Find the work done by the force field F(x,y)=x^{2i}+ye^{xj} on



Answered question


Find the work done by the force field F(x,y)=x2i+yexj on a particle that moves along the parabola x=y2+1 from (1, 0) to (2, 1)

Answer & Explanation

Jack Maxson

Jack Maxson

Beginner2022-01-07Added 25 answers

Step 1 
By definition 
The line integral along a path is the work done along that path.
That is Work done =CFdr 
Step 2 
Note that dr= dx i+ dy j 
Given that F(x,y)=x2i+yexj 
Therefore, Work done =C(x2i+yexj)( dx i+ dy j) 
Work done Cx2 dx +yex dy  
Step 3 
Given that C:x=y2+1 
Therefore, y=x1  and   dy = dx 2x1 
In order to completely express an integral in terms of x, swap out the values of y and dy.
Work done =Cx2 dx +x1ex dx 2x1 
Work done =Cx2+12ex dx  
Since x increases from 1 to 2 as we proceed from (1,0) to (2,1), we shall integrate from 1 to 2 with respect to x.
Work done =12x2+12ex dx  

Jeremy Merritt

Jeremy Merritt

Beginner2022-01-08Added 31 answers

Call the parabola P, parameterized by r(y)=y2+1,y with 0y1. Then the work done by f(x,y) along P is


Expert2022-01-11Added 613 answers

Step-by-step explanation:
By definition:
Work done along the path is the line integral along that path denoted as:
Work done =CFdr
Note: dr=dxi+dyj
Given that: F(x,y)=x2i+yexj
F(x,y) dot product with dr=x2dx+yexdy
Work done =C(x2dx+yexdy) Eq1
Given that CL y=x1
Replace the value of y and dy in Eq 1
Workdone =C(x2+ex2)dx
Limits of x are 1 to 2 respectively
Workdone =12(x2+ex2)dx
Evaluate limits to obtain
Work Done =73+e2e2

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