For $a=frac 14$, $f$ is continuous and differentiable.
$begingroup$
Let $f(x,y) =
begin{cases}
frac{xy}{(x^2+y^2)^a}, & text{if $(x,y)neq (0,0)$ } \[2ex]
0, & text{if $(x,y)= (0,0)$}
end{cases}$
Then which one of the following is TRUE for $f$ at the point $(0,0)$?
(A) For $a=1$, $f$ is continuous but not differentiable.
(B) For $a=frac 12$, $f$ is continuous and differentiable.
(C) For $a=frac 14$, $f$ is continuous and differentiable.
(D) For $a=frac 34$, $f$ is neither continuous nor differentiable
My attempt:
(A) For $a=1$ $f(x,y) =
begin{cases}
frac{xy}{(x^2+y^2)}, & text{if $(x,y)neq (0,0)$ } \[2ex]
0, & text{if $(x,y)= (0,0)$}
end{cases}$
is not continuous at $(0,0)$ as if we put $y=mx$ then $f(x,mx) =
begin{cases}
frac{m}{(1+m^2)}, & text{if $xneq 0$ } \[2ex]
0, & text{if $x= 0$}
end{cases}$
not continuous at $0$. So, (A) is not true.
(B) For $a=frac 12$ $f(x,y) =
begin{cases}
frac{xy}{(x^2+y^2)^{frac 12}}, & text{if $(x,y)neq (0,0)$ } \[2ex]
0, & text{if $(x,y)= (0,0)$}
end{cases}$
is not differentiable at $(0,0)$ as $frac {f(h,k)-f(0,0)}{sqrt{h^2+k^2}}=
frac{hk}{(h^2+k^2)}$ where $(h,k) to (0,0)$ if we put $k=mh$ then
$f(h,mh) =frac{m}{(1+m^2)} neq (0,0) $
Hence, $f$ is not differentiable at $(0,0)$. So, (B) is not true.
(C) For $a=frac 14$ $f(x,y) =
begin{cases}
frac{xy}{(x^2+y^2)^{frac 14}}, & text{if $(x,y)neq (0,0)$ } \[2ex]
0, & text{if $(x,y)= (0,0)$}
end{cases}$
Then, $f(x,y) =frac{xsqrt y}{(frac{x^2}{y^2}+1)^{frac 14}}$ Can we prove continuity of $f$ from here?
Observe, $f$ is differentiable at $(0,0)$ as $frac {f(h,k)-f(0,0)}{sqrt{h^2+k^2}}=
frac{hk}{(h^2+k^2)^frac 34}=frac{h}{(frac{h^2}{k^{4/3}}+k^{2/3})^{frac 34}} to (0,0)$ where $(h,k) to (0,0)$. So we can directly say that $f$ is differentiable and hence is continuous.
(D) For $a=frac 34$ $f(x,y) =
begin{cases}
frac{xy}{(x^2+y^2)^{frac 34}}, & text{if $(x,y)neq (0,0)$ } \[2ex]
0, & text{if $(x,y)= (0,0)$}
end{cases}$
From the observation of (C) we get $f$ is continuous at $(0,0)$ so (D) is also false.
So, my basic question is although we know that (C) will be the correct option can we prove $f$ is continuous directly there? See the highlighted portion of part (C) under My attempt section.
real-analysis calculus analysis multivariable-calculus alternative-proof
asked Jan 7 at 5:52
GimgimGimgim
1869
$endgroup$
add a comment |
$begingroup$
Let $f(x,y) =
begin{cases}
frac{xy}{(x^2+y^2)^a}, & text{if $(x,y)neq (0,0)$ } \[2ex]
0, & text{if $(x,y)= (0,0)$}
end{cases}$
Then which one of the following is TRUE for $f$ at the point $(0,0)$?
(A) For $a=1$, $f$ is continuous but not differentiable.
(B) For $a=frac 12$, $f$ is continuous and differentiable.
(C) For $a=frac 14$, $f$ is continuous and differentiable.
(D) For $a=frac 34$, $f$ is neither continuous nor differentiable
My attempt:
(A) For $a=1$ $f(x,y) =
begin{cases}
frac{xy}{(x^2+y^2)}, & text{if $(x,y)neq (0,0)$ } \[2ex]
0, & text{if $(x,y)= (0,0)$}
end{cases}$
is not continuous at $(0,0)$ as if we put $y=mx$ then $f(x,mx) =
begin{cases}
frac{m}{(1+m^2)}, & text{if $xneq 0$ } \[2ex]
0, & text{if $x= 0$}
end{cases}$
not continuous at $0$. So, (A) is not true.
(B) For $a=frac 12$ $f(x,y) =
begin{cases}
frac{xy}{(x^2+y^2)^{frac 12}}, & text{if $(x,y)neq (0,0)$ } \[2ex]
0, & text{if $(x,y)= (0,0)$}
end{cases}$
is not differentiable at $(0,0)$ as $frac {f(h,k)-f(0,0)}{sqrt{h^2+k^2}}=
frac{hk}{(h^2+k^2)}$ where $(h,k) to (0,0)$ if we put $k=mh$ then
$f(h,mh) =frac{m}{(1+m^2)} neq (0,0) $
Hence, $f$ is not differentiable at $(0,0)$. So, (B) is not true.
(C) For $a=frac 14$ $f(x,y) =
begin{cases}
frac{xy}{(x^2+y^2)^{frac 14}}, & text{if $(x,y)neq (0,0)$ } \[2ex]
0, & text{if $(x,y)= (0,0)$}
end{cases}$
Then, $f(x,y) =frac{xsqrt y}{(frac{x^2}{y^2}+1)^{frac 14}}$ Can we prove continuity of $f$ from here?
Observe, $f$ is differentiable at $(0,0)$ as $frac {f(h,k)-f(0,0)}{sqrt{h^2+k^2}}=
frac{hk}{(h^2+k^2)^frac 34}=frac{h}{(frac{h^2}{k^{4/3}}+k^{2/3})^{frac 34}} to (0,0)$ where $(h,k) to (0,0)$. So we can directly say that $f$ is differentiable and hence is continuous.
(D) For $a=frac 34$ $f(x,y) =
begin{cases}
frac{xy}{(x^2+y^2)^{frac 34}}, & text{if $(x,y)neq (0,0)$ } \[2ex]
0, & text{if $(x,y)= (0,0)$}
end{cases}$
From the observation of (C) we get $f$ is continuous at $(0,0)$ so (D) is also false.
So, my basic question is although we know that (C) will be the correct option can we prove $f$ is continuous directly there? See the highlighted portion of part (C) under My attempt section.
real-analysis calculus analysis multivariable-calculus alternative-proof
asked Jan 7 at 5:52
GimgimGimgim
1869
$endgroup$
add a comment |
$begingroup$
Let $f(x,y) =
begin{cases}
frac{xy}{(x^2+y^2)^a}, & text{if $(x,y)neq (0,0)$ } \[2ex]
0, & text{if $(x,y)= (0,0)$}
end{cases}$
Then which one of the following is TRUE for $f$ at the point $(0,0)$?
(A) For $a=1$, $f$ is continuous but not differentiable.
(B) For $a=frac 12$, $f$ is continuous and differentiable.
(C) For $a=frac 14$, $f$ is continuous and differentiable.
(D) For $a=frac 34$, $f$ is neither continuous nor differentiable
My attempt:
(A) For $a=1$ $f(x,y) =
begin{cases}
frac{xy}{(x^2+y^2)}, & text{if $(x,y)neq (0,0)$ } \[2ex]
0, & text{if $(x,y)= (0,0)$}
end{cases}$
is not continuous at $(0,0)$ as if we put $y=mx$ then $f(x,mx) =
begin{cases}
frac{m}{(1+m^2)}, & text{if $xneq 0$ } \[2ex]
0, & text{if $x= 0$}
end{cases}$
not continuous at $0$. So, (A) is not true.
(B) For $a=frac 12$ $f(x,y) =
begin{cases}
frac{xy}{(x^2+y^2)^{frac 12}}, & text{if $(x,y)neq (0,0)$ } \[2ex]
0, & text{if $(x,y)= (0,0)$}
end{cases}$
is not differentiable at $(0,0)$ as $frac {f(h,k)-f(0,0)}{sqrt{h^2+k^2}}=
frac{hk}{(h^2+k^2)}$ where $(h,k) to (0,0)$ if we put $k=mh$ then
$f(h,mh) =frac{m}{(1+m^2)} neq (0,0) $
Hence, $f$ is not differentiable at $(0,0)$. So, (B) is not true.
(C) For $a=frac 14$ $f(x,y) =
begin{cases}
frac{xy}{(x^2+y^2)^{frac 14}}, & text{if $(x,y)neq (0,0)$ } \[2ex]
0, & text{if $(x,y)= (0,0)$}
end{cases}$
Then, $f(x,y) =frac{xsqrt y}{(frac{x^2}{y^2}+1)^{frac 14}}$ Can we prove continuity of $f$ from here?
Observe, $f$ is differentiable at $(0,0)$ as $frac {f(h,k)-f(0,0)}{sqrt{h^2+k^2}}=
frac{hk}{(h^2+k^2)^frac 34}=frac{h}{(frac{h^2}{k^{4/3}}+k^{2/3})^{frac 34}} to (0,0)$ where $(h,k) to (0,0)$. So we can directly say that $f$ is differentiable and hence is continuous.
(D) For $a=frac 34$ $f(x,y) =
begin{cases}
frac{xy}{(x^2+y^2)^{frac 34}}, & text{if $(x,y)neq (0,0)$ } \[2ex]
0, & text{if $(x,y)= (0,0)$}
end{cases}$
From the observation of (C) we get $f$ is continuous at $(0,0)$ so (D) is also false.
So, my basic question is although we know that (C) will be the correct option can we prove $f$ is continuous directly there? See the highlighted portion of part (C) under My attempt section.
real-analysis calculus analysis multivariable-calculus alternative-proof
asked Jan 7 at 5:52
GimgimGimgim
1869
$endgroup$
Let $f(x,y) =
begin{cases}
frac{xy}{(x^2+y^2)^a}, & text{if $(x,y)neq (0,0)$ } \[2ex]
0, & text{if $(x,y)= (0,0)$}
end{cases}$
Then which one of the following is TRUE for $f$ at the point $(0,0)$?
(A) For $a=1$, $f$ is continuous but not differentiable.
(B) For $a=frac 12$, $f$ is continuous and differentiable.
(C) For $a=frac 14$, $f$ is continuous and differentiable.
(D) For $a=frac 34$, $f$ is neither continuous nor differentiable
My attempt:
(A) For $a=1$ $f(x,y) =
begin{cases}
frac{xy}{(x^2+y^2)}, & text{if $(x,y)neq (0,0)$ } \[2ex]
0, & text{if $(x,y)= (0,0)$}
end{cases}$
is not continuous at $(0,0)$ as if we put $y=mx$ then $f(x,mx) =
begin{cases}
frac{m}{(1+m^2)}, & text{if $xneq 0$ } \[2ex]
0, & text{if $x= 0$}
end{cases}$
not continuous at $0$. So, (A) is not true.
(B) For $a=frac 12$ $f(x,y) =
begin{cases}
frac{xy}{(x^2+y^2)^{frac 12}}, & text{if $(x,y)neq (0,0)$ } \[2ex]
0, & text{if $(x,y)= (0,0)$}
end{cases}$
is not differentiable at $(0,0)$ as $frac {f(h,k)-f(0,0)}{sqrt{h^2+k^2}}=
frac{hk}{(h^2+k^2)}$ where $(h,k) to (0,0)$ if we put $k=mh$ then
$f(h,mh) =frac{m}{(1+m^2)} neq (0,0) $
Hence, $f$ is not differentiable at $(0,0)$. So, (B) is not true.
(C) For $a=frac 14$ $f(x,y) =
begin{cases}
frac{xy}{(x^2+y^2)^{frac 14}}, & text{if $(x,y)neq (0,0)$ } \[2ex]
0, & text{if $(x,y)= (0,0)$}
end{cases}$
Then, $f(x,y) =frac{xsqrt y}{(frac{x^2}{y^2}+1)^{frac 14}}$ Can we prove continuity of $f$ from here?
Observe, $f$ is differentiable at $(0,0)$ as $frac {f(h,k)-f(0,0)}{sqrt{h^2+k^2}}=
frac{hk}{(h^2+k^2)^frac 34}=frac{h}{(frac{h^2}{k^{4/3}}+k^{2/3})^{frac 34}} to (0,0)$ where $(h,k) to (0,0)$. So we can directly say that $f$ is differentiable and hence is continuous.
(D) For $a=frac 34$ $f(x,y) =
begin{cases}
frac{xy}{(x^2+y^2)^{frac 34}}, & text{if $(x,y)neq (0,0)$ } \[2ex]
0, & text{if $(x,y)= (0,0)$}
end{cases}$
From the observation of (C) we get $f$ is continuous at $(0,0)$ so (D) is also false.
So, my basic question is although we know that (C) will be the correct option can we prove $f$ is continuous directly there? See the highlighted portion of part (C) under My attempt section.
real-analysis calculus analysis multivariable-calculus alternative-proof
real-analysis calculus analysis multivariable-calculus alternative-proof
asked Jan 7 at 5:52
GimgimGimgim
1869
asked Jan 7 at 5:52
GimgimGimgim
1869
asked Jan 7 at 5:52
GimgimGimgim
1869
asked Jan 7 at 5:52
GimgimGimgim
1869
asked Jan 7 at 5:52
GimgimGimgim
1869
1869
add a comment |
add a comment |
2 Answers
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$begingroup$
Recall that
$$|xy|leq frac{(x^2+y^2)}{2}$$
and so
$$frac{|xy|}{(x^2+y^2)^{1/4}}leq frac{(x^2+y^2)^{3/4}}{2}to 0$$
as $(x,y)to (0,0).$
Hence you have continuity at $(0,0).$
answered Jan 7 at 5:59
Hello_WorldHello_World
4,11621731
$endgroup$
add a comment |
$begingroup$
For $a=frac 1 4$ the derivative exist and is $0$: by definition of derivative this means $ frac {xy} {(x^{2}+y^{2})^{3/4}} to 0$ which follows easily from the fact that $2|xy | leq x^{2}+y^{2}$.
answered Jan 7 at 6:04
Kavi Rama MurthyKavi Rama Murthy
54.4k32055
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
Recall that
$$|xy|leq frac{(x^2+y^2)}{2}$$
and so
$$frac{|xy|}{(x^2+y^2)^{1/4}}leq frac{(x^2+y^2)^{3/4}}{2}to 0$$
as $(x,y)to (0,0).$
Hence you have continuity at $(0,0).$
answered Jan 7 at 5:59
Hello_WorldHello_World
4,11621731
$endgroup$
add a comment |
$begingroup$
Recall that
$$|xy|leq frac{(x^2+y^2)}{2}$$
and so
$$frac{|xy|}{(x^2+y^2)^{1/4}}leq frac{(x^2+y^2)^{3/4}}{2}to 0$$
as $(x,y)to (0,0).$
Hence you have continuity at $(0,0).$
answered Jan 7 at 5:59
Hello_WorldHello_World
4,11621731
$endgroup$
add a comment |
$begingroup$
Recall that
$$|xy|leq frac{(x^2+y^2)}{2}$$
and so
$$frac{|xy|}{(x^2+y^2)^{1/4}}leq frac{(x^2+y^2)^{3/4}}{2}to 0$$
as $(x,y)to (0,0).$
Hence you have continuity at $(0,0).$
answered Jan 7 at 5:59
Hello_WorldHello_World
4,11621731
$endgroup$
Recall that
$$|xy|leq frac{(x^2+y^2)}{2}$$
and so
$$frac{|xy|}{(x^2+y^2)^{1/4}}leq frac{(x^2+y^2)^{3/4}}{2}to 0$$
as $(x,y)to (0,0).$
Hence you have continuity at $(0,0).$
answered Jan 7 at 5:59
Hello_WorldHello_World
4,11621731
answered Jan 7 at 5:59
Hello_WorldHello_World
4,11621731
answered Jan 7 at 5:59
Hello_WorldHello_World
4,11621731
answered Jan 7 at 5:59
Hello_WorldHello_World
4,11621731
4,11621731
add a comment |
add a comment |
$begingroup$
For $a=frac 1 4$ the derivative exist and is $0$: by definition of derivative this means $ frac {xy} {(x^{2}+y^{2})^{3/4}} to 0$ which follows easily from the fact that $2|xy | leq x^{2}+y^{2}$.
answered Jan 7 at 6:04
Kavi Rama MurthyKavi Rama Murthy
54.4k32055
$endgroup$
add a comment |
$begingroup$
For $a=frac 1 4$ the derivative exist and is $0$: by definition of derivative this means $ frac {xy} {(x^{2}+y^{2})^{3/4}} to 0$ which follows easily from the fact that $2|xy | leq x^{2}+y^{2}$.
answered Jan 7 at 6:04
Kavi Rama MurthyKavi Rama Murthy
54.4k32055
$endgroup$
add a comment |
$begingroup$
For $a=frac 1 4$ the derivative exist and is $0$: by definition of derivative this means $ frac {xy} {(x^{2}+y^{2})^{3/4}} to 0$ which follows easily from the fact that $2|xy | leq x^{2}+y^{2}$.
answered Jan 7 at 6:04
Kavi Rama MurthyKavi Rama Murthy
54.4k32055
$endgroup$
For $a=frac 1 4$ the derivative exist and is $0$: by definition of derivative this means $ frac {xy} {(x^{2}+y^{2})^{3/4}} to 0$ which follows easily from the fact that $2|xy | leq x^{2}+y^{2}$.
answered Jan 7 at 6:04
Kavi Rama MurthyKavi Rama Murthy
54.4k32055
edited Jan 7 at 7:13
answered Jan 7 at 6:04
Kavi Rama MurthyKavi Rama Murthy
54.4k32055
answered Jan 7 at 6:04
Kavi Rama MurthyKavi Rama Murthy
54.4k32055
answered Jan 7 at 6:04
Kavi Rama MurthyKavi Rama Murthy
54.4k32055
54.4k32055
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add a comment |
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