Thread: that guy [TCTI v 8] View Single Post
#6798
February 11th, 2013 (8:32 AM).
 Gabri m8 Silver Tier Join Date: Oct 2006 Location: Portugal Age: 23 Gender: Male Posts: 3,727
If you can do simple math, tell me what is the integral of F·dr (dot product between the vectors F and dr), where F is the vectorial function whose coordinates are given by (10x^4 - 2x*y^3 ; -3x^2*y^2) and dr is the position differential vector of coordinates (dx; dy) along the path given by x^4 - 6x*y^3=4y^2, from the starting point (0;0) to the final point (2;1).

It's quite simple actually.
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The curl of the function F is zero (d/dy (10x^4 - 2x*y^3) = -6x*y^2 = d/dx (-3x^2*y^2)), and as such the path you integrate the function along doesn't matter. Find the scalar function G whose derivative with respect to x is the first coordinate of F and whose derivative with respect to y is the second coordinate of F ---> G=2x^5 - x^2*y^3. Just do G(2;1)-G(0,0) and done! The final result is 60.

Blaarh, I tried to translate the names of that stuff as best as I could.

/end show-off

216798

Also, I should do the same as Went.