12 points in a two dimensional world suppose that the...

(12 points) In a two-dimensional world, suppose that the gravitational force between mass m and mass M is given by F(x,y)=(-mMGx)/((x^(2)+y^(2))^((3)/(2)))i+(-mMGy)/((x^(2)+y^(2))^((3)/(2)))j where G is a constant and mass M is fixed at the origin. Show that F is a conservative force field. Use this fact to compute the work \int_C F*dr done by the force field in moving the mass m along a smooth curve C from (-3,4) to (1,0).

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00:01 This question we are given here so in the first subpart we can write the value of the f is equal we can write 3x square 5 x karrid plus we can write that 4x cube here y cube and jeres to 5 karrid plus we can write that 4 x cube here y cube and j x to 5 karrid plus we can write that 5 x cube here y to 4 and j -rest to 4 k karat so this all this thing we are given here this is the value of the force vector and we want to find integration of close interval that is fdr that we want to find here and then we are given a coordinate the coordinate is 0 0 and 0 to the 1 1 1 1 so data already coordinate we are given here so let's find out values and here the value of the x which is equal to is equal to t and y which is equal to t squared and that which is equal to it we are given a t q values we are given here now using this value we can write so let's find out the first let's find out the f dot d r so here checking the integration of f dot d r which is equal we can write that integration of 0 to 1 integration of 0 to 1 that is 3 x square y to 4 and z is to 5 d x we can write integration of 0 to 1 that is 4 x q here y q 4 x2 multiple y q and j is to 5 that is ty plus we can write here you can write integration of at 0 to 1 integration of 0 to 1 that is 5 x q 5 x square 5 x cube not x square y to 4 and jr is to 4 so all this value we're given here and here taking the integration through d z now we can write that's value of the integration of the integration of jettres to 5 multiple we can write 1 divided by 3 plus we can write that is 4 x cube here y to 5 here 4 x2 here jr x2 here jr x2 5 and here we can write multiple we can write 1 divided by 4 plus we can write here 5 x cube and multiply we can write 1 divided by 4 plus we can write 1 we just simplify here in this question, we just simplify and we can write that value of the y is to 4 here, y is to 4, jarish to 5 plus we can write x cube, multiple we can write x to 5 plus we can write after simplification we can invent x cube whatever we can write vash to 4.

03:33 So this is the value of the integration of close interval...

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