1. ## tangent line integral

I can't seem to get a ''nice'' solution to this question, the answer should be a simple fraction not what i get.
The question is 'calculate the line integral of the vector field
v(x,y,z)=((x-1)(z-3),xyz,x+z) along the straight line from (1,1,1) to (1,3,9)

I started by parameterising the line:
d(t)=(1,1,1)+t(0,2,8)
d(t)=(1,1+2t,1+8t)
d'(t)=(0,2,8)

thus v(d(t))=(0,(1+2t)(1+8t),(2+8t)

then i go to work out the integral between 0 and 1 of
<(v(d(t)),d'(t)> dt
and it comes out crap.
Any help would be good, pointing our error or next step
thanks

I can't seem to get a ''nice'' solution to this question, the answer should be a simple fraction not what i get.
The question is 'calculate the line integral of the vector field
v(x,y,z)=((x-1)(z-3),xyz,x+z) along the straight line from (1,1,1) to (1,3,9)

I started by parameterising the line:
d(t)=(1,1,1)+t(0,2,8)
d(t)=(1,1+2t,1+8t)
d'(t)=(0,2,8)

thus v(d(t))=(0,(1+2t)(1+8t),(2+8t)

then i go to work out the integral between 0 and 1 of
<(v(d(t)),d'(t)> dt
and it comes out crap.
Any help would be good, pointing our error or next step
thanks
So you want $\displaystyle \int_C \bold F \cdot d \bold r$, where $\displaystyle C$ is the straight line segment from (1,1,1) to (1,3,9)

You started fine. you would end up with:

$\displaystyle \int_0^1 \left< 0, (1 + 2t)(1 + 8t), 2 + 8t \right> \cdot \left< 0, 2, 8 \right> ~dt = \int_0^1 \left( 32t^2 + 84t + 18 \right)~dt$

You said you did this and it was wrong?

3. why do you seperate <(0,(1+2t)(1+8t),(2+8t) , (0,2,8)>?

$\displaystyle \int_0^1 \left< 0, (1 + 2t)(1 + 8t), 2 + 8t \right> \cdot \left< 0, 2, 8 \right> ~dt = \int_0^1 \left( 32t^2 + 84t + 18 \right)~dt$