6761hash 11 Aug 2017, 08:27
How is it done in a simpler example.
(x^2+y^2+z^2) times [xihat+yjhat+zkhat divided by sqrt (x^2+y^2+z^2)]
What is the derivative of a unit vector? - Quora
I have looked at examples and forensically concluded the result is either 1 or x, when deriving wrt x.
X
How is it done in a simpler example. (x^2+y^2+z^2) times [xihat+yjhat+zkhat divided by sqrt (x^2+y^2+z^2)]
Relevant page <a href="https://www.quora.com/What-is-the-derivative-of-a-unit-vector">What is the derivative of a unit vector? - Quora</a> What I've done so far I have looked at examples and forensically concluded the result is either 1 or x, when deriving wrt x.
Murray 11 Aug 2017, 22:54
Hello 6761
This is actually beyond the scope of the IntMath Forum, but we'll have a go.
The most common kind of differentiation problem involving vector quantities is wrt `t`. However, with respect to `x`, I don't believe we'll end up with a simple `1` or `x`.
Let's just do the simpler case before including the normalisation expressions. Can you differentiate this part wrt `x`?
`x hat i + y hat j + z hat k`
(BTW, click on the "Show code" button to see how to enter that expression so it comes out looking like "real" mathematics.)
X
Hello 6761 This is actually beyond the scope of the IntMath Forum, but we'll have a go. The most common kind of differentiation problem involving vector quantities is wrt `t`. However, with respect to `x`, I don't believe we'll end up with a simple `1` or `x`. Let's just do the simpler case before including the normalisation expressions. Can you differentiate this part wrt `x`? `x hat i + y hat j + z hat k` (BTW, click on the "Show code" button to see how to enter that expression so it comes out looking like "real" mathematics.)
6761hash 11 Aug 2017, 23:34
If deriv of 2x =2 then is the deriv of xi hat = i ?
X
If deriv of 2x =2 then is the deriv of xi hat = i ?
6761hash 12 Aug 2017, 00:04
Let's cut to the chase. What is the deriv of xi hat wrt x? Taking deriv of i,j,k is not addressed in calculus books. i,j,k are not numbers. In Cartesian coord x and xi hat are the same thing. So maybe I should just ignore the i,j,k and derive it as x,y,z.
So deriv of x,y,z wrt x is 1. Yes?
X
Let's cut to the chase. What is the deriv of xi hat wrt x? Taking deriv of i,j,k is not addressed in calculus books. i,j,k are not numbers. In Cartesian coord x and xi hat are the same thing. So maybe I should just ignore the i,j,k and derive it as x,y,z. So deriv of x,y,z wrt x is 1. Yes?
Murray 12 Aug 2017, 00:18
The answer to your final question is no (in the general sense). I get the feeling you are not sure what the expression
`x hat i + y hat j + z hat k`
actually means. Can you tell me your understanding of it? (I think this is where you are having a block with this question. Let's sort that out first.)
You are right, there is no derivative for `hat i`, since it's not a variable quantity.
You are encouraged to use the math entry system so others reading this will find the mathematics easier to follow.
X
The answer to your final question is no (in the general sense). I get the feeling you are not sure what the expression `x hat i + y hat j + z hat k` actually means. Can you tell me your understanding of it? (I think this is where you are having a block with this question. Let's sort that out first.) You are right, there is no derivative for `hat i`, since it's not a variable quantity. You are encouraged to use the math entry system so others reading this will find the mathematics easier to follow.
6761hash 12 Aug 2017, 09:55
Since i cannot find out how to do derivation of x ihat, I have come up with my own extrapolation.
d/dx (xi hat, yj hat, zk hat)=1+0+0=1
This is the only way the following derivation could be true.
A=(x^2+y^2+z^2) times [(x ihat+ y jhat+ z khat)/sqrt (x^2+y^2+z^2)]
DAx
------ =(x^2 +y^2+z^2)^1/2 + x(1/2)(x^2+y^2+z^2)^-1/2 (2x)
dx
Agreed?
X
Since i cannot find out how to do derivation of x ihat, I have come up with my own extrapolation. d/dx (xi hat, yj hat, zk hat)=1+0+0=1 This is the only way the following derivation could be true. A=(x^2+y^2+z^2) times [(x ihat+ y jhat+ z khat)/sqrt (x^2+y^2+z^2)] DAx ------ =(x^2 +y^2+z^2)^1/2 + x(1/2)(x^2+y^2+z^2)^-1/2 (2x) dx Agreed?
6761hash 12 Aug 2017, 10:10
It seems the derivative of sin(pi*x)*ihat is the same as the sin (pi*x), further proof that the derivative of xihat wrt x is 1.
Does this seem correct?
X
It seems the derivative of sin(pi*x)*ihat is the same as the sin (pi*x), further proof that the derivative of xihat wrt x is 1. Does this seem correct?
6761hash 12 Aug 2017, 13:59
Thanks.
My interpretation is it means
in Cartesian coordinates, qq i hat qq is a unit vector along the x axis, ??and x is the magnitude in that direction, and the same for j and k wrt y and z.
3 questions:
1. I'm not positive though if qq xi hat qq is a unit vector or a basis vector (or both). Could you please clarify?
2. If there is no derivative of xi,yj,zk then why have them at all in the numerator of the equation to be derived I added earlier? To me it is like saying differentiate (undifferentiable terms/sqrt(x^2+y^2+z^2) which is meaningless.
3. Why complicate matters with i,j,k when you could merely say 7x instead of 7 x i hat and they mean the exact same thing.
btw tried using the math system to type a hat but couldn't find the method.
X
Thanks. My interpretation is it means in Cartesian coordinates, qq i hat qq is a unit vector along the x axis, ??and x is the magnitude in that direction, and the same for j and k wrt y and z. 3 questions: 1. I'm not positive though if qq xi hat qq is a unit vector or a basis vector (or both). Could you please clarify? 2. If there is no derivative of xi,yj,zk then why have them at all in the numerator of the equation to be derived I added earlier? To me it is like saying differentiate (undifferentiable terms/sqrt(x^2+y^2+z^2) which is meaningless. 3. Why complicate matters with i,j,k when you could merely say 7x instead of 7 x i hat and they mean the exact same thing. btw tried using the math system to type a hat but couldn't find the method.
Murray 12 Aug 2017, 21:24
Hi again.
The expression `x hat i + y hat j + z hat k` is a position vector. The `hat i`, `hat j` and `hat k` are unit vectors in the x-, y and z directions respectively.
The terms x-, y and z in the expression stand for values multiplying each of those unit vectors.
See the 4 videos in this Khan Academy lesson for more on this.
As far as the derivative is concerned, if y and z are just constants, then, yes, the derivative of the expression will be just `hat i` (or written as a coordinate vector, `(1, 0, 0).`
But let's see what happens if (say) `y` is a function of `x` (say `y = 1- x^2`, and `z` is a different function of `x` (say `z = x^3 - 2x`).
Then we would have:
`d/dx(x hat i + y hat j + z hat k)`
`= d/dx(x hat i + (1 - x^2) hat j + (x^3 - 2x) hat k)`
`= hat i - 2x hat j + (3x^2 - 2) hat k`
So I hope you can see the derivative is not just simply `1` for all cases. If however, y and z are constants, then yes, the derivative will be a unit vector in the `x`-direction, written as `hat i`.
Hope it makes sense now.
BTW, when you are logged in, you can see at the bottom of any post a "Show code" button. When you click on that, you'll be able to see how each math expression has been entered, including hat.
X
Hi again. The expression `x hat i + y hat j + z hat k` is a <b>position</b> vector. The `hat i`, `hat j` and `hat k` are <b>unit vectors</b> in the <i>x-</i>, <i>y</i> and <i>z</i> directions respectively. The terms <i>x-</i>, <i>y</i> and <i>z</i> in the expression stand for values multiplying each of those unit vectors. See the 4 videos in this <a href="https://www.khanacademy.org/math/multivariable-calculus/multivariable-derivatives/position-vector-functions/v/position-vector-valued-functions">Khan Academy lesson</a> for more on this. As far as the derivative is concerned, if <i>y</i> and <i>z</i> are just constants, then, yes, the derivative of the expression will be just `hat i` (or written as a coordinate vector, `(1, 0, 0).` But let's see what happens if (say) `y` is a function of `x` (say `y = 1- x^2`, and `z` is a different function of `x` (say `z = x^3 - 2x`). Then we would have: `d/dx(x hat i + y hat j + z hat k)` `= d/dx(x hat i + (1 - x^2) hat j + (x^3 - 2x) hat k)` `= hat i - 2x hat j + (3x^2 - 2) hat k` So I hope you can see the derivative is not just simply `1` for all cases. If however, <i>y</i> and <i>z</i> are constants, then yes, the derivative will be a unit vector in the `x`-direction, written as `hat i`. Hope it makes sense now. BTW, when you are logged in, you can see at the bottom of any post a "Show code" button. When you click on that, you'll be able to see how each math expression has been entered, including hat.
6761hash 13 Aug 2017, 10:59
2 questions please...
1. But you see how after taking the derivative you ended up with i, j and k in your answer. In the derivative below there is no i,j, or k in the final answer.
A=(x^2+y^2+z^2) times [(x ihat+ y jhat+ z khat)/sqrt (x^2+y^2+z^2)]
DAx
------ =(x^2 +y^2+z^2)^1/2 + x(1/2)(x^2+y^2+z^2)^-1/2 (2x)
dx
What is the difference in your example of a derivative and mine whereby you have i, j or k in your answer and I do not? If you can explain that to me you effectively have answered my question.
Could it be that the i,j,k were left out of my answer by the author of the book and it could equally be written as such?
=(x^2 ihat +y^2 jhat +z^2 khat)^1/2 + xihat(x^2 ihat +y^2 jhat+z^2 khat)^)^-1/2 (2xi hat)
2. Is A (vector)=Ax (vector)+Yz (vector) any less clear to the reader as the redundandcy of A (vector)=Ax ihat (vector)+Yz jhat (vector)?
X
2 questions please... 1. But you see how after taking the derivative you ended up with i, j and k in your answer. In the derivative below there is no i,j, or k in the final answer. A=(x^2+y^2+z^2) times [(x ihat+ y jhat+ z khat)/sqrt (x^2+y^2+z^2)] DAx ------ =(x^2 +y^2+z^2)^1/2 + x(1/2)(x^2+y^2+z^2)^-1/2 (2x) dx What is the difference in your example of a derivative and mine whereby you have i, j or k in your answer and I do not? If you can explain that to me you effectively have answered my question. Could it be that the i,j,k were left out of my answer by the author of the book and it could equally be written as such? =(x^2 ihat +y^2 jhat +z^2 khat)^1/2 + xihat(x^2 ihat +y^2 jhat+z^2 khat)^)^-1/2 (2xi hat) 2. Is A (vector)=Ax (vector)+Yz (vector) any less clear to the reader as the redundandcy of A (vector)=Ax ihat (vector)+Yz jhat (vector)?
Murray 13 Aug 2017, 22:32
I think I need to see the original context to work out what the notation actually means. (You can upload an image in this forum). It's not clear to me what Ax refers to.
My derivative above is for the expression `x hat i + y hat j + z hat k`, which I said we would do first and talk about the normalized case later.
The length of the vector `x hat i + y hat j + z hat k` is `sqrt(x^2+y^2+z^2)`.
So the normalized vector is:
`(x hat i + y hat j + z hat k)/sqrt(x^2+y^2+z^2)` `= (sqrt(x^2+y^2+z^2)(x hat i + y hat j + z hat k))/(x^2+y^2+z^2)` `= (x hat i + y hat j + z hat k)xx(x^2+y^2+z^2)^(-1/2)`
I'm guessing the text book's author is using `Ax` to mean "the `x`-component of the normalized vector `A`". So that would mean
`Ax = (x)xx(x^2+y^2+z^2)^(-1/2)`
In which case,
`(dAx)/dx = x (-1/2)(x^2+y^2+z^2)^(-3/2)(2x + 2y(dy/dx + 2z dz/dx))` ` + (x^2+y^2+z^2)^(-1/2)(1)`
If `y` and `z` are constants, then this would give us:
`(dAx)/dx = x (-1/2)(x^2+y^2+z^2)^(-3/2)(2x)` ` + (x^2+y^2+z^2)^(-1/2)`
`= -x^2(x^2+y^2+z^2)^(-3/2)` ` + (x^2+y^2+z^2)^(-1/2)`
This is not the answer you put (which is hard to read - please use the math entry system), but hopefully gives you a start on following what the author is doing.
X
I think I need to see the original context to work out what the notation actually means. (You can upload an image in this forum). It's not clear to me what Ax refers to. My derivative above is for the expression `x hat i + y hat j + z hat k`, which I said we would do first and talk about the normalized case later. The length of the vector `x hat i + y hat j + z hat k` is `sqrt(x^2+y^2+z^2)`. So the normalized vector is: `(x hat i + y hat j + z hat k)/sqrt(x^2+y^2+z^2)` `= (sqrt(x^2+y^2+z^2)(x hat i + y hat j + z hat k))/(x^2+y^2+z^2)` `= (x hat i + y hat j + z hat k)xx(x^2+y^2+z^2)^(-1/2)` I'm guessing the text book's author is using `Ax` to mean "the `x`-component of the normalized vector `A`". So that would mean `Ax = (x)xx(x^2+y^2+z^2)^(-1/2)` In which case, `(dAx)/dx = x (-1/2)(x^2+y^2+z^2)^(-3/2)(2x + 2y(dy/dx + 2z dz/dx))` ` + (x^2+y^2+z^2)^(-1/2)(1)` If `y` and `z` are constants, then this would give us: `(dAx)/dx = x (-1/2)(x^2+y^2+z^2)^(-3/2)(2x)` ` + (x^2+y^2+z^2)^(-1/2)` `= -x^2(x^2+y^2+z^2)^(-3/2)` ` + (x^2+y^2+z^2)^(-1/2)` This is not the answer you put (which is hard to read - please use the math entry system), but hopefully gives you a start on following what the author is doing.
6761hash 14 Aug 2017, 05:32
Thanks.
Attached is the context regarding divergence of electric fields.
The author does get a different result than you do, although I've found errors in his text several times.
Can you tell me why i,j,k are not eliminated when you executed your derivative but is eliminated in the text book example I attached?
Also, if after reading the context you now agree with the textbook, can you please show me all the steps in the derivation in extreme detail, since I cannot understand how the book came up with its result. Thank you!
X
Thanks. Attached is the context regarding divergence of electric fields. The author does get a different result than you do, although I've found errors in his text several times. Can you tell me why i,j,k are not eliminated when you executed your derivative but is eliminated in the text book example I attached? Also, if after reading the context you now agree with the textbook, can you please show me all the steps in the derivation in extreme detail, since I cannot understand how the book came up with its result. Thank you!
6761hash 14 Aug 2017, 05:33
I attached the file but don't see it above for some reason.
X
I attached the file but don't see it above for some reason.
6761hash 14 Aug 2017, 16:30
I get an error every time I upload. Is there an email addy I can send to?
X
I get an error every time I upload. Is there an email addy I can send to?
Murray 27 Jan 2018, 02:25
Sorry, I missed your response when it came in. Were you trying to upload an image? IF so, it should work OK.
To test it I uploaded this random image just now and received no errors:
(There is a file size limit. Were you trying to upload something huge?)
X
Sorry, I missed your response when it came in. Were you trying to upload an image? IF so, it should work OK. To test it I uploaded this random image just now and received no errors: <img src="/forum/uploads/imf-2518-archimedean-spiral.png" width="220" height="237" alt="Vectors" /> (There is a file size limit. Were you trying to upload something huge?)
6761hash 27 Jan 2018, 07:33
Thanks, no. I dont remember trying to upload anything. Just had a question. Is there any difference at all in the differentiation of xihat, yjhat, zkhat than in differentiating x,y,z. Why complicate things and use xihat for instance instead of x. If you are describing a vector in 3 dimensions, x,y,z suffices perfectly well without xihat,yjhat,zkhat. Its not like someone will get confused and put the vertical component value along the wrong axis, when drawing a vector. It already indicates the z direction. No need for zkhat as far as I can see. Is the a scenario whereby one would not know what direction a y component goes unless there is a yjhat next to it? Thanks!
X
Thanks, no. I dont remember trying to upload anything. Just had a question. Is there any difference at all in the differentiation of xihat, yjhat, zkhat than in differentiating x,y,z. Why complicate things and use xihat for instance instead of x. If you are describing a vector in 3 dimensions, x,y,z suffices perfectly well without xihat,yjhat,zkhat. Its not like someone will get confused and put the vertical component value along the wrong axis, when drawing a vector. It already indicates the z direction. No need for zkhat as far as I can see. Is the a scenario whereby one would not know what direction a y component goes unless there is a yjhat next to it? Thanks!
Murray 27 Jan 2018, 19:24
I do empathise with your point - sometimes it seems these "extra bits" don't change the situation all that much, so why bother about them?
However, as Khan pointed out in the first few minutes of the first video I referred you to before, the idea behind using multiples of unit vectors is that we assume all the "action" starts at the origin, and our quantities are multiples of unit vectors in each of the `x`-, `y`- and `z`-directions (rather than being "equivalent vectors" that can start anywhere).
If you and I are working on an engineering problem involving such quantities and we both start at the origin and use the same units, I can make sense of your solutions and you can make sense of mine, because we are using a consistent reference point.
Like a lot of such things, they may not make a lot of sense when we first learn them, but the more applications we come across later, the more we are glad we know about it.
Hope that makes sense!
X
I do empathise with your point - sometimes it seems these "extra bits" don't change the situation all that much, so why bother about them? However, as Khan pointed out in the first few minutes of the first video I referred you to before, the idea behind using multiples of unit vectors is that we assume all the "action" starts at the origin, and our quantities are multiples of unit vectors in each of the `x`-, `y`- and `z`-directions (rather than being "equivalent vectors" that can start anywhere). If you and I are working on an engineering problem involving such quantities and we both start at the origin and use the same units, I can make sense of your solutions and you can make sense of mine, because we are using a consistent reference point. Like a lot of such things, they may not make a lot of sense when we first learn them, but the more applications we come across later, the more we are glad we know about it. Hope that makes sense!
6761hash 31 Jan 2018, 15:48
Thanks. Just what does `x hat j` mean in the equation `xy hat i, x hat j, y^2 hat k.` Specifically, I speak of finding the divergence of the function.
X
Thanks. Just what does `x hat j` mean in the equation `xy hat i, x hat j, y^2 hat k.` Specifically, I speak of finding the divergence of the function.
Murray 31 Jan 2018, 20:43
Divergence is somewhat beyond the scope of IntMath, but I'll try to help.
Using this notation of the Wikipedia article on Divergence, we have:
`bbF = xy hat i + x hat j + y^2 hat k`
This means the vector field varies in the `hat i`-direction by a factor of `xy`, varies in the `hat j`-direction by a factor of `x` and varies in the `hat k`-direction by a factor of `y^2.`
So for example if `x = 2` and `y=3`, the vector field would be given by:
`bbF = 6 hat i + 2 hat j + 9 hat k`
The divergence of a vector field is given by:
`"div"bbF = grad * bbF = grad ( U hat i + V hat j + W hat k)`
(In our case, `U = xy`, `V = x` and `W = y^2`.)
We use
`grad * bbF = (delU)/(del x) + (delV)/(del y) + (delW)/(del z)`
So for our case, we would have:
`grad * bbF = y + 0 + 0 = y`
The divergence for the particular values I used before would be:
`grad * bbF = 3`
Hope it makes sense.
X
Divergence is somewhat beyond the scope of IntMath, but I'll try to help. Using this notation of <a href="https://en.wikipedia.org/wiki/Divergence">the Wikipedia article on Divergence</a>, we have: `bbF = xy hat i + x hat j + y^2 hat k` This means the vector field varies in the `hat i`-direction by a factor of `xy`, varies in the `hat j`-direction by a factor of `x` and varies in the `hat k`-direction by a factor of `y^2.` So for example if `x = 2` and `y=3`, the vector field would be given by: `bbF = 6 hat i + 2 hat j + 9 hat k` The <b>divergence</b> of a vector field is given by: `"div"bbF = grad * bbF = grad ( U hat i + V hat j + W hat k)` (In our case, `U = xy`, `V = x` and `W = y^2`.) We use `grad * bbF = (delU)/(del x) + (delV)/(del y) + (delW)/(del z)` So for our case, we would have: `grad * bbF = y + 0 + 0 = y` The divergence for the particular values I used before would be: `grad * bbF = 3` Hope it makes sense.
6761hash 31 Jan 2018, 21:20
Yes, thank you. But I fear you may have left out the most important part. The x, y, z are the coordinates in the vector field from where each vector,the tail, is drawn. Take 2zy, z^2, xyz. If I want to draw a vector starting at 2,2,2. start at 2,2,2 and go right along x axis 8, then up the y axis 4, then in the z direction 8. I get the translation amounts by plugging x,y,z coordinates into the orig equation to get how much to draw the vector. I suspect people who struggle with this concept simply dont know that the XYZ are coordinates from where the tractor is drawn. Does this sound correct to you?
X
Yes, thank you. But I fear you may have left out the most important part. The x, y, z are the coordinates in the vector field from where each vector,the tail, is drawn. Take 2zy, z^2, xyz. If I want to draw a vector starting at 2,2,2. start at 2,2,2 and go right along x axis 8, then up the y axis 4, then in the z direction 8. I get the translation amounts by plugging x,y,z coordinates into the orig equation to get how much to draw the vector. I suspect people who struggle with this concept simply dont know that the XYZ are coordinates from where the tractor is drawn. Does this sound correct to you?
Murray 08 Feb 2018, 02:33
Hi again. I don't feel your interpretation is correct. A unit vector is a vector of unit length starting from the origin, in either the `x`-, `y-`- or `z`- directions. The number in front is a multiple of that unit vector. So `2 hat i` means a vector `2` units long in the `x`-direction.
X
Hi again. I don't feel your interpretation is correct. A unit vector is a vector of unit length starting from the origin, in either the `x`-, `y-`- or `z`- directions. The number in front is a multiple of that unit vector. So `2 hat i` means a vector `2` units long in the `x`-direction.
Murray 20 Dec 2018, 20:21
It appears 6761hash has disappeared, so I'll close this topic now.
X
It appears 6761hash has disappeared, so I'll close this topic now.
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