Work on a particle in an elleptic golf

A particle travels on an elliptic course x^2/4+y^2=1  (x,y>=0) from point A(0,1) to point B(2,0)
On it works the following force:
F=(1+xy)e^(xy) i + x^2e^(xy) j
Calculate work of the force .

Now I have this law:
W = Integral(AB) (1+xy)e^(xy) dx + x^2 e^(xy) dy   (see figure. beneath)
I have calculated the Integral as this:
w=1/y * e^(xy) + x*e^(xy)-e^(xy)+x*e^(xy)
the problem is when I insert y=0 I have division by zero ,so what am I doing wrong? how to go about this?
(I'm new to the subject!)
w=Integral
LVL 10
aboo_sAsked:
Who is Participating?
 
phoffricCommented:
Here's a very constructive video playlist that covers vector fields and line integrals.
https://www.khanacademy.org/math/multivariable-calculus/line_integrals_topic
0
 
aburrCommented:
suggestions for this homework type question.
remember that e^0 = 1
calculate the work done by the x component separately then the work done by the y component.
0
 
aboo_sAuthor Commented:
the problem is with 1/y when y=0.
the fact that e^0=1 never skipped my mind!
0
Free Tool: Path Explorer

An intuitive utility to help find the CSS path to UI elements on a webpage. These paths are used frequently in a variety of front-end development and QA automation tasks.

One of a set of tools we're offering as a way of saying thank you for being a part of the community.

 
aburrCommented:
Before anything else draw a rough graph of your ellipse and visulize the force. What is the force at your end points? Draw arrows. This should convince you that there are no zeros to divide by.
0
 
aburrCommented:
"the fact that e^0=1 never skipped my mind! "
good

I missed the i/y. let me think more.
0
 
aburrCommented:
the 1/y is in your work. I see that it is not in the force. I am not sure where the 1/y came from but it may be an error.
0
 
aburrCommented:
Use the equation of the ellipse to eliminate all the y in the work integral.
0
 
aboo_sAuthor Commented:
I got 2 approaches to handle this ,both are wrong!
One of which I tried your last comment.
new-doc-89-2.jpgnew-doc-89-1.jpg
0
 
aboo_sAuthor Commented:
Please advise
0
 
aboo_sAuthor Commented:
Thank you so much phoffric the videos were so helpfull (in general)  ,but I still can't figure out the answer to my problem and really need help with it.
You see when I parametrized my integral and curve I still got an unsolvable Integral!
for example what is the Integral of:  e^(cost*sint)  ?
0
 
ozoCommented:
Applying Green's Theorem may make it easier to solve the integral.
0
 
aboo_sAuthor Commented:
Well the orientation is clockwise so I cannot use Green's theorem! can I ?
0
 
aboo_sAuthor Commented:
if I am to combine the curve after it's parametrization with my second attempt that has totally disregarded the curve I will get something like this:
linearintegral3-1.jpg
Is this right or wrong !?
0
 
aboo_sAuthor Commented:
Alright ,I  am now convinced that what I did previously was wrong ,so I tried this:
integral-t.jpg
As you can see there is a slight problem with concluding this approach.
Help would be a good practice right now  :)
0
 
aboo_sAuthor Commented:
Ok I have come up with a solution with the help of some of my colleagues.
I will upload it soon.

Thank you all for your help.
0
 
aboo_sAuthor Commented:
So now this question is officially closed.

integral3-final.jpg
Thank you all.
0
 
phoffricCommented:
>> Well the orientation is clockwise so I cannot use Green's theorem! can I ?
One of the videos demonstrated that changing the direction of a vector line integral just changes the sign. So, you can just keep track of signs and still use Green's theorem if you want.

I did not check the details of your last post, but did come up with the same answer by observing that pdQ/pdx = pdP/pdy and therefore the force field is conservative. Then, choose the path A -> O and O -> B to avoid the elliptical (harder to integrate) path.
A -> O gives work = 0.
O -> B gives work = 2.
0
Question has a verified solution.

Are you are experiencing a similar issue? Get a personalized answer when you ask a related question.

Have a better answer? Share it in a comment.

All Courses

From novice to tech pro — start learning today.