# Path integral

• Feb 8th 2010, 12:37 PM
qwesl
Path integral
If $\displaystyle C$ is the curve given by $\displaystyle r(t)=(1+5sin{t}){i}+(1+2sin^2{t}){j}+(1+5sin^3{t}) {k}$ and $\displaystyle 0\leq t\leq \frac{\pi}{2}$ and $\displaystyle F$ is the radial vector field $\displaystyle F(x,y,z)=x{i}+y{j}+z{k}$, How do you compute the work done by $\displaystyle F$ on a particle moving along $\displaystyle C$
• Feb 8th 2010, 01:34 PM
HallsofIvy
Quote:

Originally Posted by qwesl
If $\displaystyle C$ is the curve given by $\displaystyle r(t)=(1+5sin{t}){i}+(1+2sin^2{t}){j}+(1+5sin^3{t}) {k}$ and $\displaystyle 0\leq t\leq \frac{\pi}{2}$ and $\displaystyle F$ is the radial vector field $\displaystyle F(x,y,z)=x{i}+y{j}+z{k}$, How do you compute the work done by $\displaystyle F$ on a particle moving along $\displaystyle C$

$\displaystyle r(t)=(1+5sin{t}){i}+(1+2sin^2{t}){j}+(1+5sin^3{t}) {k}$ so
$\displaystyle dr= \frac{dr}{dt}dt= (5 sin(t)i+ 4sin(t)cos(t)j+ 15sin^2(t)k)dt$

Integrate the dot product of F and that from t= 0 to $\displaystyle \pi/2$.