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anonymous
 one year ago
check my work
anonymous
 one year ago
check my work

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Given 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
 one year ago
Best ResponseYou've already chosen the best response.1are 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
 one year ago
Best ResponseYou've already chosen the best response.0idk, it's just a proof ques, am I right though??

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I haven't spherical coordinates before, I did with rectangular coordinate system

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1i'd need a cold towel to wade through that!!! i'll print it out and see if i can help....

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@ganeshie8 gave me a medal so it must be good, no need to print it out :)

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1this looks shorter to me and my head now hurts

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh 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
 one year ago
Best ResponseYou've already chosen the best response.0wow your method is indeed quite fast @IrishBoy123
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