Multivariable calculus, using unit vector? [duplicate]

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  Graphical intuition behind directional derivative interpreted as a(∂f/∂x)+b(∂f/∂y), where <a, b> is directional vector of unit length? [duplicate]
  
  1 answer
  
  

Suppose we want the derivative of $f(x,y)$ in the direction of $langle 1,1rangle$. We must convert this to a unit vector: $langle 1/sqrt2, 1/sqrt2rangle$ in order to use the formula. So I don’t find it intuitive that the partial $x$ must be multiplied by $1/sqrt2$ and the partial $y$ must be multiplied by $1/sqrt2$, take the sum, and that’s your derivative in the direction $langle 1, 1rangle$.

multivariable-calculus

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edited Jul 30 at 1:23

高田航

1,116318

asked Jul 30 at 1:15

King Squirrel

1,19611329

marked as duplicate by Hans Lundmark, Isaac Browne, José Carlos Santos, amWhy, Leucippus Jul 31 at 3:50

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Would you rather report speeds in miles per hour or miles per 4.26435777 hours?
  – Randall
  Jul 30 at 1:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  See a proof: tutorial.math.lamar.edu/Classes/CalcIII/DirectionalDeriv.aspx
  – é«˜ç”°èˆª
  Jul 30 at 1:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I do agree but why use a vector with magnitude 1? Why not 0.5? I completely see how directional derivative in direction <1,0> is simply 1*partial x. What I do not get is why for vector <a, b> the directional derivative is a* partial x + b* partial y.
  – King Squirrel
  Jul 30 at 1:35
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  because magnitude 1 gives you a direction, without altering the speed at which you're travelling in that direction.
  – rubikscube09
  Jul 30 at 2:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This is the third time on the same day that you ask basically the same question...
  – Hans Lundmark
  Jul 30 at 7:31
  

  
  
  
  

 | 
show 1 more comment

up vote
-1
down vote

favorite

This question already has an answer here:

• 
  Graphical intuition behind directional derivative interpreted as a(∂f/∂x)+b(∂f/∂y), where <a, b> is directional vector of unit length? [duplicate]
  
  1 answer
  
  

Suppose we want the derivative of $f(x,y)$ in the direction of $langle 1,1rangle$. We must convert this to a unit vector: $langle 1/sqrt2, 1/sqrt2rangle$ in order to use the formula. So I don’t find it intuitive that the partial $x$ must be multiplied by $1/sqrt2$ and the partial $y$ must be multiplied by $1/sqrt2$, take the sum, and that’s your derivative in the direction $langle 1, 1rangle$.

multivariable-calculus

share|cite|improve this question

edited Jul 30 at 1:23

高田航

1,116318

asked Jul 30 at 1:15

King Squirrel

1,19611329

marked as duplicate by Hans Lundmark, Isaac Browne, José Carlos Santos, amWhy, Leucippus Jul 31 at 3:50

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Would you rather report speeds in miles per hour or miles per 4.26435777 hours?
  – Randall
  Jul 30 at 1:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  See a proof: tutorial.math.lamar.edu/Classes/CalcIII/DirectionalDeriv.aspx
  – é«˜ç”°èˆª
  Jul 30 at 1:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I do agree but why use a vector with magnitude 1? Why not 0.5? I completely see how directional derivative in direction <1,0> is simply 1*partial x. What I do not get is why for vector <a, b> the directional derivative is a* partial x + b* partial y.
  – King Squirrel
  Jul 30 at 1:35
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  because magnitude 1 gives you a direction, without altering the speed at which you're travelling in that direction.
  – rubikscube09
  Jul 30 at 2:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This is the third time on the same day that you ask basically the same question...
  – Hans Lundmark
  Jul 30 at 7:31
  

  
  
  
  

 | 
show 1 more comment

up vote
-1
down vote

favorite

up vote
-1
down vote

favorite

This question already has an answer here:

• 
  Graphical intuition behind directional derivative interpreted as a(∂f/∂x)+b(∂f/∂y), where <a, b> is directional vector of unit length? [duplicate]
  
  1 answer
  
  

Suppose we want the derivative of $f(x,y)$ in the direction of $langle 1,1rangle$. We must convert this to a unit vector: $langle 1/sqrt2, 1/sqrt2rangle$ in order to use the formula. So I don’t find it intuitive that the partial $x$ must be multiplied by $1/sqrt2$ and the partial $y$ must be multiplied by $1/sqrt2$, take the sum, and that’s your derivative in the direction $langle 1, 1rangle$.

multivariable-calculus

share|cite|improve this question

edited Jul 30 at 1:23

高田航

1,116318

asked Jul 30 at 1:15

King Squirrel

1,19611329

This question already has an answer here:

• 
  Graphical intuition behind directional derivative interpreted as a(∂f/∂x)+b(∂f/∂y), where <a, b> is directional vector of unit length? [duplicate]
  
  1 answer
  
  

Suppose we want the derivative of $f(x,y)$ in the direction of $langle 1,1rangle$. We must convert this to a unit vector: $langle 1/sqrt2, 1/sqrt2rangle$ in order to use the formula. So I don’t find it intuitive that the partial $x$ must be multiplied by $1/sqrt2$ and the partial $y$ must be multiplied by $1/sqrt2$, take the sum, and that’s your derivative in the direction $langle 1, 1rangle$.

This question already has an answer here:

• 
  Graphical intuition behind directional derivative interpreted as a(∂f/∂x)+b(∂f/∂y), where <a, b> is directional vector of unit length? [duplicate]
  
  1 answer
  
  

multivariable-calculus

share|cite|improve this question

edited Jul 30 at 1:23

高田航

1,116318

asked Jul 30 at 1:15

King Squirrel

1,19611329

share|cite|improve this question

share|cite|improve this question

edited Jul 30 at 1:23

高田航

1,116318

edited Jul 30 at 1:23

高田航

1,116318

edited Jul 30 at 1:23

高田航

1,116318

1,116318

asked Jul 30 at 1:15

King Squirrel

1,19611329

asked Jul 30 at 1:15

King Squirrel

1,19611329

asked Jul 30 at 1:15

King Squirrel

1,19611329

1,19611329

marked as duplicate by Hans Lundmark, Isaac Browne, José Carlos Santos, amWhy, Leucippus Jul 31 at 3:50

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

marked as duplicate by Hans Lundmark, Isaac Browne, José Carlos Santos, amWhy, Leucippus Jul 31 at 3:50

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Would you rather report speeds in miles per hour or miles per 4.26435777 hours?
  – Randall
  Jul 30 at 1:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  See a proof: tutorial.math.lamar.edu/Classes/CalcIII/DirectionalDeriv.aspx
  – é«˜ç”°èˆª
  Jul 30 at 1:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I do agree but why use a vector with magnitude 1? Why not 0.5? I completely see how directional derivative in direction <1,0> is simply 1*partial x. What I do not get is why for vector <a, b> the directional derivative is a* partial x + b* partial y.
  – King Squirrel
  Jul 30 at 1:35
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  because magnitude 1 gives you a direction, without altering the speed at which you're travelling in that direction.
  – rubikscube09
  Jul 30 at 2:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This is the third time on the same day that you ask basically the same question...
  – Hans Lundmark
  Jul 30 at 7:31
  

  
  
  
  

 | 
show 1 more comment

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Would you rather report speeds in miles per hour or miles per 4.26435777 hours?
  – Randall
  Jul 30 at 1:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  See a proof: tutorial.math.lamar.edu/Classes/CalcIII/DirectionalDeriv.aspx
  – é«˜ç”°èˆª
  Jul 30 at 1:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I do agree but why use a vector with magnitude 1? Why not 0.5? I completely see how directional derivative in direction <1,0> is simply 1*partial x. What I do not get is why for vector <a, b> the directional derivative is a* partial x + b* partial y.
  – King Squirrel
  Jul 30 at 1:35
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  because magnitude 1 gives you a direction, without altering the speed at which you're travelling in that direction.
  – rubikscube09
  Jul 30 at 2:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This is the third time on the same day that you ask basically the same question...
  – Hans Lundmark
  Jul 30 at 7:31
  

  
  
  
  

1

1

Would you rather report speeds in miles per hour or miles per 4.26435777 hours?
– Randall
Jul 30 at 1:20

Would you rather report speeds in miles per hour or miles per 4.26435777 hours?
– Randall
Jul 30 at 1:20

See a proof: tutorial.math.lamar.edu/Classes/CalcIII/DirectionalDeriv.aspx
– é«˜ç”°èˆª
Jul 30 at 1:23

See a proof: tutorial.math.lamar.edu/Classes/CalcIII/DirectionalDeriv.aspx
– é«˜ç”°èˆª
Jul 30 at 1:23

I do agree but why use a vector with magnitude 1? Why not 0.5? I completely see how directional derivative in direction <1,0> is simply 1*partial x. What I do not get is why for vector <a, b> the directional derivative is a* partial x + b* partial y.
– King Squirrel
Jul 30 at 1:35

I do agree but why use a vector with magnitude 1? Why not 0.5? I completely see how directional derivative in direction <1,0> is simply 1*partial x. What I do not get is why for vector <a, b> the directional derivative is a* partial x + b* partial y.
– King Squirrel
Jul 30 at 1:35

1

1

because magnitude 1 gives you a direction, without altering the speed at which you're travelling in that direction.
– rubikscube09
Jul 30 at 2:21

because magnitude 1 gives you a direction, without altering the speed at which you're travelling in that direction.
– rubikscube09
Jul 30 at 2:21

This is the third time on the same day that you ask basically the same question...
– Hans Lundmark
Jul 30 at 7:31

This is the third time on the same day that you ask basically the same question...
– Hans Lundmark
Jul 30 at 7:31

 | 
show 1 more comment

1 Answer
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1
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The directional derivative is the rate of change of your function at a given point when you move along the given direction. You want to find the linear change when you move in the given direction one unit of length. That is why you normalize your direction vector to make its length $1$.

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answered Jul 30 at 2:18

Mohammad Riazi-Kermani

27.3k41851

add a comment | 

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote

The directional derivative is the rate of change of your function at a given point when you move along the given direction. You want to find the linear change when you move in the given direction one unit of length. That is why you normalize your direction vector to make its length $1$.

share|cite|improve this answer

answered Jul 30 at 2:18

Mohammad Riazi-Kermani

27.3k41851

add a comment | 

up vote
1
down vote

The directional derivative is the rate of change of your function at a given point when you move along the given direction. You want to find the linear change when you move in the given direction one unit of length. That is why you normalize your direction vector to make its length $1$.

share|cite|improve this answer

answered Jul 30 at 2:18

Mohammad Riazi-Kermani

27.3k41851

add a comment | 

up vote
1
down vote

up vote
1
down vote

The directional derivative is the rate of change of your function at a given point when you move along the given direction. You want to find the linear change when you move in the given direction one unit of length. That is why you normalize your direction vector to make its length $1$.

share|cite|improve this answer

answered Jul 30 at 2:18

Mohammad Riazi-Kermani

27.3k41851

The directional derivative is the rate of change of your function at a given point when you move along the given direction. You want to find the linear change when you move in the given direction one unit of length. That is why you normalize your direction vector to make its length $1$.

share|cite|improve this answer

answered Jul 30 at 2:18

Mohammad Riazi-Kermani

27.3k41851

share|cite|improve this answer

share|cite|improve this answer

answered Jul 30 at 2:18

Mohammad Riazi-Kermani

27.3k41851

answered Jul 30 at 2:18

Mohammad Riazi-Kermani

27.3k41851

answered Jul 30 at 2:18

Mohammad Riazi-Kermani

27.3k41851

27.3k41851

add a comment | 

add a comment |