Question

Rephrase of my previous question regarding line integrals:

Answer 1

Integrate h(x, y) = yi + xj over the circle of radius 5 centered at the origin traversed clockwise.

Answer 2

Ive worked out:
f(x,y)=(5sin(t), 5cos(t))
f'(x,y)=(5cos(t), -5sin(t))
Do I now dot these?

Answer 3

Im confused because in the formula it says something like you parameterize f(x,y) then plug that in for your formula.

So you'd have:
\(\int\limits_{0}^{1}f(x(t), y(t))ds\)

Where ds = \(\sqrt{(\frac{ dx }{ dt })^2+(\frac{ dy }{ dt })^2}\)

Im confused because everyone keeps telling me to dot the function with its derivative and I dont really see how that has anything to do with this formula...

Answer 4

I forgot to add above:

"integrating with respect to t"

Answer 5

I realize the answer is 0 as well, but I dont understand the process

Answer 6

The line integral is
\[\int_C \mathbf F\cdot \mathrm d\vec{r}=\int_Ch(x(t),y(t))\cdot\frac{\mathrm df(x(t),y(t))}{\mathrm dt}\,\mathrm dt\]That is to say, you're "dotting" \(h\) with the derivative of \(f\).

Answer 7

So for:
\[h(x,y)=xy^2i+2j\]
\[r(u)=e^ui+e^{-u}j\]
\[u \epsilon [0,1]\]

\[\int_C \mathbf F\cdot \mathrm d\vec{r}=\int_Ch(x(t),y(t))\cdot\frac{\mathrm df(x(t),y(t))}{\mathrm dt}\,\mathrm dt\]

\[(e^ue^{-2u}, 2) \cdot (?)\]

Heres where its throwing me off a bit, im not sure how to do the derivatives.

Answer 8

Do I take the derivative of h(x,y) then substitute my t parameters in and multiply by the derivative of the t parameters I plugged in?

Answer 9

This looks like a completely different problem. Let's just stick to the first for a moment.

You have a vector field \(h(x,y)=y\,\vec{i}+x\,\vec{j}\) to be integrated along the circular path of radius \(5\), so a sufficient parameterization might be \(f(t)=\sqrt5\cos t\,\vec{i}+\sqrt5\sin t\,\vec{j}\), with derivative \(\dfrac{\mathrm d}{\mathrm dt}f(t)=-\sqrt5\sin t\,\mathrm dt\,\vec{i}+\sqrt5\cos t\,\mathrm dt\,\vec{j}\). Then you can also write \(h(t)=\sqrt5\sin t\,\vec{i}+\sqrt5\cos t\,\vec{j}\).

Now the line integral would be
\[\begin{align*}
\int_C h(x,y)\cdot\mathrm df&=-\int_0^{2\pi}(\sqrt5\sin t,\sqrt5\cos t)\cdot(-\sqrt5\sin t,\sqrt5\cos t)\,\mathrm dt\\[1ex]
&=-5\int_0^{2\pi}(-\sin^2t+\cos^2t)\,\mathrm dt\\[1ex]
&=-5\int_0^{2\pi}\cos2t\,\mathrm dt\\[1ex]
&=-\frac{5}{2}\int_0^\pi \cos t\,\mathrm dt=\color{red}0
\end{align*}\]Note the negative sign, since \(C\) is traversed in the clockwise direction, though the ultimate result is independent of that.

Does all this make sense?

Answer 10

(Actually, ignore the \(\mathrm dt\)s in the derivative of \(f\), they shouldn't be there.)

Answer 11

Are you using the same path for the second problem?

Answer 12

"This looks like a completely different problem. Let's just stick to the first for a moment.

You have a vector field \(h(x,y)=y\,\vec{i}+x\,\vec{j}\) to be integrated along the circular path of radius \(5\), so a sufficient parameterization might be \(f(t)=\sqrt5\cos t\,\vec{i}+\sqrt5\sin t\,\vec{j}\), with derivative \(\dfrac{\mathrm d}{\mathrm dt}f(t)=-\sqrt5\sin t\,\mathrm dt\,\vec{i}+\sqrt5\cos t\,\mathrm dt\,\vec{j}\). Then you can also write \(h(t)=\sqrt5\sin t\,\vec{i}+\sqrt5\cos t\,\vec{j}\)."

Sure lets stick to the first problem.

why is f(t)=\(\sqrt{5}\cos(t)i+\sqrt{5}\sin(t)j\) if the function is \(yi+xj\) and x=rcos(t), and y=rsin(t). Am I thinking about this the wrong way?

Answer 13

Sorry about the delayed response

Answer 14

theres no root 5

Answer 15

one possible paramet for circle rad 5
x=5cost
y=5sint
-pi<t<pi

Answer 16

Oops, that's correct. I somehow misinterpreted the question as saying the path was \(x^2+y^2=5\), not \(x^2+y^2=5^2\). Luckily this doesn't change the outcome, though the coefficient in front of the integral would be \(25\) instead of a \(5\).

Answer 17

Basically replace every instance of \(\sqrt5\) with \(5\).

Answer 18

Just to address your initial confusion, the reason we're using the formula \(\int_C\mathbf F\cdot\mathrm d\vec{r}\) is because this is the line integral formulation over a vector field \(\mathbf F\). The formula you were intent on using, \(\int_Cf\,\mathrm ds\), only applies for scalar fields.

Answer 19

And for your second problem, the derivative you would be using is the one of the vector function \(\vec{r}(u)\).
\[\vec{r}(u)=e^u\,\vec{i}+e^{-u}\,\vec{j}\implies\frac{\mathrm d\vec{r}(u)}{\mathrm du}=e^u\,\vec{i}-e^{-u}\,\vec{j}\]Given the vector field \(h(x,y)=xy^2\,\vec{i}+2\,\vec{j}\), the line integral over whatever the path you're using is would be
\[\int_Ch\cdot\mathrm d\vec{r}=\int_C(e^ue^{-2u},2)\cdot\color{red}{(e^u,-e^{-u})}\,\mathrm du=\int_C(e^{-u},2)\cdot(e^u,-e^{-u})\,\mathrm du\]

Answer 20

i would recommend that you do your best to see what that vector \(\vec {h}(x, y) = yi + xj\) is actually doing over the path it is following. When does it ever point along the path it is following? Plot it at some key points....

And despite that stuff i posted on your other question, viz that the curl is not infallible, it is always worth checking out as a heads up. \(\partial_x \vec F - \partial_y \vec F = \mathbf{????}\). So you might be able to do the answer in yr head before you even start :-))

In terms of doing the integral, namely the first one .....

\(\vec {h}(x, y) = yi + xj\)

So you can write

\(\oint \color{red}{\left(\begin{matrix}y \\ x\end{matrix}\right)} \cdot \color{blue}{\left(\begin{matrix}dx \\ dy\end{matrix}\right) }= \oint y ~ dx + \oint x ~ dy\), if you like.


Or if we are going to parameterise, we should go polar as it is motion in a circle.

To polarise the red vector, we say \(x = \cos t, y = \sin t\), so the red vector is \(\left(\begin{matrix} \sin t \\ \cos t\end{matrix}\right)\)



for the blue vector, a circle radius 5, we should say \(x = 5 \cos t, y = 5 \sin t\).

This means: \(dx = -5\sin t ~ dt, dy = 5 \cos t ~ dt\), and it becomes:

\(\oint \color{red}{\left(\begin{matrix} \sin t \\ \cos t\end{matrix}\right)} \cdot 5 \color{blue}{\left(\begin{matrix} - \sin t \\ cos t\end{matrix}\right) } ~ dt \)

\(= 5 \oint -\sin^2 t + \cos^2 t ~ dt \)

\(= 5 \oint \cos 2t ~ dt , \qquad t \in [0, -2 \pi ]\)

this is where we kick in CW motion, ie in the \(\color{red}{-}2 \pi\) thing


\( = \frac{5}{2} \left[\sin 2t \right]_{0}^ {-2 \pi }\)

\(= 0\)