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Question
Find the work done by the force field F in moving an object from P to Q.
F(x,y)=e^-y i-xe^-y j
P(0,1), Q(2,0)
Hello Mandy, The definition of work is ∫F(x(t),y(t)).r'(t)dt.
Where r(t) is the path from point a to point b.
Let's 1st make parametrization from our line from <0,1> to <2,0> :
r(t) = <0,1> + [<2,0> - <0,1>]t = <2t,1-t>. This is the line
equation. Here's another way to find parametrization form the line:
Since this line passes through the points (0,1) & (2,0) , then it's
equation will be y=-(1/2)x+1 so the trajectory describing the line
will be r(t)=[t,1-(1/2)t]. We will get the same work result form
both. Now, as for the work :
r'(t)=<2,-1>. Now, let's perform the scalar product between r'(t) &
F(x,y) : F(x,y).r'(t)= <e^-y,-xe^-y>.<2,-1>=2e^(-y)+xe^(-y).
We substitute our parametrization in x & y we get :
[x(t),y(t)]=[2t,1-t] , then our integral will become :
∫2e^(t-1)+2te^(t-1)dt {t goes from 0 to 1}.
∫2e^(t-1)+2te^(t-1)dt = 2∫e^(t-1)+2∫te^(t-1)dt.
I'll leave it to you as an exercise to proceed.
Alon.
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Alon Mandes
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