Anwar Saiah
asked on
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!)
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!)
ASKER
the problem is with 1/y when y=0.
the fact that e^0=1 never skipped my mind!
the fact that e^0=1 never skipped my mind!
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.
"the fact that e^0=1 never skipped my mind! "
good
I missed the i/y. let me think more.
good
I missed the i/y. let me think more.
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.
Use the equation of the ellipse to eliminate all the y in the work integral.
ASKER
Please advise
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ASKER
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) ?
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) ?
Applying Green's Theorem may make it easier to solve the integral.
ASKER
Well the orientation is clockwise so I cannot use Green's theorem! can I ?
ASKER
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ASKER
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.
I will upload it soon.
Thank you all for your help.
>> 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.
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.
remember that e^0 = 1
calculate the work done by the x component separately then the work done by the y component.