# Calculate the work done by the force F along the path C.

• Apr 29th 2010, 12:45 PM
dizizviet
Calculate the work done by the force F along the path C.
Calculate the work done by the force F along the path C.
F= xe^(4x^2) i + e^(6y) j + e^(5z) k; the path is c1 to c2 where c1 is the straight line from (0,0,0) to (1,1,0) and c2 is the straight line from (1,1,0) to (1,1,1).

I just need someone to explain what exactly is it asking for because I'm just lost to what to with the two lines.
• Apr 29th 2010, 01:08 PM
piglet
Quote:

Originally Posted by dizizviet
Calculate the work done by the force F along the path C.
F= xe^(4x)^x i + e^(6y) j + e^(5z) k; the path is c1 to c2 where c1 is the straight line from (0,0,0) to (1,1,0) and c2 is the straight line from (1,1,0) to (1,1,1).
just lost to what to with the two lines.

Notice on c1 that the x, and y value is the same.

let $\displaystyle x = y = t$ where $\displaystyle 0<=t<=1$ and as $\displaystyle x = y= t$ => $\displaystyle dy= dt$ and $\displaystyle dx = dt$. Notice how the z value on path c1 is the same, therefore you can say $\displaystyle z= 0$ => $\displaystyle dz = 0$

so your first line integral is, using work integral formula $\displaystyle w = \int{F.dr}$

$\displaystyle \int^{1}_{0}(t.e^{4t}{^{^t}} , e^{6t} , e^{5(0)}).(dt,dt,0)$

Solve for c2 in the same way.

Then sum c1 and c2
• Apr 29th 2010, 02:48 PM
dizizviet
OK, i think I'm integrating the i component wrong, isn't it the derivative of the exponent goes on the bottom, then the integration of the base? I'm getting (xe^(4x^2))/16 but my calculator is giving me (e^(4x^2))/8. Can somebody tell me where i went wrong with my integration? Oh and how to input equations so i won't have to type it out like i am now?
• Apr 29th 2010, 02:50 PM
dizizviet
Quote:

Originally Posted by piglet
Notice on c1 that the x, and y value is the same.

let $\displaystyle x = y = t$ where $\displaystyle 0<=t<=1$ and as $\displaystyle x = y= t$ => $\displaystyle dy= dt$ and $\displaystyle dx = dt$. Notice how the z value on path c1 is the same, therefore you can say $\displaystyle z= 0$ => $\displaystyle dz = 0$

so your first line integral is, using work integral formula $\displaystyle w = \int{F.dr}$

$\displaystyle \int^{1}_{0}(t.e^{4t}{^{^t}} , e^{6t} , e^{5(0)}).(dt,dt,0)$

Solve for c2 in the same way.

Then sum c1 and c2

I just don't even understand what you are doing there, aren't you suppose to find the potential function for the three components then integrate. If so, i'm having trouble setting up the limits on what i should integrate with...
• Apr 30th 2010, 01:53 PM
dizizviet
I can't seem to understand how to set up the limits, I have the potential function for the three components but find the integration limit I'm having trouble. And if find the potential function is the wrong way to do it then can i get corrected because i don't understand the method up top.
• Apr 30th 2010, 02:38 PM
mr fantastic
Quote:

Originally Posted by dizizviet
I just don't even understand what you are doing there, aren't you suppose to find the potential function for the three components then integrate. If so, i'm having trouble setting up the limits on what i should integrate with...

Not all vector fields have a potential function. only those which are conservative. What do you think the potential is (assuming there is a potential).

Post #2 tells you what to do. At this stage, I think your best plan is to go back and review this material in your class notes and textbook.
• Apr 30th 2010, 03:15 PM
zzzoak
$\displaystyle r(t)=(t,t,0)$
$\displaystyle r'(t)=(1,1,0)$
$\displaystyle \int_{0}^{1}F(r(t))r'(t)dt=\int_{0}^{1}(te^{4t^2}+ e^{6t})\:dt=1/8e^{4t^2}|_{0}^{1}+1/6e^{6t}|_{0}^{1}=1/8(e^4-1)+1/6(e^6-1)$