Find r(t): Given Initial Conditions

mamma_mia66
Messages
51
Reaction score
0

Homework Statement


Given \vec{}r"(t)= 6i-4cos (2t)j+ 9e3tk,
r'\vec{} (0)= 4i +3k and r\vec{}(0)=j+k, find r\vec{}(t).

Homework Equations





The Attempt at a Solution



I will appreciate any ideas how to start this problem. Thank you.
 
Physics news on Phys.org
mamma_mia66 said:

Homework Statement


Given \vec{}r"(t)= 6i-4cos (2t)j+ 9e3tk,
r'\vec{} (0)= 4i +3k and r\vec{}(0)=j+k, find r\vec{}(t).

Homework Equations





The Attempt at a Solution



I will appreciate any ideas how to start this problem. Thank you.
If r(t) = x(t)i + y(t)j + z(t)k, then r''(t) = x''(t)i + y''(t)j + z''(t)k.

How you would you normally solve x''(t)=6 for x(t), given x'(0) and x(0) ?

What is troubling you?
 
I get it now. I need to integrate the given. Thank you.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top