Fdtxjr

Do I really need to have a message in a novel to appeal to readers?

When the Haste spell ends on a creature, do attackers have advantage against that creature?

Is grep documentation wrong?

Compare a given version number in the form major.minor.build.patch and see if one is less than the other

Is it common practice to audition new musicians one-on-one before rehearsing with the entire band?

What do you call a floor made of glass so you can see through the floor?

Do I really need recursive chmod to restrict access to a folder?

Why didn't Eitri join the fight?

Around usage results

First console to have temporary backward compatibility

Most bit efficient text communication method?

How come Sam didn't become Lord of Horn Hill?

old style "caution" boxes

How does the math work when buying airline miles?

Crossing US/Canada Border for less than 24 hours

How to compare two different files line by line in unix?

Wu formula for manifolds with boundary

How do I stop a creek from eroding my steep embankment?

Is there such thing as an Availability Group failover trigger?

How do I make this wiring inside cabinet safer? (Pic)

How to tell that you are a giant?

Can a new player join a group only when a new campaign starts?

Maximum summed powersets with non-adjacent items

What is homebrew?

Finding the direction in which the derivative is exactly $2$

1

$begingroup$

I found the gradient taking partial derivatives and got $nabla f = (1,2)$. I know that the directional derivative has to equal $2$ so if I set $(1,2)cdot (x,y) = 2$ , where $(x,y)$ is a unit vector, I get $x + 2y = 2$. I'm not sure how to proceed from here.

multivariable-calculus partial-derivative

share|cite|improve this question

edited Mar 27 at 9:19

Ernie060

2,940719

asked Mar 27 at 4:29

krauser126krauser126

636

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Unit vector implies $x^2 + y^2 = 1$. Now you've got two equations for your two unknowns.
  $endgroup$
  – JonathanZ
  Mar 27 at 4:34
  

  
  
  
  

add a comment | 

1

$begingroup$

I found the gradient taking partial derivatives and got $nabla f = (1,2)$. I know that the directional derivative has to equal $2$ so if I set $(1,2)cdot (x,y) = 2$ , where $(x,y)$ is a unit vector, I get $x + 2y = 2$. I'm not sure how to proceed from here.

multivariable-calculus partial-derivative

share|cite|improve this question

edited Mar 27 at 9:19

Ernie060

2,940719

asked Mar 27 at 4:29

krauser126krauser126

636

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Unit vector implies $x^2 + y^2 = 1$. Now you've got two equations for your two unknowns.
  $endgroup$
  – JonathanZ
  Mar 27 at 4:34
  

  
  
  
  

add a comment | 

$begingroup$

I found the gradient taking partial derivatives and got $nabla f = (1,2)$. I know that the directional derivative has to equal $2$ so if I set $(1,2)cdot (x,y) = 2$ , where $(x,y)$ is a unit vector, I get $x + 2y = 2$. I'm not sure how to proceed from here.

multivariable-calculus partial-derivative

share|cite|improve this question

edited Mar 27 at 9:19

Ernie060

2,940719

asked Mar 27 at 4:29

krauser126krauser126

636

$endgroup$

share|cite|improve this question

edited Mar 27 at 9:19

Ernie060

2,940719

asked Mar 27 at 4:29

krauser126krauser126

636

share|cite|improve this question

edited Mar 27 at 9:19

Ernie060

2,940719

asked Mar 27 at 4:29

krauser126krauser126

636

edited Mar 27 at 9:19

Ernie060

2,940719

edited Mar 27 at 9:19

Ernie060

2,940719

asked Mar 27 at 4:29

krauser126krauser126

636

asked Mar 27 at 4:29

krauser126krauser126

636

2 Answers
2

active

oldest

votes

1

$begingroup$

A "direction" is nothing but a unit vector, since the way in which a vector is oriented depends only upon the way in which its unit vector is oriented.

Therefore, you may assume that $(x,y)$ has unit length i.e. $x^2+y^2 = 1$, and then find $x,y$ , being two equations of two unknowns.

(Note : the computations performed by you are correct till the end of your question)

share|cite|improve this answer

answered Mar 27 at 5:13

астон вілла олоф мэллбэргастон вілла олоф мэллбэрг

40.6k33678

$endgroup$

add a comment | 

1

$begingroup$

You are right all along. Now, it is a bit of time solving the following system of variables:$$begincasesx+2y=2\x^2+y^2=1endcases$$which has two answers $$(x,y)=(0,1)\(x,y)=(0.8,0.6)$$

share|cite|improve this answer

answered Mar 27 at 8:10

Mostafa AyazMostafa Ayaz

18.1k31040

$endgroup$

add a comment | 

Your Answer

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

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3164106%2ffinding-the-direction-in-which-the-derivative-is-exactly-2%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

1

$begingroup$

A "direction" is nothing but a unit vector, since the way in which a vector is oriented depends only upon the way in which its unit vector is oriented.

Therefore, you may assume that $(x,y)$ has unit length i.e. $x^2+y^2 = 1$, and then find $x,y$ , being two equations of two unknowns.

(Note : the computations performed by you are correct till the end of your question)

share|cite|improve this answer

answered Mar 27 at 5:13

астон вілла олоф мэллбэргастон вілла олоф мэллбэрг

40.6k33678

$endgroup$

add a comment | 

1

$begingroup$

A "direction" is nothing but a unit vector, since the way in which a vector is oriented depends only upon the way in which its unit vector is oriented.

Therefore, you may assume that $(x,y)$ has unit length i.e. $x^2+y^2 = 1$, and then find $x,y$ , being two equations of two unknowns.

(Note : the computations performed by you are correct till the end of your question)

share|cite|improve this answer

answered Mar 27 at 5:13

астон вілла олоф мэллбэргастон вілла олоф мэллбэрг

40.6k33678

$endgroup$

add a comment | 

$begingroup$

A "direction" is nothing but a unit vector, since the way in which a vector is oriented depends only upon the way in which its unit vector is oriented.

Therefore, you may assume that $(x,y)$ has unit length i.e. $x^2+y^2 = 1$, and then find $x,y$ , being two equations of two unknowns.

(Note : the computations performed by you are correct till the end of your question)

share|cite|improve this answer

answered Mar 27 at 5:13

астон вілла олоф мэллбэргастон вілла олоф мэллбэрг

40.6k33678

$endgroup$

share|cite|improve this answer

answered Mar 27 at 5:13

астон вілла олоф мэллбэргастон вілла олоф мэллбэрг

40.6k33678

answered Mar 27 at 5:13

астон вілла олоф мэллбэргастон вілла олоф мэллбэрг

40.6k33678

answered Mar 27 at 5:13

астон вілла олоф мэллбэргастон вілла олоф мэллбэрг

40.6k33678

1

$begingroup$

You are right all along. Now, it is a bit of time solving the following system of variables:$$begincasesx+2y=2\x^2+y^2=1endcases$$which has two answers $$(x,y)=(0,1)\(x,y)=(0.8,0.6)$$

share|cite|improve this answer

answered Mar 27 at 8:10

Mostafa AyazMostafa Ayaz

18.1k31040

$endgroup$

add a comment | 

1

$begingroup$

You are right all along. Now, it is a bit of time solving the following system of variables:$$begincasesx+2y=2\x^2+y^2=1endcases$$which has two answers $$(x,y)=(0,1)\(x,y)=(0.8,0.6)$$

share|cite|improve this answer

answered Mar 27 at 8:10

Mostafa AyazMostafa Ayaz

18.1k31040

$endgroup$

add a comment | 

$begingroup$

You are right all along. Now, it is a bit of time solving the following system of variables:$$begincasesx+2y=2\x^2+y^2=1endcases$$which has two answers $$(x,y)=(0,1)\(x,y)=(0.8,0.6)$$

share|cite|improve this answer

answered Mar 27 at 8:10

Mostafa AyazMostafa Ayaz

18.1k31040

$endgroup$

share|cite|improve this answer

answered Mar 27 at 8:10

Mostafa AyazMostafa Ayaz

18.1k31040

answered Mar 27 at 8:10

Mostafa AyazMostafa Ayaz

18.1k31040

answered Mar 27 at 8:10

Mostafa AyazMostafa Ayaz

18.1k31040