00:01
In this problem of line integrals, we have to find the work run y, a force field given here, which is f of x, y and z is equals to yzi plus x z j plus x, y, k, and along the line, i'm showing you the figure.
00:25
So this is the required figure and curve c is such that c is a line from 0 ,000 to, and 2 this is 5 3 and 2 so this point is 5 3 and 2 now we have to parameterize so c is just line straight line from 0 0 to 5 3 2 so we can say that rt would be equals to 5 t so this is 5 t plus 3 t plus 2 t this is i j and k respectively so this this is i j and k 2 tk this is 2 tk and the value of t is varying from 0 to 1 so this is value of t is varying from 0 to 1 now we have to find the value of d r so d r is equal to when we differentiate it so this is 5 i plus 3 j plus 2 k and this is d t now we have to calculate f f so, f of t, f of t would be here.
01:42
From here we say that x is equals to 5 t, y is equals to 3 t and z is equal to 2 t.
01:49
Putting the value y z, so this is 3 t multiplied with 2 t which is equals to 6 t square i, x that is 5 t multiplied with 2 t that is 10 t squared j plus x y that is 5 multiplied by 3 that is 15 t square k now we have to take the dot product for finding the work so work is f dot d r along the path c taking the dot product so this is equal to 60 square is multiplied with 5 so which is equal to 30 t square 10 t squared is multiplied with 3 so which is again 30 t square and 15 t square is multiplied with 2 t so this is only 2 so this value would be again 30 t square and d t now integration from 0 to 1.
02:38
So when we add all the terms, so this is equal to 90 t squared d t, integration from 0 to 1...