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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

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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 | 

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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