1. ## [SOLVED] Work

I was able to get this question correct but now I can't.

Let F be a constant unit force that is parallel to the vector $\displaystyle <-1,0,1>$ in xyz-space. What is the work done by F on a particle that moves along the path given by $\displaystyle (t,t^2,t^3)$ between time $\displaystyle t=0$ and time $\displaystyle t=1$?

$\displaystyle W=cos(\theta)\parallel F\parallel \parallel PQ\parallel$

$\displaystyle cos(\theta)=\frac{<,u,v>}{\parallel u\parallel \parallel v\parallel}$

simplifies down to:
$\displaystyle W=<u,v>=<F,PQ>$

For PQ, I have (but something is wrong since the answer is 0) $\displaystyle 0\leq x\leq 1, t\mathbf{i}+t^2\mathbf{j}+xt^3\mathbf{k}$

$\displaystyle W=-t+t^3\neq 0$

2. You're using the wrong expression for Work. It should be the line integral $\displaystyle \int_C {\bf F}\cdot d {\bf r}$.

3. Originally Posted by ojones
You're using the wrong expression for Work. It should be the line integral $\displaystyle \int_C {\bf F}\cdot d {\bf r}$.
I don't remember how to setup the integral with what is giving.

F is the force, C is the path, and ds is ??

Since this path is in 3-space, I thinking this will be an iterated integral?

5. Originally Posted by ojones
6. $\displaystyle F=(-1,0,1)$
$\displaystyle r(t)=(t,t^2,t^3)$
$\displaystyle F(r(t))=(-1,0,1)$
$\displaystyle r'(t)=(1,2t,3t^2)$
$\displaystyle F(r(t))r'(t)=-1+3t^2$
$\displaystyle \int F(r(t))r'(t) \: dt=\int (-1+3t^2) \: dt \: = \: -t+t^3 \: |_0^1 \: = \: 0$