Maximizing directional derivatives [closed]












0












$begingroup$


Find the angle that maximizes the directional derivatives of $f(x,y)=x^2 - y^2$ at point $(3,4)$.



I don't know the angle is formed by which lines, vectors , or plane ?






multivariable-calculus partial-derivative







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$endgroup$



closed as off-topic by RRL, John Douma, Lord Shark the Unknown, onurcanbektas, Lee David Chung Lin Jan 22 at 8:31


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, John Douma, Lord Shark the Unknown, onurcanbektas, Lee David Chung Lin

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 2




    $begingroup$
    "The angle" in a two dimensional xy-coordinate system is typically the angle the line makes with the positive x-axis.
    $endgroup$
    – user247327
    Jan 22 at 1:18










  • $begingroup$
    so how can I find that angle to maximize the directional derivatives?
    $endgroup$
    – Andy Lam
    Jan 22 at 1:26
















0












$begingroup$


Find the angle that maximizes the directional derivatives of $f(x,y)=x^2 - y^2$ at point $(3,4)$.



I don't know the angle is formed by which lines, vectors , or plane ?






multivariable-calculus partial-derivative







share|cite|improve this question











$endgroup$



closed as off-topic by RRL, John Douma, Lord Shark the Unknown, onurcanbektas, Lee David Chung Lin Jan 22 at 8:31


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, John Douma, Lord Shark the Unknown, onurcanbektas, Lee David Chung Lin

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 2




    $begingroup$
    "The angle" in a two dimensional xy-coordinate system is typically the angle the line makes with the positive x-axis.
    $endgroup$
    – user247327
    Jan 22 at 1:18










  • $begingroup$
    so how can I find that angle to maximize the directional derivatives?
    $endgroup$
    – Andy Lam
    Jan 22 at 1:26














0












0








0





$begingroup$


Find the angle that maximizes the directional derivatives of $f(x,y)=x^2 - y^2$ at point $(3,4)$.



I don't know the angle is formed by which lines, vectors , or plane ?






multivariable-calculus partial-derivative







share|cite|improve this question











$endgroup$




Find the angle that maximizes the directional derivatives of $f(x,y)=x^2 - y^2$ at point $(3,4)$.



I don't know the angle is formed by which lines, vectors , or plane ?






multivariable-calculus partial-derivative




multivariable-calculus partial-derivative






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 22 at 1:50









Gnumbertester

670114




670114










asked Jan 22 at 1:15









Andy LamAndy Lam

63




63




closed as off-topic by RRL, John Douma, Lord Shark the Unknown, onurcanbektas, Lee David Chung Lin Jan 22 at 8:31


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, John Douma, Lord Shark the Unknown, onurcanbektas, Lee David Chung Lin

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by RRL, John Douma, Lord Shark the Unknown, onurcanbektas, Lee David Chung Lin Jan 22 at 8:31


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, John Douma, Lord Shark the Unknown, onurcanbektas, Lee David Chung Lin

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 2




    $begingroup$
    "The angle" in a two dimensional xy-coordinate system is typically the angle the line makes with the positive x-axis.
    $endgroup$
    – user247327
    Jan 22 at 1:18










  • $begingroup$
    so how can I find that angle to maximize the directional derivatives?
    $endgroup$
    – Andy Lam
    Jan 22 at 1:26














  • 2




    $begingroup$
    "The angle" in a two dimensional xy-coordinate system is typically the angle the line makes with the positive x-axis.
    $endgroup$
    – user247327
    Jan 22 at 1:18










  • $begingroup$
    so how can I find that angle to maximize the directional derivatives?
    $endgroup$
    – Andy Lam
    Jan 22 at 1:26








2




2




$begingroup$
"The angle" in a two dimensional xy-coordinate system is typically the angle the line makes with the positive x-axis.
$endgroup$
– user247327
Jan 22 at 1:18




$begingroup$
"The angle" in a two dimensional xy-coordinate system is typically the angle the line makes with the positive x-axis.
$endgroup$
– user247327
Jan 22 at 1:18












$begingroup$
so how can I find that angle to maximize the directional derivatives?
$endgroup$
– Andy Lam
Jan 22 at 1:26




$begingroup$
so how can I find that angle to maximize the directional derivatives?
$endgroup$
– Andy Lam
Jan 22 at 1:26










1 Answer
1






active

oldest

votes


















1












$begingroup$

The gradient vector $nabla f(x, y)$ always points in the direction of fastest increase- and its length is that rate of increase. Here $f(x,y)= x^2- y^2$ so $nabla f(x, y)= 2xvec{i}- 2yvec{j}$. At (3, 4) that is $nabla f(3, 4)= 2(3)vec{i}- 2(4)vec{k}= 6vec{i}+ 8vec{j}$. Taking $theta$ to be that angle that vector makes with the positive x- axis, we have $tan(theta)= frac{4}{3}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    can i ask why it is 4/3 but not -4/3?
    $endgroup$
    – Andy Lam
    Jan 22 at 3:49










  • $begingroup$
    Just a silly arithmetic error on my part! $nabla f(3,4)= 2(3)vec{i}- 2(4)vec{j}= 6vec{i}- 8vec{j}$ where I had "+ 8" before. Yes, $tan(theta)= frac{-8}{6}= -frac{4}{3}$. Thanks for catching that.
    $endgroup$
    – user247327
    Jan 22 at 12:52












  • $begingroup$
    thank you for your help
    $endgroup$
    – Andy Lam
    Jan 22 at 17:26


















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

The gradient vector $nabla f(x, y)$ always points in the direction of fastest increase- and its length is that rate of increase. Here $f(x,y)= x^2- y^2$ so $nabla f(x, y)= 2xvec{i}- 2yvec{j}$. At (3, 4) that is $nabla f(3, 4)= 2(3)vec{i}- 2(4)vec{k}= 6vec{i}+ 8vec{j}$. Taking $theta$ to be that angle that vector makes with the positive x- axis, we have $tan(theta)= frac{4}{3}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    can i ask why it is 4/3 but not -4/3?
    $endgroup$
    – Andy Lam
    Jan 22 at 3:49










  • $begingroup$
    Just a silly arithmetic error on my part! $nabla f(3,4)= 2(3)vec{i}- 2(4)vec{j}= 6vec{i}- 8vec{j}$ where I had "+ 8" before. Yes, $tan(theta)= frac{-8}{6}= -frac{4}{3}$. Thanks for catching that.
    $endgroup$
    – user247327
    Jan 22 at 12:52












  • $begingroup$
    thank you for your help
    $endgroup$
    – Andy Lam
    Jan 22 at 17:26
















1












$begingroup$

The gradient vector $nabla f(x, y)$ always points in the direction of fastest increase- and its length is that rate of increase. Here $f(x,y)= x^2- y^2$ so $nabla f(x, y)= 2xvec{i}- 2yvec{j}$. At (3, 4) that is $nabla f(3, 4)= 2(3)vec{i}- 2(4)vec{k}= 6vec{i}+ 8vec{j}$. Taking $theta$ to be that angle that vector makes with the positive x- axis, we have $tan(theta)= frac{4}{3}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    can i ask why it is 4/3 but not -4/3?
    $endgroup$
    – Andy Lam
    Jan 22 at 3:49










  • $begingroup$
    Just a silly arithmetic error on my part! $nabla f(3,4)= 2(3)vec{i}- 2(4)vec{j}= 6vec{i}- 8vec{j}$ where I had "+ 8" before. Yes, $tan(theta)= frac{-8}{6}= -frac{4}{3}$. Thanks for catching that.
    $endgroup$
    – user247327
    Jan 22 at 12:52












  • $begingroup$
    thank you for your help
    $endgroup$
    – Andy Lam
    Jan 22 at 17:26














1












1








1





$begingroup$

The gradient vector $nabla f(x, y)$ always points in the direction of fastest increase- and its length is that rate of increase. Here $f(x,y)= x^2- y^2$ so $nabla f(x, y)= 2xvec{i}- 2yvec{j}$. At (3, 4) that is $nabla f(3, 4)= 2(3)vec{i}- 2(4)vec{k}= 6vec{i}+ 8vec{j}$. Taking $theta$ to be that angle that vector makes with the positive x- axis, we have $tan(theta)= frac{4}{3}$.






share|cite|improve this answer









$endgroup$



The gradient vector $nabla f(x, y)$ always points in the direction of fastest increase- and its length is that rate of increase. Here $f(x,y)= x^2- y^2$ so $nabla f(x, y)= 2xvec{i}- 2yvec{j}$. At (3, 4) that is $nabla f(3, 4)= 2(3)vec{i}- 2(4)vec{k}= 6vec{i}+ 8vec{j}$. Taking $theta$ to be that angle that vector makes with the positive x- axis, we have $tan(theta)= frac{4}{3}$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 22 at 1:27









user247327user247327

11.4k1516




11.4k1516












  • $begingroup$
    can i ask why it is 4/3 but not -4/3?
    $endgroup$
    – Andy Lam
    Jan 22 at 3:49










  • $begingroup$
    Just a silly arithmetic error on my part! $nabla f(3,4)= 2(3)vec{i}- 2(4)vec{j}= 6vec{i}- 8vec{j}$ where I had "+ 8" before. Yes, $tan(theta)= frac{-8}{6}= -frac{4}{3}$. Thanks for catching that.
    $endgroup$
    – user247327
    Jan 22 at 12:52












  • $begingroup$
    thank you for your help
    $endgroup$
    – Andy Lam
    Jan 22 at 17:26


















  • $begingroup$
    can i ask why it is 4/3 but not -4/3?
    $endgroup$
    – Andy Lam
    Jan 22 at 3:49










  • $begingroup$
    Just a silly arithmetic error on my part! $nabla f(3,4)= 2(3)vec{i}- 2(4)vec{j}= 6vec{i}- 8vec{j}$ where I had "+ 8" before. Yes, $tan(theta)= frac{-8}{6}= -frac{4}{3}$. Thanks for catching that.
    $endgroup$
    – user247327
    Jan 22 at 12:52












  • $begingroup$
    thank you for your help
    $endgroup$
    – Andy Lam
    Jan 22 at 17:26
















$begingroup$
can i ask why it is 4/3 but not -4/3?
$endgroup$
– Andy Lam
Jan 22 at 3:49




$begingroup$
can i ask why it is 4/3 but not -4/3?
$endgroup$
– Andy Lam
Jan 22 at 3:49












$begingroup$
Just a silly arithmetic error on my part! $nabla f(3,4)= 2(3)vec{i}- 2(4)vec{j}= 6vec{i}- 8vec{j}$ where I had "+ 8" before. Yes, $tan(theta)= frac{-8}{6}= -frac{4}{3}$. Thanks for catching that.
$endgroup$
– user247327
Jan 22 at 12:52






$begingroup$
Just a silly arithmetic error on my part! $nabla f(3,4)= 2(3)vec{i}- 2(4)vec{j}= 6vec{i}- 8vec{j}$ where I had "+ 8" before. Yes, $tan(theta)= frac{-8}{6}= -frac{4}{3}$. Thanks for catching that.
$endgroup$
– user247327
Jan 22 at 12:52














$begingroup$
thank you for your help
$endgroup$
– Andy Lam
Jan 22 at 17:26




$begingroup$
thank you for your help
$endgroup$
– Andy Lam
Jan 22 at 17:26



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