This function fails the conservative test, yet it is path independent?

flyingpig
Messages
2,574
Reaction score
1

Homework Statement



Show that the line integral is path independent and evaluate the integral

\int_C tan(y) dx + x sec^2 (y) dy

The Attempt at a Solution



\frac{\partial }{\partial x}(tan(y)) = 0 \neq 2xsec^2(y)tan(y) = x\frac{\partial }{\partial y}(sec^2(y))
 
Physics news on Phys.org
I believe your problem is stemming from the presence of the differentials, dx and dy. The general form of a line integral is the following:
<br /> \int_C M dx + N dy<br />

Where M & N relate to the quantities in your problem such that:

M = tan(y)
N = x sec^2 (y)

Now we know for a force to be conservative it must satisfy the following relationship:

\frac{\partial M }{\partial y} = \frac{\partial N }{\partial x}

Given this information, is your force conservative?

Hope this helps.

Joe
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top