anonymous
  • anonymous
check my work
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Given a scalar point function \[f(r)\] where \[r=\sqrt{x^2+y^2+z^2}\] Prove that \[\nabla^{2}f(r)=f''(r)+\frac{2}{r}f'(r)\] \[\nabla^{2}f(r)=\frac{\partial^2f}{\partial x^2}+\frac{\partial^2f}{\partial y^2}=\frac{\partial^2f}{\partial z^2}\] \[\therefore \frac{\partial}{\partial x}(\frac{\partial f}{\partial r}\frac{\partial r}{\partial x})+\frac{\partial}{\partial y}(\frac{\partial f}{\partial r}\frac{\partial r}{\partial y})+\frac{\partial}{\partial z}(\frac{\partial f}{\partial r}\frac{\partial r}{\partial z})\] Now \[r^2=x^2+y^2+z^2\] \[\implies 2r\frac{\partial r}{\partial x}=2x\]\[\implies \frac{\partial r}{\partial x}=\frac{x}{r}\] \[\implies \frac{\partial r}{\partial y}=\frac{y}{r}\]\[\implies \frac{\partial y}{\partial z}=\frac{z}{r}\] Now LHS again, \[\frac{\partial}{\partial x}(f'(r)\frac{x}{r})+\frac{\partial}{\partial y}(f'(r)\frac{y}{r})+\frac{\partial }{\partial z}(f'(r)\frac{z}{r})\] \[\frac{x}{r}.\frac{\partial f'}{\partial x}+\frac{1}{r}.f'(r)+xf'(r).\frac{\partial}{\partial x}(r^{-1})\] \[+\frac{y}{r}.\frac{\partial f'}{\partial y}+\frac{1}{r}.f'(r)+yf'(r)\frac{\partial}{\partial y}(r^{-1})\] \[+\frac{z}{r}\frac{\partial f'}{\partial z}+\frac{1}{r}.f'(r)+yf'(r).\frac{\partial }{\partial z}(r^{-1})\] Applying chain rule \[\frac{x}{r}.\frac{\partial f'}{\partial r}.\frac{\partial r}{\partial x}+xf'(r).\frac{\partial }{\partial r}(r^{-1}).\frac{\partial r}{\partial x}\] \[+\frac{y}{r}.\frac{\partial f'}{\partial r}.\frac{\partial r}{\partial y}+yf'(r).\frac{\partial }{\partial r}(r^{-1}).\frac{\partial r}{\partial y}\] \[+\frac{z}{r}.\frac{\partial f'}{\partial r}.\frac{\partial r}{\partial z}+zf'(r).\frac{\partial }{\partial r}(r^{-1}).\frac{\partial r}{\partial z}\] \[+\frac{3}{r}.f'(r)\] Now we have \[\frac{x^2}{r^2}.f''(r)-\frac{x^2}{r^3}.f'(r)+\frac{y^2}{r^2}.f''(r)-\frac{y^2}{r^3}.f'(r)+\frac{z^2}{r^2}.f''(r)-\frac{z^2}{r^3}.f'(r)+\frac{3}{r}.f'(r)\] \[\frac{3}{r}.f'(r)+f''(r)(\frac{x^2+y^2+z^2}{r^2})-\frac{1}{r}.f'(r)(\frac{x^2+y^2+z^2}{r^2})\] \[\implies \frac{3}{r}.f'(r)+f''(r)-\frac{1}{r}.f'(r)=f''(r)+\frac{2}{r}.f'(r)\]
IrishBoy123
  • IrishBoy123
are you allowed spherical? https://gyazo.com/abacaf4b950f6e57251b8fdd887162b8 if so, it comes down to expanding out the first term on the RHS of this as everything else = 0
anonymous
  • anonymous
idk, it's just a proof ques, am I right though??

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anonymous
  • anonymous
I haven't spherical coordinates before, I did with rectangular coordinate system
IrishBoy123
  • IrishBoy123
i'd need a cold towel to wade through that!!! i'll print it out and see if i can help....
anonymous
  • anonymous
haven't used*
anonymous
  • anonymous
@ganeshie8 gave me a medal so it must be good, no need to print it out :)
IrishBoy123
  • IrishBoy123
this looks shorter to me and my head now hurts
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anonymous
  • anonymous
Oh you used the quotient rule, I used triple product rule \[\frac{d}{dx}(u(x).v(x).w(x))=\frac{du}{dx}.v.w+u.\frac{dv}{dx}.w+u.v.\frac{dw}{dx}\]
anonymous
  • anonymous
wow your method is indeed quite fast @IrishBoy123

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