# Thread: How i fond curl F at point (2, 3, 2)

1. ## How i fond curl F at point (2, 3, 2)

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2. Originally Posted by king imran
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personally, i would have used Soroban's method to do it...but CaptainBlack's method was so cool! and it seems to have much wider applications. I don't know what Wallis' formula is though, so you should be able to make a better choice than i can

3. Originally Posted by king imran
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Here's question 1

Recall that for a vector field, $\bold{F} = P \bold {i} + Q \bold {j} + R \bold {k}$, $\mbox{curl} \bold{F}$ is given by:

$\mbox{curl} \bold{F} = \nabla \times \bold{F}$

where $\times$ means the cross-product, and $\nabla = \frac {\partial}{\partial x} + \frac {\partial}{\partial y} + \frac {\partial}{\partial z}$

That is, $\mbox {curl} \bold{F} = \left| \begin{array}{ccc} \bold{i}&\bold{j}&\bold{k} \\ \partial / \partial x & \partial / \partial y & \partial / \partial z \\ P & Q & R \end{array} \right|$ $= \left( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z}\right)\bold{i}+ \left( \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} \right) \bold{j} + \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \bold{k}$

Here we have $\bold {F} = z e^{xy} \bold{i} + 2x^2 z \sin y \bold{j} + \left( 2x^2 + 3y^2 \right) \bold{k}$

and we want the curl at $(2,3,2)$

So, $\mbox{curl} \bold {F} = \left( \frac{\partial \left( 2x^2 + 3y^2 \right)}{\partial y} - \frac{\partial \left( 2x^2 z \sin y \right)}{\partial z}\right)\bold{i}+$ $\left( \frac{\partial \left(z e^{xy} \right)}{\partial z} - \frac{\partial \left( 2x^2 + 3y^2 \right)}{\partial x} \right) \bold{j} + \left( \frac{\partial \left( 2x^2 z \sin y \right)}{\partial x} - \frac{\partial \left( z e^{xy}\right)}{\partial y} \right) \bold{k}$

.............. $= \left( 6y - 2x^2 \sin y \right) \bold{i} + \left( e^{xy} - 4x \right) \bold{j} + \left( 4x z \sin y - xz e^{xy} \right) \bold {k}$

Since we want the curl at $(2,3,2)$, we simply plug in $x = 2 \mbox { , } y = 3 \mbox { , and } z = 2$. I leave that to you

4. Originally Posted by Jhevon
Here's question 1 ....
I hope you learned that from my incomplete tutorial.

5. Originally Posted by ThePerfectHacker
I hope you learned that from my incomplete tutorial.
i did actually