Line integral to find r(t)

[nasi112]

Can I use this integral to verify?

$\displaystyle \int \cos(x - 3y) \ dx - 3\cos(x - 3y) \ dy$

If solve the integral, I will get

$\displaystyle \int_{\pi/4}^{3\pi/4} \cos(x) \ dx - 0 = 0$

For $\displaystyle C_1$, I still did not find $\displaystyle \mathbf{r}(t)$. How can I find it?

 

[blamocur]

Can I use this integral to verify?

$\displaystyle \int \cos(x - 3y) \ dx - 3\cos(x - 3y) \ dy$

If solve the integral, I will get

$\displaystyle \int_{\pi/4}^{3\pi/4} \cos(x) \ dx - 0 = 0$

For $\displaystyle C_1$, I still did not find $\displaystyle \mathbf{r}(t)$. How can I find it?

I meant reducing the integral along the curve, which you are yet to define, to a 1D case with $t$ as independent variable.
You have successfully computed 2 points in -- can you define a straight line which connects them?
I.e. define $\mathbf r(t)$ such that $\mathbf r(0)=\mathbf p$ and $\mathbf r(1) = \mathbf q$. You can start by rewriting the last identity in $x,y$ coordinates

 

[nasi112]

$\displaystyle \mathbf{r}(t) = \frac{\pi}{2}t + \frac{\pi}{4}$

If this is correct, do you mean by using this straight line equation I can verify the integral. How?

 

[blamocur]

$\displaystyle \mathbf{r}(t) = \frac{\pi}{2}t + \frac{\pi}{4}$

If this is correct, do you mean by using this straight line equation I can verify the integral. How?

You equation for $\mathbf r(t)$ does not specify a curve on a plane (i.e. a 2D function of 't'), but only a 1D function. You need to specify both X and Y components for $\mathbf r$.

 

[nasi112]

I tried. I couldn't figure it out. If I want to define a straight line connecting the two points I found, I get the slope zero.

 

[nasi112]

Should I do this?

$\displaystyle \mathbf{r}(t) = (\frac{\pi}{2}t + \pi, \ \frac{\pi}{4})$

 

[blamocur]

Looks perfect to me. BTW, nothing wrong with zero slope. Can you now plug it in into the original expression. I.e., can you translate $\mathbf F \cdot d\mathbf r$ to an expression containing $t$ and $dt$?

 

[nasi112]

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


Since $\displaystyle y = \frac{\pi}{4}, dy = 0$. So the integral will be


$\displaystyle \int_{C_1} \mathbf{F} \cdot d\mathbf{r} = \int_0^1 \cos(\frac{\pi}{2}t + \pi - \frac{3\pi}{4}) \ \frac{\pi}{2} \ dt = \frac{\pi}{2}\int_0^1 \cos(\frac{\pi}{2}t + \frac{\pi}{4}) \ dt$

 

[blamocur]

Neat! Can you work out the last integral to make sure it is 0? After that you are ready to solve the $C_2$ curve.

 

[nasi112]

$\displaystyle \frac{\pi}{2}\int_0^1 \cos(\frac{\pi}{2}t + \frac{\pi}{4}) =$ zero, perfectly.

Thank you a lot blamocur.

In the middle of our discussion, I was about to quit solving because I felt like my brain was spinning lol
I didn't imagine I will reach this far.

Let's find $\displaystyle C_2$, then I have a few comments.

$\displaystyle \sin(90^{\circ}) - \sin(180^{\circ}) = 1$

$\displaystyle x_1 - 3y_1 = \frac{\pi}{2}$

$\displaystyle x_2 - 3y_2 = \pi$

$\displaystyle x_1 = 2\pi$ and $\displaystyle y_1 = \frac{\pi}{2}$

$\displaystyle x_2 = 4\pi$ and $\displaystyle y_2 = \pi$

$\displaystyle \mathbf{r}(t) = (2\pi t + 2\pi, \ \frac{\pi}{2}t + \frac{\pi}{2})$

Verifying the integral.

$\displaystyle \int_{C_2} \mathbf{F} \cdot d\mathbf{r} = \int \cos(x - 3y) \ dx - 3\cos(x - 3y) \ dy$

$\displaystyle = \int_0^1 \left[\ \cos(2\pi t + 2\pi - \frac{3\pi}{2}t - \frac{3\pi}{2}) \ 2\pi - 3\cos(2\pi t + 2\pi - \frac{3\pi}{2}t - \frac{3\pi}{2}) \ \frac{\pi}{2} \ \right] \ dt$

$\displaystyle = \int_0^1 \left[ \ 2\pi\cos(\frac{\pi}{2}t - \frac{\pi}{2}) - \frac{3\pi}{2}\cos(\frac{\pi}{2}t - \frac{\pi}{2}) \ \right] \ dt =$ one, perfectly.

??

 

[blamocur]

$\displaystyle \frac{\pi}{2}\int_0^1 \cos(\frac{\pi}{2}t + \frac{\pi}{4}) =$ zero, perfectly.

Thank you a lot blamocur.

In the middle of our discussion, I was about to quit solving because I felt like my brain was spinning lol
I didn't imagine I will reach this far.

Let's find $\displaystyle C_2$, then I have a few comments.

$\displaystyle \sin(90^{\circ}) - \sin(180^{\circ}) = 1$

$\displaystyle x_1 - 3y_1 = \frac{\pi}{2}$

$\displaystyle x_2 - 3y_2 = \pi$

$\displaystyle x_1 = 2\pi$ and $\displaystyle y_1 = \frac{\pi}{2}$

$\displaystyle x_2 = 4\pi$ and $\displaystyle y_2 = \pi$

$\displaystyle \mathbf{r}(t) = (2\pi t + 2\pi, \ \frac{\pi}{2}t + \frac{\pi}{2})$

Verifying the integral.

$\displaystyle \int_{C_2} \mathbf{F} \cdot d\mathbf{r} = \int \cos(x - 3y) \ dx - 3\cos(x - 3y) \ dy$

$\displaystyle = \int_0^1 \left[\ \cos(2\pi t + 2\pi - \frac{3\pi}{2}t - \frac{3\pi}{2}) \ 2\pi - 3\cos(2\pi t + 2\pi - \frac{3\pi}{2}t - \frac{3\pi}{2}) \ \frac{\pi}{2} \ \right] \ dt$

$\displaystyle = \int_0^1 \left[ \ 2\pi\cos(\frac{\pi}{2}t - \frac{\pi}{2}) - \frac{3\pi}{2}\cos(\frac{\pi}{2}t - \frac{\pi}{2}) \ \right] \ dt =$ one, perfectly.

??

Looks good to me - congrats!

 

[nasi112]

Thank you blamocur. We would have saved a lot of time if you gave me the fish.

Now I have a few comments regarding this line integral.

1.
$\displaystyle \mathbf{r}(t)$ is bold, and it must be a vector. What is the correct notation to write it?
$\displaystyle \mathbf{r}(t) = (\frac{\pi}{2}t + \pi, \ \frac{\pi}{4})$
or
$\displaystyle \mathbf{r}(t) = < \ \frac{\pi}{2}t + \pi, \ \frac{\pi}{4} \ >$
or
$\displaystyle \mathbf{r}(t) = (\frac{\pi}{2}t + \pi)i + (\frac{\pi}{4})j$



2.
The question asked us to find non-closed path. Would it be wrong to choose these?
$\displaystyle \sin(405^{\circ}) - \sin(45^{\circ}) = 0$
or
$\displaystyle \sin(495^{\circ}) - \sin(45^{\circ}) = 0$
or
$\displaystyle \sin(360^{\circ}) - \sin(0^{\circ}) = 0$



3.
We chose two values for $\displaystyle x$ and $\displaystyle y$ and got $\displaystyle \mathbf{r_1}(t)$. If we choose again different values, we will get $\displaystyle \mathbf{r_2}(t)$. They are different. Can we consider them both as a valid solutions?



4.
We have taken our path orientation counterclockwise. Would it matter if we took it instead clockwise (Why?), like this:
$\displaystyle \sin(45^{\circ}) - \sin(135^{\circ}) = 0$.

 

[blamocur]

Thank you blamocur. We would have saved a lot of time if you gave me the fish.

Now I have a few comments regarding this line integral.

1.
$\displaystyle \mathbf{r}(t)$ is bold, and it must be a vector. What is the correct notation to write it?
$\displaystyle \mathbf{r}(t) = (\frac{\pi}{2}t + \pi, \ \frac{\pi}{4})$
or
$\displaystyle \mathbf{r}(t) = < \ \frac{\pi}{2}t + \pi, \ \frac{\pi}{4} \ >$
or
$\displaystyle \mathbf{r}(t) = (\frac{\pi}{2}t + \pi)i + (\frac{\pi}{4})j$



2.
The question asked us to find non-closed path. Would it be wrong to choose these?
$\displaystyle \sin(405^{\circ}) - \sin(45^{\circ}) = 0$
or
$\displaystyle \sin(495^{\circ}) - \sin(45^{\circ}) = 0$
or
$\displaystyle \sin(360^{\circ}) - \sin(0^{\circ}) = 0$



3.
We chose two values for $\displaystyle x$ and $\displaystyle y$ and got $\displaystyle \mathbf{r_1}(t)$. If we choose again different values, we will get $\displaystyle \mathbf{r_2}(t)$. They are different. Can we consider them both as a valid solutions?



4.
We have taken our path orientation counterclockwise. Would it matter if we took it instead clockwise (Why?), like this:
$\displaystyle \sin(45^{\circ}) - \sin(135^{\circ}) = 0$.

Speaking of "line integral": the problem asks to find a curve, of which a straight line is a special case. We picked a straight line simply to make verification easier.

1. I personally prefer the first notation, but different teachers might have different preferences. The 2nd notation might cause confusion with inequalities, in the 3rd you don't need parentheses around $\frac{\pi}{4}$.
2. I don't see anything wrong with any of your choices there.
3. Yes, they would both be valid as long as $f(x_1,y_1) - f(x_2,y_2)$ returns the required value. Moreover, you can choose different curves, i.e. $\mathbf r(t)$, for the same end points and still have the correct answer.
4. It would not matter.
Glad you got curious about this problem.

 

[nasi112]

Thanks blamocur.

in number 2, you said you don't see anything wrong. Isn't starting at $\displaystyle 45^{\circ}$ and ending at $\displaystyle 45^{\circ}$ ($\displaystyle 405^{\circ}$) consider a closed path? The same thing for starting at $\displaystyle 0^{\circ}$ and ending at $\displaystyle 360^{\circ}$. While starting at $\displaystyle 45^{\circ}$ and ending at $\displaystyle 495^{\circ}$ has passed through the starting point twice. I thought this is meant by closed path!

 

[blamocur]

Thanks blamocur.

in number 2, you said you don't see anything wrong. Isn't starting at $\displaystyle 45^{\circ}$ and ending at $\displaystyle 45^{\circ}$ ($\displaystyle 405^{\circ}$) consider a closed path? The same thing for starting at $\displaystyle 0^{\circ}$ and ending at $\displaystyle 360^{\circ}$. While starting at $\displaystyle 45^{\circ}$ and ending at $\displaystyle 495^{\circ}$ has passed through the starting point twice. I thought this is meant by closed path!

I would agree with you if the angle defined the point, i.e., (x,y) pair, uniquely. But from the statement of your problem there can be infinitely many (x,y) pairs all corresponding to the same angle: $x-3y = \alpha$. And for the curve to be closed its end points must be the same.

 

[nasi112]

Look at the question again. Find curves $\displaystyle C_1$ and $\displaystyle C_2$ that are not closed and satisfy the equation.

The statement "are not closed" here would be useless because any values we choose would be valid. Then, why did he alert us about it?

 

[blamocur]

Look at the question again. Find curves $\displaystyle C_1$ and $\displaystyle C_2$ that are not closed and satisfy the equation.

The statement "are not closed" here would be useless because any values we choose would be valid. Then, why did he alert us about it?

Can you tell me what is your definition of a closed point?

 

[nasi112]

Closed point or closed path? My own definition or the book definition? Anyway, I will suppose you mean a closed path and my own definition.

A closed path is when you start at a point regardless of the orientation, and move around while returning to the same starting point.

 

[blamocur]

Closed point or closed path? My own definition or the book definition? Anyway, I will suppose you mean a closed path and my own definition.

A closed path is when you start at a point regardless of the orientation, and move around while returning to the same starting point.

Agree. But you can have different starting and ending points for the same angle in your expression $f(x,y) = \sin (x-3y)$.

 

[nasi112]

Ineed, I can have. The question is again why they alerted us to have an open path while we will never have a closed path?