calculus lll how to prove this question?
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it is known that $f(0,0) = 0$ and $f$ is differentiable at $(0,0)$. It is also known that for every $t>0$ we have $f(cos(t)/t , sin(t)/t) > 0$.
Show that necessarily $operatorname{Gradient}(0,0) = (0,0)$.
I have no clue how to solve this. I have tried to show through limits, I have tried to get something from the definition ( that with $varepsilon, :(x/y) to 0$ at the limit)
I am sorry for the bad latex language really.
multivariable-calculus
edited Nov 18 at 15:02
David K
51.4k340113
asked Nov 18 at 14:40
Razi Awad
186
|
show 6 more comments
up vote
-2
down vote
favorite
it is known that $f(0,0) = 0$ and $f$ is differentiable at $(0,0)$. It is also known that for every $t>0$ we have $f(cos(t)/t , sin(t)/t) > 0$.
Show that necessarily $operatorname{Gradient}(0,0) = (0,0)$.
I have no clue how to solve this. I have tried to show through limits, I have tried to get something from the definition ( that with $varepsilon, :(x/y) to 0$ at the limit)
I am sorry for the bad latex language really.
multivariable-calculus
edited Nov 18 at 15:02
David K
51.4k340113
asked Nov 18 at 14:40
Razi Awad
186
please help i have asked this question before (a week ago) but no one try to help thank you
– Razi Awad
Nov 18 at 14:41
nevermind i'll do it myself remove this post.
– Razi Awad
Nov 18 at 14:49
1
Is your function defined as $$f: mathbb{R} to mathbb{R}^2, t mapsto begin{cases} left(frac{cos(t)}{t}, frac{sin(t)}{t}right) & t > 0 \ 0 & t = 0 end{cases}$$?
– Viktor Glombik
Nov 18 at 14:49
yes exactly like that
– Razi Awad
Nov 18 at 14:50
do you know how the gradient is defined?
– Viktor Glombik
Nov 18 at 14:52
|
show 6 more comments
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
it is known that $f(0,0) = 0$ and $f$ is differentiable at $(0,0)$. It is also known that for every $t>0$ we have $f(cos(t)/t , sin(t)/t) > 0$.
Show that necessarily $operatorname{Gradient}(0,0) = (0,0)$.
I have no clue how to solve this. I have tried to show through limits, I have tried to get something from the definition ( that with $varepsilon, :(x/y) to 0$ at the limit)
I am sorry for the bad latex language really.
multivariable-calculus
edited Nov 18 at 15:02
David K
51.4k340113
asked Nov 18 at 14:40
Razi Awad
186
it is known that $f(0,0) = 0$ and $f$ is differentiable at $(0,0)$. It is also known that for every $t>0$ we have $f(cos(t)/t , sin(t)/t) > 0$.
Show that necessarily $operatorname{Gradient}(0,0) = (0,0)$.
I have no clue how to solve this. I have tried to show through limits, I have tried to get something from the definition ( that with $varepsilon, :(x/y) to 0$ at the limit)
I am sorry for the bad latex language really.
multivariable-calculus
multivariable-calculus
edited Nov 18 at 15:02
David K
51.4k340113
asked Nov 18 at 14:40
Razi Awad
186
edited Nov 18 at 15:02
David K
51.4k340113
asked Nov 18 at 14:40
Razi Awad
186
edited Nov 18 at 15:02
David K
51.4k340113
edited Nov 18 at 15:02
David K
51.4k340113
edited Nov 18 at 15:02
David K
51.4k340113
51.4k340113
asked Nov 18 at 14:40
Razi Awad
186
asked Nov 18 at 14:40
Razi Awad
186
asked Nov 18 at 14:40
Razi Awad
186
186
please help i have asked this question before (a week ago) but no one try to help thank you
– Razi Awad
Nov 18 at 14:41
nevermind i'll do it myself remove this post.
– Razi Awad
Nov 18 at 14:49
1
Is your function defined as $$f: mathbb{R} to mathbb{R}^2, t mapsto begin{cases} left(frac{cos(t)}{t}, frac{sin(t)}{t}right) & t > 0 \ 0 & t = 0 end{cases}$$?
– Viktor Glombik
Nov 18 at 14:49
yes exactly like that
– Razi Awad
Nov 18 at 14:50
do you know how the gradient is defined?
– Viktor Glombik
Nov 18 at 14:52
|
show 6 more comments
please help i have asked this question before (a week ago) but no one try to help thank you
– Razi Awad
Nov 18 at 14:41
nevermind i'll do it myself remove this post.
– Razi Awad
Nov 18 at 14:49
1
Is your function defined as $$f: mathbb{R} to mathbb{R}^2, t mapsto begin{cases} left(frac{cos(t)}{t}, frac{sin(t)}{t}right) & t > 0 \ 0 & t = 0 end{cases}$$?
– Viktor Glombik
Nov 18 at 14:49
yes exactly like that
– Razi Awad
Nov 18 at 14:50
do you know how the gradient is defined?
– Viktor Glombik
Nov 18 at 14:52
please help i have asked this question before (a week ago) but no one try to help thank you
– Razi Awad
Nov 18 at 14:41
please help i have asked this question before (a week ago) but no one try to help thank you
– Razi Awad
Nov 18 at 14:41
nevermind i'll do it myself remove this post.
– Razi Awad
Nov 18 at 14:49
nevermind i'll do it myself remove this post.
– Razi Awad
Nov 18 at 14:49
1
1
Is your function defined as $$f: mathbb{R} to mathbb{R}^2, t mapsto begin{cases} left(frac{cos(t)}{t}, frac{sin(t)}{t}right) & t > 0 \ 0 & t = 0 end{cases}$$?
– Viktor Glombik
Nov 18 at 14:49
Is your function defined as $$f: mathbb{R} to mathbb{R}^2, t mapsto begin{cases} left(frac{cos(t)}{t}, frac{sin(t)}{t}right) & t > 0 \ 0 & t = 0 end{cases}$$?
– Viktor Glombik
Nov 18 at 14:49
yes exactly like that
– Razi Awad
Nov 18 at 14:50
yes exactly like that
– Razi Awad
Nov 18 at 14:50
do you know how the gradient is defined?
– Viktor Glombik
Nov 18 at 14:52
do you know how the gradient is defined?
– Viktor Glombik
Nov 18 at 14:52
|
show 6 more comments
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please help i have asked this question before (a week ago) but no one try to help thank you
– Razi Awad
Nov 18 at 14:41
nevermind i'll do it myself remove this post.
– Razi Awad
Nov 18 at 14:49
1
Is your function defined as $$f: mathbb{R} to mathbb{R}^2, t mapsto begin{cases} left(frac{cos(t)}{t}, frac{sin(t)}{t}right) & t > 0 \ 0 & t = 0 end{cases}$$?
– Viktor Glombik
Nov 18 at 14:49
yes exactly like that
– Razi Awad
Nov 18 at 14:50
do you know how the gradient is defined?
– Viktor Glombik
Nov 18 at 14:52