# A(t), v(t), r(t) converting, integration and derivatives

## Homework Statement

Given that the acceleration vector is
a(t) = (-16cos(-4t))i + (-16sin(-4t))j + -2tk,
the initial velocity is v(0) = i + k
and the initial position vector is r(0) = i + j + k, compute:
The velocity vector v(t) = ___i + ____j + ____k
The position vector r(t) = ___i + ____j + ____k

## Homework Equations

v(t) = r'(t) = ∫a(t)dt

## The Attempt at a Solution

v(t) = ∫a(t)dt = ∫a(t) = (-16cos(-4t)) i + (-16sin(-4t)) j + -2t k dt
v(t) = 4sin(-4t) i + (-4cos(-4t) j + (-t^2) k

r(t) = ∫v(t) = ∫4sin(-4t) i + (-4cos(-4t)) j + (-t^2) k dt
r(t) = -cos(-4t) i + sin(-4t) j + (-1/3)t^3 k

I've double checked my integration but can't figure out what I did wrong. None of the parts of my answers are correct.
Thanks!

Dick
Homework Helper
No, you left out the arbitrary constants of integration,

v(t) = ∫a(t)dt = ∫a(t) = (-16cos(-4t) +C1) i + (-16sin(-4t) +C2) j + (-2t +C3) k dt

Use the value of v(0) they gave you to figure out what the constants are.

No, you left out the arbitrary constants of integration,

v(t) = ∫a(t)dt = ∫a(t) = (-16cos(-4t) +C1) i + (-16sin(-4t) +C2) j + (-2t +C3) k dt

Use the value of v(0) they gave you to figure out what the constants are.

Ok, thanks so much! I got it now.