Is the integral ſ A. dr path independent if A = (2x - y)i + (x + y)j? Evaluate A dr around a unit circle centered at the origin. What is t

value of the integral if we take a clockwise direction?

The integral is path-independent if we can find a scalar function f such that grad(f ) = A. This requires

Take the first PDE and integrate both sides with respect to x to get

where g is assumed to be a function of y alone. Differentiating this with respect to x gives

which would mean g is *not* a function of only y, but also x, contradicting our assumption. So the integral is path-dependent.

Parameterize the unit circle (call it C) by the vector function,

with t between 0 and 2π.

Note that this parameterization takes C to have counter-clockwise orientation; if we compute the line integral of A over C, we can multiply the result by -1 to get the value of the integral in the opposite, clockwise direction.

Then

and the (counter-clockwise) integral over C is

and so the integral in the direction we want is -2π.

By the way, that the integral doesn't have a value of 0 is more evidence of the fact that the integral is path-dependent.

correct answer is infinitely many solutions

step-by-step explanation:

2(2x + 3) = -4 + 2(2x + 5)

solving the above equation:

4x + 6 = -4 +4x +10

combining like terms

4x -4x = -4+10-6

0=0

since both sides are 0=0, we end up into true statement it means it has infinite many solutions.

so correct answer is infinitely many solutions

yes. we do proportions in probabilities.