# Conservative fields

Determine whether the vector field is conservative. F(x,y)=(e^x sin(y) + tan(y)i + (e^x cos(y) + sec^2(y)j.
f(x1,y1) = [inte]A to B (F * T) ds = [inte]A to B(e^x sin(y) + tan(y)dx + (e^x cos(y) + sec^2(y)dy = [inte]1 to 0 (e^x sin(y) + tan(y)dx + (e^x cos(y) + sec^2(y)dy

I am lost from here can anyone help me solve from here plz?
Dx

## Answers and Replies

Homework Helper
You are given that F=(e^x sin(y) + tan(y))i + (e^x cos(y) + sec^2(y))j.

The rest of the formula you gave is not necessary. F is the vector field. saying that it is "conservative" (that's really a physics term- the mathematics term is "exact") means that it is the gradient of some scalar function. In other words, is there a function f(x,y) such that F= grad(f)= (df/dx)i+ (df/dy)j.

If df/dx= e^x sin(y)+ tan(y) then

df/dxdy= e^x cos(y)+ sec^2(y)

If df/dy= e^x cos(y)+ sec^2(y)

df/dydx= e^x cos(y). Notice that df/dxdy is not the same as df/dy/dx: but mixed derivatives have to be equal! F is not the gradient of any function, this vector field is not conservative (exact).

By the way, if it were F= (e^x sin(y) + tan(y))i + (e^x cos(y) + xsec^2(y))j then it would be conservative. Do you see the difference?

Antiproton
Or even more directly, if you take the curl of the vector field (something we physicsts do all the time), you'll find conservative field curls are equal to 0. Of course, the curl is little computationally intensive, but the answer is unmistakeable and there is a great deal of satisfaction when you get through the thing 