Mixing
gradient of f at (1,1) from directional derivatives and vectors
$begingroup$
Image of the problem
Find the derivative (gradient) of $f$ at $(1,1)$ given the following directional derivatives where $u=2i+2j$ and $v=3i+j$
$D_uf(1,1)=frac{3}{sqrt2}$,
$D_uf(-1,1)=frac{7}{sqrt2}$,
$D_vf(1,1)=-frac{1}{sqrt10}$,
$D_vf(-1,1)=-frac{5}{sqrt10}$.
multivariable-calculus partial-derivative
asked Jan 31 at 17:03
ZaltahZaltah
74
$endgroup$
add a comment |
$begingroup$
Image of the problem
Find the derivative (gradient) of $f$ at $(1,1)$ given the following directional derivatives where $u=2i+2j$ and $v=3i+j$
$D_uf(1,1)=frac{3}{sqrt2}$,
$D_uf(-1,1)=frac{7}{sqrt2}$,
$D_vf(1,1)=-frac{1}{sqrt10}$,
$D_vf(-1,1)=-frac{5}{sqrt10}$.
multivariable-calculus partial-derivative
asked Jan 31 at 17:03
ZaltahZaltah
74
$endgroup$
$begingroup$
For the task of finding the gradient at (1, 1), information about the function and its derivatives at (-1, -1) is irrelevant. The function in a neighborhood of (-1, -1) might be completely different from the function in a neighborhood of (1, 1).
$endgroup$
– user247327
Jan 31 at 20:03
$begingroup$
Also there must be more information about f. f can be continuous at (1,1) and have those directional derivatives and NOT have a gradient. You need that the function is "differentiable in a neighborhood of (1, 1)".
$endgroup$
– user247327
Jan 31 at 20:08
add a comment |
$begingroup$
Image of the problem
Find the derivative (gradient) of $f$ at $(1,1)$ given the following directional derivatives where $u=2i+2j$ and $v=3i+j$
$D_uf(1,1)=frac{3}{sqrt2}$,
$D_uf(-1,1)=frac{7}{sqrt2}$,
$D_vf(1,1)=-frac{1}{sqrt10}$,
$D_vf(-1,1)=-frac{5}{sqrt10}$.
multivariable-calculus partial-derivative
asked Jan 31 at 17:03
ZaltahZaltah
74
$endgroup$
Image of the problem
Find the derivative (gradient) of $f$ at $(1,1)$ given the following directional derivatives where $u=2i+2j$ and $v=3i+j$
$D_uf(1,1)=frac{3}{sqrt2}$,
$D_uf(-1,1)=frac{7}{sqrt2}$,
$D_vf(1,1)=-frac{1}{sqrt10}$,
$D_vf(-1,1)=-frac{5}{sqrt10}$.
multivariable-calculus partial-derivative
multivariable-calculus partial-derivative
asked Jan 31 at 17:03
ZaltahZaltah
74
asked Jan 31 at 17:03
ZaltahZaltah
74
edited Jan 31 at 20:24
Zaltah
asked Jan 31 at 17:03
ZaltahZaltah
74
asked Jan 31 at 17:03
ZaltahZaltah
74
asked Jan 31 at 17:03
ZaltahZaltah
74
74
$begingroup$
For the task of finding the gradient at (1, 1), information about the function and its derivatives at (-1, -1) is irrelevant. The function in a neighborhood of (-1, -1) might be completely different from the function in a neighborhood of (1, 1).
$endgroup$
– user247327
Jan 31 at 20:03
$begingroup$
Also there must be more information about f. f can be continuous at (1,1) and have those directional derivatives and NOT have a gradient. You need that the function is "differentiable in a neighborhood of (1, 1)".
$endgroup$
– user247327
Jan 31 at 20:08
add a comment |
$begingroup$
For the task of finding the gradient at (1, 1), information about the function and its derivatives at (-1, -1) is irrelevant. The function in a neighborhood of (-1, -1) might be completely different from the function in a neighborhood of (1, 1).
$endgroup$
– user247327
Jan 31 at 20:03
$begingroup$
Also there must be more information about f. f can be continuous at (1,1) and have those directional derivatives and NOT have a gradient. You need that the function is "differentiable in a neighborhood of (1, 1)".
$endgroup$
– user247327
Jan 31 at 20:08
$begingroup$
For the task of finding the gradient at (1, 1), information about the function and its derivatives at (-1, -1) is irrelevant. The function in a neighborhood of (-1, -1) might be completely different from the function in a neighborhood of (1, 1).
$endgroup$
– user247327
Jan 31 at 20:03
$begingroup$
For the task of finding the gradient at (1, 1), information about the function and its derivatives at (-1, -1) is irrelevant. The function in a neighborhood of (-1, -1) might be completely different from the function in a neighborhood of (1, 1).
$endgroup$
– user247327
Jan 31 at 20:03
$begingroup$
Also there must be more information about f. f can be continuous at (1,1) and have those directional derivatives and NOT have a gradient. You need that the function is "differentiable in a neighborhood of (1, 1)".
$endgroup$
– user247327
Jan 31 at 20:08
$begingroup$
Also there must be more information about f. f can be continuous at (1,1) and have those directional derivatives and NOT have a gradient. You need that the function is "differentiable in a neighborhood of (1, 1)".
$endgroup$
– user247327
Jan 31 at 20:08
add a comment |
1 Answer
1
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Assuming that this function has a gradient (see my comments) we can write it as $nabla f= g(x,y)vec{i}+ h(x,y)vec{j}$. Now, its derivative in the direction of $vec{u}= vec{i}+ 2vec{j}$ is $2g(1,1)+ 2h(1,1)= frac{3}{sqrt{2}}$ and its derivative in the direction of $vec{v}= 3vec{i}+ vec{j}$ is $3g(1,1)+ h(1,1)= frac{7}{sqrt{2}}$. That gives two equations that can be solved for g(1, 1) and h(1, 1). But as I implied in my comments, the gradient is local. Knowing the value of g and h at (1, 1) does not tell us what g and h (and so f) are for other points.
Were you given other information about f, like it being linear of a conic?
answered Jan 31 at 20:22
user247327user247327
11.5k1516
$endgroup$
$begingroup$
I have included a picture of the problem I've been given
$endgroup$
– Zaltah
Jan 31 at 20:26
$begingroup$
Ah, so the problem is just asking for the gradient at (1, 1). For some reason I missed that when I first read the problem. All you need to do is solve $2g(1, 1)+ 2h(1,1)= frac{3}{sqrt{2}}$ and $3g(1.1)+ h(1,1)= frac{7}{sqrt{2}}$ for g(1,1) and h(1,1). Then $nabla f(1,1)= g(1, 1)vec{i}+ h(1,1)vec{j}$.
$endgroup$
– user247327
Jan 31 at 20:32
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
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votes
$begingroup$
Assuming that this function has a gradient (see my comments) we can write it as $nabla f= g(x,y)vec{i}+ h(x,y)vec{j}$. Now, its derivative in the direction of $vec{u}= vec{i}+ 2vec{j}$ is $2g(1,1)+ 2h(1,1)= frac{3}{sqrt{2}}$ and its derivative in the direction of $vec{v}= 3vec{i}+ vec{j}$ is $3g(1,1)+ h(1,1)= frac{7}{sqrt{2}}$. That gives two equations that can be solved for g(1, 1) and h(1, 1). But as I implied in my comments, the gradient is local. Knowing the value of g and h at (1, 1) does not tell us what g and h (and so f) are for other points.
Were you given other information about f, like it being linear of a conic?
answered Jan 31 at 20:22
user247327user247327
11.5k1516
$endgroup$
$begingroup$
I have included a picture of the problem I've been given
$endgroup$
– Zaltah
Jan 31 at 20:26
$begingroup$
Ah, so the problem is just asking for the gradient at (1, 1). For some reason I missed that when I first read the problem. All you need to do is solve $2g(1, 1)+ 2h(1,1)= frac{3}{sqrt{2}}$ and $3g(1.1)+ h(1,1)= frac{7}{sqrt{2}}$ for g(1,1) and h(1,1). Then $nabla f(1,1)= g(1, 1)vec{i}+ h(1,1)vec{j}$.
$endgroup$
– user247327
Jan 31 at 20:32
add a comment |
$begingroup$
Assuming that this function has a gradient (see my comments) we can write it as $nabla f= g(x,y)vec{i}+ h(x,y)vec{j}$. Now, its derivative in the direction of $vec{u}= vec{i}+ 2vec{j}$ is $2g(1,1)+ 2h(1,1)= frac{3}{sqrt{2}}$ and its derivative in the direction of $vec{v}= 3vec{i}+ vec{j}$ is $3g(1,1)+ h(1,1)= frac{7}{sqrt{2}}$. That gives two equations that can be solved for g(1, 1) and h(1, 1). But as I implied in my comments, the gradient is local. Knowing the value of g and h at (1, 1) does not tell us what g and h (and so f) are for other points.
Were you given other information about f, like it being linear of a conic?
answered Jan 31 at 20:22
user247327user247327
11.5k1516
$endgroup$
$begingroup$
I have included a picture of the problem I've been given
$endgroup$
– Zaltah
Jan 31 at 20:26
$begingroup$
Ah, so the problem is just asking for the gradient at (1, 1). For some reason I missed that when I first read the problem. All you need to do is solve $2g(1, 1)+ 2h(1,1)= frac{3}{sqrt{2}}$ and $3g(1.1)+ h(1,1)= frac{7}{sqrt{2}}$ for g(1,1) and h(1,1). Then $nabla f(1,1)= g(1, 1)vec{i}+ h(1,1)vec{j}$.
$endgroup$
– user247327
Jan 31 at 20:32
add a comment |
$begingroup$
Assuming that this function has a gradient (see my comments) we can write it as $nabla f= g(x,y)vec{i}+ h(x,y)vec{j}$. Now, its derivative in the direction of $vec{u}= vec{i}+ 2vec{j}$ is $2g(1,1)+ 2h(1,1)= frac{3}{sqrt{2}}$ and its derivative in the direction of $vec{v}= 3vec{i}+ vec{j}$ is $3g(1,1)+ h(1,1)= frac{7}{sqrt{2}}$. That gives two equations that can be solved for g(1, 1) and h(1, 1). But as I implied in my comments, the gradient is local. Knowing the value of g and h at (1, 1) does not tell us what g and h (and so f) are for other points.
Were you given other information about f, like it being linear of a conic?
answered Jan 31 at 20:22
user247327user247327
11.5k1516
$endgroup$
Assuming that this function has a gradient (see my comments) we can write it as $nabla f= g(x,y)vec{i}+ h(x,y)vec{j}$. Now, its derivative in the direction of $vec{u}= vec{i}+ 2vec{j}$ is $2g(1,1)+ 2h(1,1)= frac{3}{sqrt{2}}$ and its derivative in the direction of $vec{v}= 3vec{i}+ vec{j}$ is $3g(1,1)+ h(1,1)= frac{7}{sqrt{2}}$. That gives two equations that can be solved for g(1, 1) and h(1, 1). But as I implied in my comments, the gradient is local. Knowing the value of g and h at (1, 1) does not tell us what g and h (and so f) are for other points.
Were you given other information about f, like it being linear of a conic?
answered Jan 31 at 20:22
user247327user247327
11.5k1516
answered Jan 31 at 20:22
user247327user247327
11.5k1516
answered Jan 31 at 20:22
user247327user247327
11.5k1516
answered Jan 31 at 20:22
user247327user247327
11.5k1516
11.5k1516
$begingroup$
I have included a picture of the problem I've been given
$endgroup$
– Zaltah
Jan 31 at 20:26
$begingroup$
Ah, so the problem is just asking for the gradient at (1, 1). For some reason I missed that when I first read the problem. All you need to do is solve $2g(1, 1)+ 2h(1,1)= frac{3}{sqrt{2}}$ and $3g(1.1)+ h(1,1)= frac{7}{sqrt{2}}$ for g(1,1) and h(1,1). Then $nabla f(1,1)= g(1, 1)vec{i}+ h(1,1)vec{j}$.
$endgroup$
– user247327
Jan 31 at 20:32
add a comment |
$begingroup$
I have included a picture of the problem I've been given
$endgroup$
– Zaltah
Jan 31 at 20:26
$begingroup$
Ah, so the problem is just asking for the gradient at (1, 1). For some reason I missed that when I first read the problem. All you need to do is solve $2g(1, 1)+ 2h(1,1)= frac{3}{sqrt{2}}$ and $3g(1.1)+ h(1,1)= frac{7}{sqrt{2}}$ for g(1,1) and h(1,1). Then $nabla f(1,1)= g(1, 1)vec{i}+ h(1,1)vec{j}$.
$endgroup$
– user247327
Jan 31 at 20:32
$begingroup$
I have included a picture of the problem I've been given
$endgroup$
– Zaltah
Jan 31 at 20:26
$begingroup$
I have included a picture of the problem I've been given
$endgroup$
– Zaltah
Jan 31 at 20:26
$begingroup$
Ah, so the problem is just asking for the gradient at (1, 1). For some reason I missed that when I first read the problem. All you need to do is solve $2g(1, 1)+ 2h(1,1)= frac{3}{sqrt{2}}$ and $3g(1.1)+ h(1,1)= frac{7}{sqrt{2}}$ for g(1,1) and h(1,1). Then $nabla f(1,1)= g(1, 1)vec{i}+ h(1,1)vec{j}$.
$endgroup$
– user247327
Jan 31 at 20:32
$begingroup$
Ah, so the problem is just asking for the gradient at (1, 1). For some reason I missed that when I first read the problem. All you need to do is solve $2g(1, 1)+ 2h(1,1)= frac{3}{sqrt{2}}$ and $3g(1.1)+ h(1,1)= frac{7}{sqrt{2}}$ for g(1,1) and h(1,1). Then $nabla f(1,1)= g(1, 1)vec{i}+ h(1,1)vec{j}$.
$endgroup$
– user247327
Jan 31 at 20:32
add a comment |
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$begingroup$
For the task of finding the gradient at (1, 1), information about the function and its derivatives at (-1, -1) is irrelevant. The function in a neighborhood of (-1, -1) might be completely different from the function in a neighborhood of (1, 1).
$endgroup$
– user247327
Jan 31 at 20:03
$begingroup$
Also there must be more information about f. f can be continuous at (1,1) and have those directional derivatives and NOT have a gradient. You need that the function is "differentiable in a neighborhood of (1, 1)".
$endgroup$
– user247327
Jan 31 at 20:08