Vector Calculus curl's and divergence? Watch

CammieInfinity
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In the context of these questions what is r? Is it the sqrt(x^2+y^2+z^2)?
Also wherever you see an x it should be underlined/ highlighted because it's meant to be a vector.


. If f(r) is a differentiable function of r = ||x ||, show that
(a) grad f(r) = f′(r) x /r,
(b) curl [f(r)x ]=0 .

15. Let x be the position vector in three dimensions, with r = ||x ||, and let a be a constant vector. Show that
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TeeEm
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(Original post by CammieInfinity)
In the context of these questions what is r? Is it the sqrt(x^2+y^2+z^2)?
Also wherever you see an x it should be underlined/ highlighted because it's meant to be a vector.


. If f(r) is a differentiable function of r = ||x ||, show that
(a) grad f(r) = f′(r) x /r,
(b) curl [f(r)x ]=0 .

15. Let x be the position vector in three dimensions, with r = ||x ||, and let a be a constant vector. Show that
please post a photo of the question
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CammieInfinity
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Here it is
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CammieInfinity
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I actually just can't do 14a
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TeeEm
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(Original post by CammieInfinity)
I actually just can't do 14a
assuming x is not constant, i.e. x =(a1,a2,a3)

then write down |x| in terms of a1, a2 and a3

use the definition of grad
(maybe easier to do component by component.

It should be just a standard Chain rule with a tidy up at the end
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nerak99
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Well since you have had no other answer (well you didn't when I started typing this, TeeEm feel free to correct me.) I think you need a combination of the chain rule and partial differentiation. Taking the right hand side, the following should help short of doing the whole question and taking note that it is a very long time since I did vector calculus.

 r=\sqrt {x^2+y^2+z^2} \n
Example \displaystyle \frac{\partial r}{\partial x}  = \frac{2x}{2}(x^2+y^2+z^2)^{\frac  {-1}{2}}
Chain rule
\displaystyle \frac{\partial f}{\partial x}  = \frac{\partial f}{\partial r}\times \frac{\partial r}{\partial x}

The trick then is to write the differential of f using the chain rule for each of x, y and z and take out the common factor of
\displaystyle \frac{\partial f}{\partial r}
Which would leave
\displaystyle \frac{\partial f}{\partial r} \left ( \frac{\partial r}{\partial x} \bold i + \dots \right )
note that ​ \displaystyle (x^2+y^2+z^2)^{\frac{-1}{2}}= \frac {1}{r}

Apols for my Latex however I can't say I am impressed by the typesetting in your book. Note to self, Shift-Cmd V for paste without nonsense.



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TeeEm
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(Original post by nerak99)
Well since you have had no other answer (well you didn't when I started typing this, TeeEm feel free to correct me.) I think you need a combination of the chain rule and partial differentiation. Taking the right hand side, the following should help short of doing the whole question and taking note that it is a very long time since I did vector calculus.

 r=\sqrt {x^2+y^2+z^2} \n
Example \displaystyle \frac{\partial r}{\partial x}  = \frac{2x}{2}(x^2+y^2+z^2)^{\frac  {-1}{2}}
Chain rule
\displaystyle \frac{\partial f}{\partial x}  = \frac{\partial f}{\partial r}\times \frac{\partial r}{\partial x}

The trick then is to write the differential of f using the chain rule for each of x, y and z and take out the common factor of
\displaystyle \frac{\partial f}{\partial r}
Which would leave
\displaystyle \frac{\partial f}{\partial r} \left ( \frac{\partial r}{\partial x} \bold i + \dots \right )
note that ​ \displaystyle (x^2+y^2+z^2)^{\frac{-1}{2}}= \frac {1}{r}

Apols for my Latex however I can't say I am impressed by the typesetting in your book. Note to self, Shift-Cmd V for paste without nonsense.



no correction/ I admire you latex skills.

Btw Chain rule whether in differentiation of one variable or several variables is still call it chain rule but then again I did this stuff a very long time ago so notation and vocabulary unfortunately is subject to fashion trends (even in Mathematics)
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TeeEm
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(Original post by nerak99)

Apols for my Latex however I can't say I am impressed by the typesetting in your book. Note to self, Shift-Cmd V for paste without nonsense.





???
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nerak99
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??? ?
Assume you refer to my Note to self.
It took ages to realise that when copying and pasting a complex bit of latex, if I just did cmd V to paste, I got a whole host of html as well as the bit of latex. That when I stumbled on Shift-Cmd V to do a plain text copy.
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TeeEm
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(Original post by nerak99)
??? ?
Assume you refer to my Note to self.
It took ages to realise that when copying and pasting a complex bit of latex, if I just did cmd V to paste, I got a whole host of html as well as the bit of latex. That when I stumbled on Shift-Cmd V to do a plain text copy.
I see ...
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