Function whose minimum is independent of the input?I have a question that asks me to show that the functionalI(x)=∫{x1}x2(x2+3y2)y′+2xydxhas an extremal that is independent of the choice of y that joins two arbitrary points (x1,y1),(x2,y2).So I tried solving for the extremal, which give me the euler lagrange equation:6yy′+2x=2x+6yy′That's not an error it seems that ddx(∂F∂y′)=∂F∂y.Ok then no wonder that the path doesn't matter. But now I am trying to compute the value of the functional for the two points, I tried a straight line between the 2, but this is giving me a messy expression, it's computable but ugly:J(y)=∫{x1}x2s(x2+3(sx+k)2)+2x(sx+k)dx(where s and k are the linear coefficients of the line connecting the two points).

Question

Function whose minimum is independent of the input?I have a question that asks me to show that the functionalI(x)=∫{x1}x2(x2+3y2)y′+2xydxhas an extremal that is independent of the choice of y that joins two arbitrary points (x1,y1),(x2,y2).So I tried solving for the extremal, which give me the euler lagrange equation:6yy′+2x=2x+6yy′That&#039;s not an error it seems that ddx(∂F∂y′)=∂F∂y.Ok then no wonder that the path doesn&#039;t matter. But now I am trying to compute the value of the functional for the two points, I tried a straight line between the 2, but this is giving me a messy expression, it&#039;s computable but ugly:J(y)=∫{x1}x2s(x2+3(sx+k)2)+2x(sx+k)dx(where s and k are the linear coefficients of the line connecting the two points).

Melody Gamble · Accepted Answer

You can actually bypass certain variation considerations for this one, but its good to treat it both ways.Step 1Notice I is a line integral. Multiplied out, you are integrating (x2+3y2)dy+(2xy)dx. This has the form of a vector field times a distance element, F→⋅ds→. By the Fundamental Theorem of Line Integrals, the integral is path independent if F→=∇V for some scalar V. A quick test for that is to take the x derivative of the term being multipled by dy and the y derivative of the term being multiplied by dx. If these derivatives are equal, then you can find the necessary scalar potential and the Theorem applies. In the background this follows because the curl of a gradient is zero and Stoke&#039;s Law.Step 2To find the potential, integrate the term multiplied by dy with respect to y. To that integral add an arbitrary function g(x), then take the x derivative of the resulting sum. Subtract the term being multiplied by dx and set what remains to zero. Solve the resulting differential equation for g(x).In this case g(x)=0 and V(x,y)=x2y+y3 as pointed out by one of the comments, so the integral is always V(x2,y2)−V(x1,y1) regardless of path.

Function whose minimum is independent of the input? I

Answered question

Answer & Explanation

New Questions in Calculus 2