're heading it looks like :)

Answer 6

I kept getting \[v(t) = <\sin t, -2\cos t + 1>\] and don't know what I'm doing wrong.

Answer 7

Oh my gosh, I see it now. Hah

Answer 8

Oh subtracting 2 from each side by mistake? :) You silly billy!

Answer 9

Something like that?

Answer 10

So at time, t=0, plug everything in and solve for the C's.

Answer 11

yes.

Answer 12

Lol thank you so much!