- #1

joshmccraney

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Does anyone know of a formula for taking the dot product in other non-rectangular coordinate systems, or should I always convert to rectangular?

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- #1

joshmccraney

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Does anyone know of a formula for taking the dot product in other non-rectangular coordinate systems, or should I always convert to rectangular?

- #2

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How are you defining the dot product?Let's take two orthogonal curves in polar coordinates of the form ##\langle r,\theta \rangle##, say ##\langle r,0\rangle## and ##\langle r,\pi/2\rangle##. Cleary both lines are orthogonal, but the dot product is not zero. This must be since I do not have these vectors in the form ##\langle x,y\rangle##, but can someone explain conceptually where my mistake is?

- #3

joshmccraney

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Sorry, I changed the question stem slightly. I'm defining it as ##(a \hat x + b\hat y)\cdot (c \hat x + d\hat y) = ac + bd##.How are you defining the dot product?

- #4

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I'd convert to rectangular. It's fairly easy to generate the formula for polar coordinates by doing this in general.

Does anyone know of a formula for taking the dot product in other non-rectangular coordinate systems, or should I always convert to rectangular?

- #5

joshmccraney

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Thanks!

- #6

joshmccraney

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Lets say I have a vector of the form ##r \hat r + \theta \hat \theta##. We know ##\hat r = \cos\theta \hat x + \sin\theta\hat y## and ##\hat\theta = -\sin\theta\hat x + \cos\theta\hat y##. Substituting these expressions into ##r \hat r + \theta \hat \theta## yields ##(r\cos\theta-\theta\sin\theta)\hat x + (r\sin\theta+\theta\cos\theta)\hat y##. But then if I take the two lines ##r \hat r ## and ##r \hat r + \frac{\pi}{2} \hat \theta## and dot them using the above formula, I do not get zero. What am I doing wrong?

- #7

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A vector in polar coordinates is given by ##r \hat{r}##. There's no ##\hat{\theta}## involved.

Lets say I have a vector of the form ##r \hat r + \theta \hat \theta##. We know ##\hat r = \cos\theta \hat x + \sin\theta\hat y## and ##\hat\theta = -\sin\theta\hat x + \cos\theta\hat y##. Substituting these expressions into ##r \hat r + \theta \hat \theta## yields ##(r\cos\theta-\theta\sin\theta)\hat x + (r\sin\theta+\theta\cos\theta)\hat y##. But then if I take the two lines ##r \hat r ## and ##r \hat r + \frac{\pi}{2} \hat \theta## and dot them using the above formula, I do not get zero. What am I doing wrong?

You don't need the formula for ##\hat{r}##. Instead, you should start with:

##\vec{a} \cdot \vec{b} = a_x b_x + a_y b_y##

Now, can you relate ##a_x## etc. to your polar coordinates?

- #8

joshmccraney

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Wow, obviously!!!!! Thanks so much, I'm good to go! I can finish what you've written if you'd like but it makes perfect sense now.A vector in polar coordinates is given by ##r \hat{r}##. There's no ##\hat{\theta}## involved.

- #9

Chestermiller

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$$\hat{r}(0)=\hat{i}_x$$

$$\hat{r}(\pi/2)=\hat{i}_y$$

$$\hat{r}(\pi/2)=\hat{i}_y$$

- #10

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By the way, hopefully, once you get the answer it might look familiar.Wow, obviously!!!!! Thanks so much, I'm good to go! I can finish what you've written if you'd like but it makes perfect sense now.

- #11

joshmccraney

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Yep, it all makes perfect sense! Can't believe I was parameterizing a vector in polar incorrectly!!!By the way, hopefully, once you get the answer it might look familiar.

- #12

jedishrfu

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The polar caps have frozen many an unprepared explorer!

- #13

joshmccraney

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At this point can't I just directly take $$\langle \sin\beta,-\cos\beta \rangle \cdot \left( 2r \theta^3 (\cos\theta \hat x + \sin\theta\hat y) + 3r\theta^2 (-\sin\theta\hat x + \cos\theta\hat y) \right)$$ and then set ##\theta = \beta##?

- #14

Chestermiller

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You left off the unit vectors on n.

At this point can't I just directly take $$\langle \sin\beta,-\cos\beta \rangle \cdot \left( 2r \theta^3 (\cos\theta \hat x + \sin\theta\hat y) + 3r\theta^2 (-\sin\theta\hat x + \cos\theta\hat y) \right)$$ and then set ##\theta = \beta##?

- #15

joshmccraney

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I think the unit normal vector is correct but good call, the "unit" tangent vector I list is actually a tangent vector.You left off the unit vectors on n.

- #16

Chestermiller

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I don't know what you are saying.I think the unit normal vector is correct but good call, the "unit" tangent vector I list is actually a tangent vector.

- #17

joshmccraney

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I mentioned a unit tangent vector in post 13 but it's actually just a tangent vector (not with norm 1).I don't know what you are saying.

- #18

Chestermiller

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You’re not trying to het the normal derivative of f, are you?I mentioned a unit tangent vector in post 13 but it's actually just a tangent vector (not with norm 1).

- #19

joshmccraney

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No, ##\hat n## is a unit normal to a line with slope ##\tan \beta##.You’re not trying to het the normal derivative of f, are you?

- #20

mathwonk

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just (r1r2)(cos(t1)cos(t2 + sin(t1)sin(t2)) = r1r2.cos(t2 - t1) = the product of the lengths of the vectors by the cosine of the angle t2-t1 between them, which is always the dot product, in any coordinate system.

- #21

FactChecker

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Regardless of how two vectors are represented, their dot product is defined as the product of their magnitudes times the cosine of the angle between them. The dot product of your example using their polar coordinate form is ## r^2 \cos \frac \pi 2 = 0 ##. The result agrees with the fact that the vectors are orthogonal.

Does anyone know of a formula for taking the dot product in other non-rectangular coordinate systems, or should I always convert to rectangular?

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