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critical points of $f(x,y) = sqrt{x^2+y^2}$
The function is $f(x,y) = sqrt{x^2+y^2}$ and I need to find its critical points.
The gradient is ($frac{x}{sqrt{x^2+y^2}}$, $frac{y}{sqrt{x^2+y^2}}$) but the only points where it equals 0 is the point (0,0), where the partial derivatives don't exist.
The calculation of $lim_{hto 0}frac{f(h,0)-f(0,0)}{h}$ yields to: $lim_{hto 0}frac{sqrt{h^2}}{h} = lim_{hto 0}frac{|h|}{h}$ which doesn't exist.
So what does that means about the critical points?
calculus real-analysis multivariable-calculus
asked Nov 21 '18 at 2:12
Gabi G
36218
add a comment |
The function is $f(x,y) = sqrt{x^2+y^2}$ and I need to find its critical points.
The gradient is ($frac{x}{sqrt{x^2+y^2}}$, $frac{y}{sqrt{x^2+y^2}}$) but the only points where it equals 0 is the point (0,0), where the partial derivatives don't exist.
The calculation of $lim_{hto 0}frac{f(h,0)-f(0,0)}{h}$ yields to: $lim_{hto 0}frac{sqrt{h^2}}{h} = lim_{hto 0}frac{|h|}{h}$ which doesn't exist.
So what does that means about the critical points?
calculus real-analysis multivariable-calculus
asked Nov 21 '18 at 2:12
Gabi G
36218
1
It has no critical points in $mathbb R^{2}setminus {(0,0}$.
– Kavi Rama Murthy
Nov 21 '18 at 9:53
add a comment |
The function is $f(x,y) = sqrt{x^2+y^2}$ and I need to find its critical points.
The gradient is ($frac{x}{sqrt{x^2+y^2}}$, $frac{y}{sqrt{x^2+y^2}}$) but the only points where it equals 0 is the point (0,0), where the partial derivatives don't exist.
The calculation of $lim_{hto 0}frac{f(h,0)-f(0,0)}{h}$ yields to: $lim_{hto 0}frac{sqrt{h^2}}{h} = lim_{hto 0}frac{|h|}{h}$ which doesn't exist.
So what does that means about the critical points?
calculus real-analysis multivariable-calculus
asked Nov 21 '18 at 2:12
Gabi G
36218
The function is $f(x,y) = sqrt{x^2+y^2}$ and I need to find its critical points.
The gradient is ($frac{x}{sqrt{x^2+y^2}}$, $frac{y}{sqrt{x^2+y^2}}$) but the only points where it equals 0 is the point (0,0), where the partial derivatives don't exist.
The calculation of $lim_{hto 0}frac{f(h,0)-f(0,0)}{h}$ yields to: $lim_{hto 0}frac{sqrt{h^2}}{h} = lim_{hto 0}frac{|h|}{h}$ which doesn't exist.
So what does that means about the critical points?
calculus real-analysis multivariable-calculus
calculus real-analysis multivariable-calculus
asked Nov 21 '18 at 2:12
Gabi G
36218
asked Nov 21 '18 at 2:12
Gabi G
36218
asked Nov 21 '18 at 2:12
Gabi G
36218
asked Nov 21 '18 at 2:12
Gabi G
36218
asked Nov 21 '18 at 2:12
Gabi G
36218
36218
1
It has no critical points in $mathbb R^{2}setminus {(0,0}$.
– Kavi Rama Murthy
Nov 21 '18 at 9:53
add a comment |
1
It has no critical points in $mathbb R^{2}setminus {(0,0}$.
– Kavi Rama Murthy
Nov 21 '18 at 9:53
1
1
It has no critical points in $mathbb R^{2}setminus {(0,0}$.
– Kavi Rama Murthy
Nov 21 '18 at 9:53
It has no critical points in $mathbb R^{2}setminus {(0,0}$.
– Kavi Rama Murthy
Nov 21 '18 at 9:53
add a comment |
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It has no critical points in $mathbb R^{2}setminus {(0,0}$.
– Kavi Rama Murthy
Nov 21 '18 at 9:53