Directional derivatives at certain points












1












$begingroup$



Consider the function $f(x,y,z) = |x+y+z|$. If ${bf a} =
(a_1,a_2,a_3)$ satisfies $a_1+a_2+a_3=0$ find those directions ${bf
v}$ in which the derivative at ${bf a}$ exists.




Attempt



Note that $f(a_1,a_2,a_3) = 0$. So, if ${bf v } = (v_1,v_2,v_3)$ is any vector we have



$$ D_{bf v} f( {bf a} ) = lim_{s to 0} frac{ f( a_i + s v_i ) - f(a_i ) }{s} $$



but $f(a_i) = 0$ and $f(a_i+sv_i) = |a_1 + sv_1 + a_2 + sv_2 + a_3 + sv_3| = s |v_1+v_2+v_3|$



So,



$$ D_{bf v} f( {bf a}) = lim_{s to 0} frac{ s |v_1+v_2+v_3| }{s} = |v_1+v_2+v_3| = f ({bf v})$$



but inst this hold for any $v$ and so the derivative of $f$ at ${bf a}$ exists at all directions. Is this correct?






calculus







share|cite|improve this question









$endgroup$












  • $begingroup$
    This is "close" to being correct but there is an error.
    $endgroup$
    – RRL
    Jan 12 at 5:44


















1












$begingroup$



Consider the function $f(x,y,z) = |x+y+z|$. If ${bf a} =
(a_1,a_2,a_3)$ satisfies $a_1+a_2+a_3=0$ find those directions ${bf
v}$ in which the derivative at ${bf a}$ exists.




Attempt



Note that $f(a_1,a_2,a_3) = 0$. So, if ${bf v } = (v_1,v_2,v_3)$ is any vector we have



$$ D_{bf v} f( {bf a} ) = lim_{s to 0} frac{ f( a_i + s v_i ) - f(a_i ) }{s} $$



but $f(a_i) = 0$ and $f(a_i+sv_i) = |a_1 + sv_1 + a_2 + sv_2 + a_3 + sv_3| = s |v_1+v_2+v_3|$



So,



$$ D_{bf v} f( {bf a}) = lim_{s to 0} frac{ s |v_1+v_2+v_3| }{s} = |v_1+v_2+v_3| = f ({bf v})$$



but inst this hold for any $v$ and so the derivative of $f$ at ${bf a}$ exists at all directions. Is this correct?






calculus







share|cite|improve this question









$endgroup$












  • $begingroup$
    This is "close" to being correct but there is an error.
    $endgroup$
    – RRL
    Jan 12 at 5:44
















1












1








1





$begingroup$



Consider the function $f(x,y,z) = |x+y+z|$. If ${bf a} =
(a_1,a_2,a_3)$ satisfies $a_1+a_2+a_3=0$ find those directions ${bf
v}$ in which the derivative at ${bf a}$ exists.




Attempt



Note that $f(a_1,a_2,a_3) = 0$. So, if ${bf v } = (v_1,v_2,v_3)$ is any vector we have



$$ D_{bf v} f( {bf a} ) = lim_{s to 0} frac{ f( a_i + s v_i ) - f(a_i ) }{s} $$



but $f(a_i) = 0$ and $f(a_i+sv_i) = |a_1 + sv_1 + a_2 + sv_2 + a_3 + sv_3| = s |v_1+v_2+v_3|$



So,



$$ D_{bf v} f( {bf a}) = lim_{s to 0} frac{ s |v_1+v_2+v_3| }{s} = |v_1+v_2+v_3| = f ({bf v})$$



but inst this hold for any $v$ and so the derivative of $f$ at ${bf a}$ exists at all directions. Is this correct?






calculus







share|cite|improve this question









$endgroup$





Consider the function $f(x,y,z) = |x+y+z|$. If ${bf a} =
(a_1,a_2,a_3)$ satisfies $a_1+a_2+a_3=0$ find those directions ${bf
v}$ in which the derivative at ${bf a}$ exists.




Attempt



Note that $f(a_1,a_2,a_3) = 0$. So, if ${bf v } = (v_1,v_2,v_3)$ is any vector we have



$$ D_{bf v} f( {bf a} ) = lim_{s to 0} frac{ f( a_i + s v_i ) - f(a_i ) }{s} $$



but $f(a_i) = 0$ and $f(a_i+sv_i) = |a_1 + sv_1 + a_2 + sv_2 + a_3 + sv_3| = s |v_1+v_2+v_3|$



So,



$$ D_{bf v} f( {bf a}) = lim_{s to 0} frac{ s |v_1+v_2+v_3| }{s} = |v_1+v_2+v_3| = f ({bf v})$$



but inst this hold for any $v$ and so the derivative of $f$ at ${bf a}$ exists at all directions. Is this correct?






calculus




calculus






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 12 at 4:02









Jimmy SabaterJimmy Sabater

2,551320




2,551320












  • $begingroup$
    This is "close" to being correct but there is an error.
    $endgroup$
    – RRL
    Jan 12 at 5:44




















  • $begingroup$
    This is "close" to being correct but there is an error.
    $endgroup$
    – RRL
    Jan 12 at 5:44


















$begingroup$
This is "close" to being correct but there is an error.
$endgroup$
– RRL
Jan 12 at 5:44






$begingroup$
This is "close" to being correct but there is an error.
$endgroup$
– RRL
Jan 12 at 5:44












1 Answer
1






active

oldest

votes


















4












$begingroup$

The directional derivative exists if limits as $s to 0+$ from the right and as $s to 0-$ from the left both exist and are equal.



You overlooked the fact that $|s v_1 + sv_2 + sv_3| = |s||v_1 + v_2 +v_3| neq s|v_1 + v_2 +v_3|$ if $s < 0$.



Thus,



$$frac{f(mathbf{a}+ smathbf{v}) - f(mathbf{a})}{s} = frac{|s|}{s}|v_1+v_2 + v_3| = text{sgn}(s)|v_1 + v_2 + v_3|,$$



and left- and right-limits are unequal (due to opposite sign) unless $|v_1 + v_2 +v_3| = 0$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks so much. So this means that the derivative exist only on the direction of vector $$ (x,y,z) = (-1,1,0) alpha + (-1,0,1) beta $$ where $alpha, beta in mathbb{R}$
    $endgroup$
    – Jimmy Sabater
    Jan 12 at 17:33










  • $begingroup$
    @JimmySabater: You're welcome and you are correct -- the two-dimensional subspace of vectors with components that sum to 0.
    $endgroup$
    – RRL
    Jan 12 at 17:47











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3070588%2fdirectional-derivatives-at-certain-points%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

The directional derivative exists if limits as $s to 0+$ from the right and as $s to 0-$ from the left both exist and are equal.



You overlooked the fact that $|s v_1 + sv_2 + sv_3| = |s||v_1 + v_2 +v_3| neq s|v_1 + v_2 +v_3|$ if $s < 0$.



Thus,



$$frac{f(mathbf{a}+ smathbf{v}) - f(mathbf{a})}{s} = frac{|s|}{s}|v_1+v_2 + v_3| = text{sgn}(s)|v_1 + v_2 + v_3|,$$



and left- and right-limits are unequal (due to opposite sign) unless $|v_1 + v_2 +v_3| = 0$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks so much. So this means that the derivative exist only on the direction of vector $$ (x,y,z) = (-1,1,0) alpha + (-1,0,1) beta $$ where $alpha, beta in mathbb{R}$
    $endgroup$
    – Jimmy Sabater
    Jan 12 at 17:33










  • $begingroup$
    @JimmySabater: You're welcome and you are correct -- the two-dimensional subspace of vectors with components that sum to 0.
    $endgroup$
    – RRL
    Jan 12 at 17:47
















4












$begingroup$

The directional derivative exists if limits as $s to 0+$ from the right and as $s to 0-$ from the left both exist and are equal.



You overlooked the fact that $|s v_1 + sv_2 + sv_3| = |s||v_1 + v_2 +v_3| neq s|v_1 + v_2 +v_3|$ if $s < 0$.



Thus,



$$frac{f(mathbf{a}+ smathbf{v}) - f(mathbf{a})}{s} = frac{|s|}{s}|v_1+v_2 + v_3| = text{sgn}(s)|v_1 + v_2 + v_3|,$$



and left- and right-limits are unequal (due to opposite sign) unless $|v_1 + v_2 +v_3| = 0$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks so much. So this means that the derivative exist only on the direction of vector $$ (x,y,z) = (-1,1,0) alpha + (-1,0,1) beta $$ where $alpha, beta in mathbb{R}$
    $endgroup$
    – Jimmy Sabater
    Jan 12 at 17:33










  • $begingroup$
    @JimmySabater: You're welcome and you are correct -- the two-dimensional subspace of vectors with components that sum to 0.
    $endgroup$
    – RRL
    Jan 12 at 17:47














4












4








4





$begingroup$

The directional derivative exists if limits as $s to 0+$ from the right and as $s to 0-$ from the left both exist and are equal.



You overlooked the fact that $|s v_1 + sv_2 + sv_3| = |s||v_1 + v_2 +v_3| neq s|v_1 + v_2 +v_3|$ if $s < 0$.



Thus,



$$frac{f(mathbf{a}+ smathbf{v}) - f(mathbf{a})}{s} = frac{|s|}{s}|v_1+v_2 + v_3| = text{sgn}(s)|v_1 + v_2 + v_3|,$$



and left- and right-limits are unequal (due to opposite sign) unless $|v_1 + v_2 +v_3| = 0$






share|cite|improve this answer











$endgroup$



The directional derivative exists if limits as $s to 0+$ from the right and as $s to 0-$ from the left both exist and are equal.



You overlooked the fact that $|s v_1 + sv_2 + sv_3| = |s||v_1 + v_2 +v_3| neq s|v_1 + v_2 +v_3|$ if $s < 0$.



Thus,



$$frac{f(mathbf{a}+ smathbf{v}) - f(mathbf{a})}{s} = frac{|s|}{s}|v_1+v_2 + v_3| = text{sgn}(s)|v_1 + v_2 + v_3|,$$



and left- and right-limits are unequal (due to opposite sign) unless $|v_1 + v_2 +v_3| = 0$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 12 at 5:59

























answered Jan 12 at 5:38









RRLRRL

50.3k42573




50.3k42573












  • $begingroup$
    Thanks so much. So this means that the derivative exist only on the direction of vector $$ (x,y,z) = (-1,1,0) alpha + (-1,0,1) beta $$ where $alpha, beta in mathbb{R}$
    $endgroup$
    – Jimmy Sabater
    Jan 12 at 17:33










  • $begingroup$
    @JimmySabater: You're welcome and you are correct -- the two-dimensional subspace of vectors with components that sum to 0.
    $endgroup$
    – RRL
    Jan 12 at 17:47


















  • $begingroup$
    Thanks so much. So this means that the derivative exist only on the direction of vector $$ (x,y,z) = (-1,1,0) alpha + (-1,0,1) beta $$ where $alpha, beta in mathbb{R}$
    $endgroup$
    – Jimmy Sabater
    Jan 12 at 17:33










  • $begingroup$
    @JimmySabater: You're welcome and you are correct -- the two-dimensional subspace of vectors with components that sum to 0.
    $endgroup$
    – RRL
    Jan 12 at 17:47
















$begingroup$
Thanks so much. So this means that the derivative exist only on the direction of vector $$ (x,y,z) = (-1,1,0) alpha + (-1,0,1) beta $$ where $alpha, beta in mathbb{R}$
$endgroup$
– Jimmy Sabater
Jan 12 at 17:33




$begingroup$
Thanks so much. So this means that the derivative exist only on the direction of vector $$ (x,y,z) = (-1,1,0) alpha + (-1,0,1) beta $$ where $alpha, beta in mathbb{R}$
$endgroup$
– Jimmy Sabater
Jan 12 at 17:33












$begingroup$
@JimmySabater: You're welcome and you are correct -- the two-dimensional subspace of vectors with components that sum to 0.
$endgroup$
– RRL
Jan 12 at 17:47




$begingroup$
@JimmySabater: You're welcome and you are correct -- the two-dimensional subspace of vectors with components that sum to 0.
$endgroup$
– RRL
Jan 12 at 17:47


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3070588%2fdirectional-derivatives-at-certain-points%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Mario Kart Wii

Image
Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro...
Read more

The Binding of Isaac: Rebirth/Afterbirth

Image
Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! The Binding of Isaac: Rebirth/Afterbirth From The Cutting Room Floor < The Binding of Isaac: Rebirth Jump to: navigation, search This is a sub-page of The Binding of Isaac: Rebirth . Afterbirth is the first expansion pack for Rebirth and features a variety of improvements. Contents 1 Unused Graphics 1.1 Old Man Keeper 1.2 Skinless Hush 1.3 Miscellaneous Unused Hush Sprites 1.4 Generosity achievement 1.5 Afterbirth+ Room Editor 1.6 Bearded Ultra Greed 1.7 Early Ultra Greed 1.8 Unused Brownie Expression 1.9 Frail's Spinning Head 1.10 Mom's Dead Hand Alt 2 Unused Audio 2.1 Unused Dead Womb Layer 2.2 Unused Hush Battle Outro 2.3 The Stain Grunt Unused Graphics Old Man Keeper Image file: gfxuimain me...
Read more

What does “Dominus providebit” mean?

Image
5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1 ...
Read more