# Initial value problem-application (multivariable calculus)

1. Dec 9, 2015

### mmont012

1. The problem statement, all variables and given/known data
This is a solution to a problem that was on a quiz, and I am confused about how to do it. Especially lines
two (<0,1,1>=v(0)=<C1, 1+C2, C3> --> C1=0, C2=0, C3=1 and
five (<1,0,0>=r(0)=<1+ K1, K2, K3> -->K1=0, K2=0, K3=0
How do you do these steps? Can someone walk me through this process?

I'm studying for my final, and I KNOW that this will be one there.

2. Dec 9, 2015

### RUber

Since $v(t) = \int -\cos t \hat i -\sin t \hat j dt,$ you get $v(t) = -\sin t \hat i + \cos t \hat j + \vec C$.
Note that C can be any constant vector.
Plug in t = 0 and compare with your v(0) term to solve for vector C.
$v(0) = -\sin 0 \hat i + \cos 0 \hat j + \vec C = 0 \hat i + 1 \hat j + \vec C = \hat j + \hat k$
This gives you $\vec C = \hat k$. Put this back into your equation for v(t) and you get $v(t) = -\sin t \hat i + \cos t \hat j + \hat k$.
Next, you integrate velocity to get position.
$r(t) = \int -\sin t \hat i + \cos t \hat j + \hat k dt = \cos t \hat i + \sin t \hat j + t \hat k + \vec K.$
Where, again, vector K is any constant vector.
As before, put in t = 0 and compare with initial position $\hat i$ to solve for the constant vector K.
$r(0) = \hat i = \cos 0 \hat i + \sin 0 \hat j + 0 \hat k + \vec K =1 \hat i + 0 \hat j + 0 \hat k + \vec K .$
This shows you that vector K is the zero vector, so you can write r(t) as
$r(t) = \cos t \hat i + \sin t \hat j + t \hat k .$