Line integral to find r(t)

[nasi112]

Let $\displaystyle \mathbf{F} = \bigtriangledown f$, where $\displaystyle f(x, y) = \sin(x - 3y)$. Find curves $\displaystyle C_1$ and $\displaystyle C_2$ that are not closed and satisfy the equation.



(a)

$\displaystyle \int_{C_1} \mathbf{F} \cdot d\mathbf{r} = 0$

$\displaystyle C_1: \mathbf{r}(t) \ = \ \ ?, \ 0 \leq t \leq 1$




(b)

$\displaystyle \int_{C_2} \mathbf{F} \cdot d\mathbf{r} = 1$

$\displaystyle C_2: \mathbf{r}(t) \ = \ \ ?, \ 0 \leq t \leq 1$




I know that $\displaystyle \mathbf{F} = \bigtriangledown f = \ <f_x, \ f_y> \ = \ <\cos(x - 3y), \ -3\cos(x - 3y)>$ and $\displaystyle d\mathbf{r} = \ <dx, \ dy>$

$\displaystyle \int_{C_1} \mathbf{F} \cdot d\mathbf{r} = \int \cos(x - 3y) \ dx - 3\cos(x - 3y) \ dy$ = 0

What should I do now?

 

[blamocur]

Actually, the integral of a gradient of a function along a curve has a nice property which usually allows to avoid explicit integration.
Hint: it is similar to integrating a derivative of a function in 1D.

 

[nasi112]

Hi blamocur.

Do you mean I should write directly $\displaystyle \sin(x - 3y) = 0$ and $\displaystyle \sin(x - 3y) = 1?$

What is the property?

I have a doubt that the property is to double the function $\displaystyle f(x,y)$ like this

$\displaystyle 2\sin(x - 3y) = 0$ and $\displaystyle 2\sin(x - 3y) = 1$

 

[blamocur]

Gradient theorem - Wikipedia

en.wikipedia.org

 

[nasi112]

Thank you blamocur.

According to the website you sent, the answer will be

$\displaystyle \int \bigtriangledown f(x,y) \ \cdot <dx, \ dy> \ = \ f(x,y)\Bigg|_{(x,y)_1}^{(x,y)_2}$

The points $\displaystyle (x,y)_1$ and $\displaystyle (x,y)_2$ are missing. What should I do now?

 

[blamocur]

Thank you blamocur.

According to the website you sent, the answer will be

$\displaystyle \int \bigtriangledown f(x,y) \ \cdot <dx, \ dy> \ = \ f(x,y)\Bigg|_{(x,y)_1}^{(x,y)_2}$

The points $\displaystyle (x,y)_1$ and $\displaystyle (x,y)_2$ are missing. What should I do now?

I like the first formula on that website -- can you rewrite this in the notation of your original post? I.e., using $\mathbf F, \mathbf r, C$.

 

[nasi112]

Do you mean this formula?

$\displaystyle \int_{\gamma} \nabla \varphi (\mathbf{r}) \cdot d\mathbf{r} = \varphi (\mathbf{q}) - \varphi (\mathbf{p})$

I would write it like this

$\displaystyle \int_{C_1} \mathbf{F} \cdot d\mathbf{r} = f (\mathbf{q}) - f (\mathbf{p}) = 0$

Is it correct?

 

[blamocur]

Do you mean this formula?

$\displaystyle \int_{\gamma} \nabla \varphi (\mathbf{r}) \cdot d\mathbf{r} = \varphi (\mathbf{q}) - \varphi (\mathbf{p})$

I would write it like this

$\displaystyle \int_{C_1} \mathbf{F} \cdot d\mathbf{r} = f (\mathbf{q}) - f (\mathbf{p}) = 0$

Is it correct?

Correct. Now, what is the relation between $C_1$, $\mathbf p$ and $\mathbf q$?

 

[nasi112]

$\displaystyle \mathbf{p}$ is the beginning of $\displaystyle C_1$ and $\displaystyle \mathbf{q}$ is the end of $\displaystyle C_1$.

 

[blamocur]

$\displaystyle \mathbf{p}$ is the beginning of $\displaystyle C_1$ and $\displaystyle \mathbf{q}$ is the end of $\displaystyle C_1$.

Exactly! Does this help you with finding $C_1$ and $C_2$ ? Hint: I believe it does

 

[blamocur]

You might have noticed this already, but I cannot resist commenting on the elegance of the Gradient Theorem: if you integrate the gradient of a function along a curve then the result depends only on the end points of the curve! Even better, it only depends on the function's (whose gradient is being integrated) values at the end points.
And here are some other (useless in the context of your current problem) notes:

• In physics, if $\mathbf F$ describes an electric field then $f$ describes that field's potential.
• If you study calculus even further you might learn that the Gradient Theorem is a special case of a more general Stokes Theorem.

 

[nasi112]

Exactly! Does this help you with finding $C_1$ and $C_2$ ? Hint: I believe it does

It should help me to find $\displaystyle C_1$, but I still didn't figure it out.

$\displaystyle f(\mathbf{p})$ and $\displaystyle f(\mathbf{q})$ are functions of $\displaystyle x$ and $\displaystyle y$

the beginning and the end of the path are given in terms of $\displaystyle t$, $\displaystyle 0 < t < 1.$

How to change the interval to $\displaystyle x$ and $\displaystyle y$? If I don't need to change it, what I should do?

 

[nasi112]

I have an idea, but I am not sure if it is correct.

$\displaystyle \sin(x_1 - 3y_1) = \sin(x_2 - 3y_2)$

$\displaystyle x_1 - 3y_1 = x_2 - 3y_2$

Now, I have an equation of four unknowns. I am stuck!

 

[blamocur]

It should help me to find $\displaystyle C_1$, but I still didn't figure it out.

$\displaystyle f(\mathbf{p})$ and $\displaystyle f(\mathbf{q})$ are functions of $\displaystyle x$ and $\displaystyle y$

the beginning and the end of the path are given in terms of $\displaystyle t$, $\displaystyle 0 < t < 1.$

How to change the interval to $\displaystyle x$ and $\displaystyle y$? If I don't need to change it, what I should do?

$f(\mathbf p)$ and $f(\mathbf q)$ are just values, not functions. You need to pick $\mathbf q$ and $\mathbf p$ so that $f(\mathbf q) - f(\mathbf p)$ matches the required result (0 for $C_1$, 1 for $C_2$), then remember that the actual curve does not matter as long as it connects your points.

 

[blamocur]

I have an idea, but I am not sure if it is correct.

$\displaystyle \sin(x_1 - 3y_1) = \sin(x_2 - 3y_2)$

$\displaystyle x_1 - 3y_1 = x_2 - 3y_2$

Now, I have an equation of four unknowns. I am stuck!

This equation has multiple solutions (infinite number in fact), you just have to pick any that work. But you have to remember that the curves cannot be closed.

 

[blamocur]

This equation has multiple solutions (inifinite number in fact), you just have to pick any that work. But you have to remember that the curves cannot be closed.

BTW, when two sines are equal it is enough but not necessary for their arguments to be equal. But equal arguments will work too as long as the points are different, i.e., the curve is not closed.

 

[nasi112]

This equation has multiple solutions (infinite number in fact), you just have to pick any that work. But you have to remember that the curves cannot be closed.

$\displaystyle \sin(135^{\circ}) - \sin(45^{\circ}) = 0$

Now I know the beginning point of the curve is $\displaystyle x_1 - 3y_1 = \frac{\pi}{4}$ and the end point is $\displaystyle x_2 - 3y_2 = \frac{3\pi}{4}$. But I still have four unknowns. How to proceed?

 

[blamocur]

$\displaystyle \sin(135^{\circ}) - \sin(45^{\circ}) = 0$

Now I know the beginning point of the curve is $\displaystyle x_1 - 3y_1 = \frac{\pi}{4}$ and the end point is $\displaystyle x_2 - 3y_2 = \frac{3\pi}{4}$. But I still have four unknowns. How to proceed?

Can you find any pairs of (x,y) which satisfy the equation? Just because there is more than one solution does not mean one should be difficult to find. E.g., set $x_1$ and $x_2$ to some simple values and compute $y_1$ and $y_2$. Conversely, set $y_1,y_2$ to something simple, like 0, and compute $x_1,x_2$.

 

[nasi112]

$\displaystyle x_1 = \pi$ and $\displaystyle y_1 = \frac{\pi}{4}$ will satisfy $\displaystyle \frac{\pi}{4}$


$\displaystyle x_2 = \frac{3\pi}{2}$ and $\displaystyle y_2 = \frac{\pi}{4}$ will satisfy $\displaystyle \frac{3\pi}{4}$

Is this correct? What can I do next?

 

[blamocur]

Looks good to me. For the next you can find an answer for $C_2$.

Also, if I were you I'd try to verify the result. For example, if for $C_1$ you use points $\left(\frac{\pi}{4},0\right)$ and $\left(\frac{3\pi}{4},0\right)$ you can then connect them by a simple straight line and compute the corresponding integral explicitly. If you get 0 you know you got it right. You can do the same for your answer, but the line is slightly more complex.