Closed curves and Line Integrals

[flyingpig]

Homework Statement

Given $$\mathbf{F} = \nabla f\; where \;f(x,y) = sin(x-2y)$$

Find a curve C that is not closed and satisfy the equation

$$\int_C \mathbf{F}\cdot dr = 0$$

The Attempt at a Solution

$$\nabla f = \;<cos(x - 2y),-2cos(x-2y)>$$

So to satisfy the dot product being 0 (I am hoping I can do this)

cos(x - 2y)dx = -2cos(x-2y)dy

dx = -2dy

$$y = \frac{-t}{2}+K$$

$$x = t$$

$$t \in [a,b]$$

I am just wondering, am I doing this correctly...?

Solutions

My book just took $$r(t) = t<\pi,\pi>$$

 

[Mark44]

Homework Statement

Given $$\mathbf{F} = \nabla f\; where \;f(x,y) = sin(x-2y)$$

Find a curve C that is not closed and satisfy the equation

$$\int_C \mathbf{F}\cdot dr = 0$$

The Attempt at a Solution

$$\nabla f = \;<cos(x - 2y),-2cos(x-2y)>$$

So to satisfy the dot product being 0 (I am hoping I can do this)

cos(x - 2y)dx = -2cos(x-2y)dy

You're not too far off, but you have a sign error in the line above. If you set the dot product to zero, you get cos(x - 2y) dx - 2cos(x - 2y)dy = 0, which is equivalent to cos(x - 2y)dx = +2cos(x-2y)dy.

Rather than solve a differential equation, I took a different tack. The integral can be zero for several reasons, one of which is along a path for which F = 0. You have F = <cos(x - 2y),-2cos(x-2y)> = cos(x - 2y) <1, -2>
cos(x - 2y) = 0 <==> x - 2y = π/2 + kπ, with k an integer
<==> 2y = x - π/2 - kπ
This is the same as saying
2y = x + π/2 + kπ
or
y = x/2 + π/4 + kπ/2

Since we're interested in just one path, we can take k = 0.

I don't understand where they came up with the answer you showed as the textbook answer. Possibly it's wrong.

dx = -2dy

$$y = \frac{-t}{2}+K$$

$$x = t$$

$$t \in [a,b]$$

I am just wondering, am I doing this correctly...?

Solutions

My book just took $$r(t) = t<\pi,\pi>$$

 

[flyingpig]

t ranges from 0 to 1

I show you from my book

[PLAIN]http://img593.imageshack.us/img593/6121/unledne.jpg