Mixing
Directional derivative and the gradient
$begingroup$
I couldn't find the solution of this question, can you help me, please?
Let the function f(x,y,z) be differentiable at a point P. The maximum value of the directional derivative of f at P is 6 and the direction of v=(1,1,-1). Find the the gradient vector of f at P.
multivariable-calculus
asked Jan 12 at 19:27
Kaan IzmirKaan Izmir
61
$endgroup$
add a comment |
$begingroup$
I couldn't find the solution of this question, can you help me, please?
Let the function f(x,y,z) be differentiable at a point P. The maximum value of the directional derivative of f at P is 6 and the direction of v=(1,1,-1). Find the the gradient vector of f at P.
multivariable-calculus
asked Jan 12 at 19:27
Kaan IzmirKaan Izmir
61
$endgroup$
1
$begingroup$
Do you know the relation between the directional derivative and the gradient?
$endgroup$
– Ian
Jan 12 at 19:32
$begingroup$
Yes, I know the relation between them.
$endgroup$
– Kaan Izmir
Jan 12 at 19:34
1
$begingroup$
Knowing that and the bound $-| x | | y | leq x cdot y leq | x | | y |$ (called the Cauchy-Schwarz inequality), you can finish the problem.
$endgroup$
– Ian
Jan 12 at 19:36
$begingroup$
Yeah, thank you for that
$endgroup$
– Kaan Izmir
Jan 12 at 19:44
add a comment |
$begingroup$
I couldn't find the solution of this question, can you help me, please?
Let the function f(x,y,z) be differentiable at a point P. The maximum value of the directional derivative of f at P is 6 and the direction of v=(1,1,-1). Find the the gradient vector of f at P.
multivariable-calculus
asked Jan 12 at 19:27
Kaan IzmirKaan Izmir
61
$endgroup$
I couldn't find the solution of this question, can you help me, please?
Let the function f(x,y,z) be differentiable at a point P. The maximum value of the directional derivative of f at P is 6 and the direction of v=(1,1,-1). Find the the gradient vector of f at P.
multivariable-calculus
multivariable-calculus
asked Jan 12 at 19:27
Kaan IzmirKaan Izmir
61
asked Jan 12 at 19:27
Kaan IzmirKaan Izmir
61
asked Jan 12 at 19:27
Kaan IzmirKaan Izmir
61
asked Jan 12 at 19:27
Kaan IzmirKaan Izmir
61
asked Jan 12 at 19:27
Kaan IzmirKaan Izmir
61
61
1
$begingroup$
Do you know the relation between the directional derivative and the gradient?
$endgroup$
– Ian
Jan 12 at 19:32
$begingroup$
Yes, I know the relation between them.
$endgroup$
– Kaan Izmir
Jan 12 at 19:34
1
$begingroup$
Knowing that and the bound $-| x | | y | leq x cdot y leq | x | | y |$ (called the Cauchy-Schwarz inequality), you can finish the problem.
$endgroup$
– Ian
Jan 12 at 19:36
$begingroup$
Yeah, thank you for that
$endgroup$
– Kaan Izmir
Jan 12 at 19:44
add a comment |
1
$begingroup$
Do you know the relation between the directional derivative and the gradient?
$endgroup$
– Ian
Jan 12 at 19:32
$begingroup$
Yes, I know the relation between them.
$endgroup$
– Kaan Izmir
Jan 12 at 19:34
1
$begingroup$
Knowing that and the bound $-| x | | y | leq x cdot y leq | x | | y |$ (called the Cauchy-Schwarz inequality), you can finish the problem.
$endgroup$
– Ian
Jan 12 at 19:36
$begingroup$
Yeah, thank you for that
$endgroup$
– Kaan Izmir
Jan 12 at 19:44
1
1
$begingroup$
Do you know the relation between the directional derivative and the gradient?
$endgroup$
– Ian
Jan 12 at 19:32
$begingroup$
Do you know the relation between the directional derivative and the gradient?
$endgroup$
– Ian
Jan 12 at 19:32
$begingroup$
Yes, I know the relation between them.
$endgroup$
– Kaan Izmir
Jan 12 at 19:34
$begingroup$
Yes, I know the relation between them.
$endgroup$
– Kaan Izmir
Jan 12 at 19:34
1
1
$begingroup$
Knowing that and the bound $-| x | | y | leq x cdot y leq | x | | y |$ (called the Cauchy-Schwarz inequality), you can finish the problem.
$endgroup$
– Ian
Jan 12 at 19:36
$begingroup$
Knowing that and the bound $-| x | | y | leq x cdot y leq | x | | y |$ (called the Cauchy-Schwarz inequality), you can finish the problem.
$endgroup$
– Ian
Jan 12 at 19:36
$begingroup$
Yeah, thank you for that
$endgroup$
– Kaan Izmir
Jan 12 at 19:44
$begingroup$
Yeah, thank you for that
$endgroup$
– Kaan Izmir
Jan 12 at 19:44
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The gradient is the vector pointed in the direction of maximum ascent, and its size is the rate at which the function is ascending. With that definition, you can see that we have all the information we need.
We have the direction of maximum ascent: $$(1,1,-1).$$
A unit vector in that direction is $$(1,1,-1)/sqrt{3}.$$ And we know the rate of ascent is $6$; therefore, the gradient is $$left(6/sqrt{3}right)(1,1,-1).$$
answered Jan 12 at 19:32
NicNic8NicNic8
4,50531123
$endgroup$
$begingroup$
Give a man a fish, ....
$endgroup$
– irchans
Jan 12 at 19:33
$begingroup$
Is that it? When the question says the gradient vector I thought it would be different. I got confused by the word vector.
$endgroup$
– Kaan Izmir
Jan 12 at 19:36
$begingroup$
@irchans Please concern yourself with own methods of instruction. I will concern myself with mine.
$endgroup$
– NicNic8
Jan 12 at 19:36
1
$begingroup$
@NicNic8 Sorry. For what it's worth, I found your explanation to be very clear.
$endgroup$
– irchans
Jan 12 at 19:46
1
$begingroup$
And I found your solution very clear, thanks for the answer, it helped a lot
$endgroup$
– Kaan Izmir
Jan 12 at 19:53
|
show 2 more comments
Your Answer
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1 Answer
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oldest
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1 Answer
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oldest
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active
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active
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votes
$begingroup$
The gradient is the vector pointed in the direction of maximum ascent, and its size is the rate at which the function is ascending. With that definition, you can see that we have all the information we need.
We have the direction of maximum ascent: $$(1,1,-1).$$
A unit vector in that direction is $$(1,1,-1)/sqrt{3}.$$ And we know the rate of ascent is $6$; therefore, the gradient is $$left(6/sqrt{3}right)(1,1,-1).$$
answered Jan 12 at 19:32
NicNic8NicNic8
4,50531123
$endgroup$
$begingroup$
Give a man a fish, ....
$endgroup$
– irchans
Jan 12 at 19:33
$begingroup$
Is that it? When the question says the gradient vector I thought it would be different. I got confused by the word vector.
$endgroup$
– Kaan Izmir
Jan 12 at 19:36
$begingroup$
@irchans Please concern yourself with own methods of instruction. I will concern myself with mine.
$endgroup$
– NicNic8
Jan 12 at 19:36
1
$begingroup$
@NicNic8 Sorry. For what it's worth, I found your explanation to be very clear.
$endgroup$
– irchans
Jan 12 at 19:46
1
$begingroup$
And I found your solution very clear, thanks for the answer, it helped a lot
$endgroup$
– Kaan Izmir
Jan 12 at 19:53
|
show 2 more comments
$begingroup$
The gradient is the vector pointed in the direction of maximum ascent, and its size is the rate at which the function is ascending. With that definition, you can see that we have all the information we need.
We have the direction of maximum ascent: $$(1,1,-1).$$
A unit vector in that direction is $$(1,1,-1)/sqrt{3}.$$ And we know the rate of ascent is $6$; therefore, the gradient is $$left(6/sqrt{3}right)(1,1,-1).$$
answered Jan 12 at 19:32
NicNic8NicNic8
4,50531123
$endgroup$
$begingroup$
Give a man a fish, ....
$endgroup$
– irchans
Jan 12 at 19:33
$begingroup$
Is that it? When the question says the gradient vector I thought it would be different. I got confused by the word vector.
$endgroup$
– Kaan Izmir
Jan 12 at 19:36
$begingroup$
@irchans Please concern yourself with own methods of instruction. I will concern myself with mine.
$endgroup$
– NicNic8
Jan 12 at 19:36
1
$begingroup$
@NicNic8 Sorry. For what it's worth, I found your explanation to be very clear.
$endgroup$
– irchans
Jan 12 at 19:46
1
$begingroup$
And I found your solution very clear, thanks for the answer, it helped a lot
$endgroup$
– Kaan Izmir
Jan 12 at 19:53
|
show 2 more comments
$begingroup$
The gradient is the vector pointed in the direction of maximum ascent, and its size is the rate at which the function is ascending. With that definition, you can see that we have all the information we need.
We have the direction of maximum ascent: $$(1,1,-1).$$
A unit vector in that direction is $$(1,1,-1)/sqrt{3}.$$ And we know the rate of ascent is $6$; therefore, the gradient is $$left(6/sqrt{3}right)(1,1,-1).$$
answered Jan 12 at 19:32
NicNic8NicNic8
4,50531123
$endgroup$
The gradient is the vector pointed in the direction of maximum ascent, and its size is the rate at which the function is ascending. With that definition, you can see that we have all the information we need.
We have the direction of maximum ascent: $$(1,1,-1).$$
A unit vector in that direction is $$(1,1,-1)/sqrt{3}.$$ And we know the rate of ascent is $6$; therefore, the gradient is $$left(6/sqrt{3}right)(1,1,-1).$$
answered Jan 12 at 19:32
NicNic8NicNic8
4,50531123
edited Jan 12 at 19:33
answered Jan 12 at 19:32
NicNic8NicNic8
4,50531123
answered Jan 12 at 19:32
NicNic8NicNic8
4,50531123
answered Jan 12 at 19:32
NicNic8NicNic8
4,50531123
4,50531123
$begingroup$
Give a man a fish, ....
$endgroup$
– irchans
Jan 12 at 19:33
$begingroup$
Is that it? When the question says the gradient vector I thought it would be different. I got confused by the word vector.
$endgroup$
– Kaan Izmir
Jan 12 at 19:36
$begingroup$
@irchans Please concern yourself with own methods of instruction. I will concern myself with mine.
$endgroup$
– NicNic8
Jan 12 at 19:36
1
$begingroup$
@NicNic8 Sorry. For what it's worth, I found your explanation to be very clear.
$endgroup$
– irchans
Jan 12 at 19:46
1
$begingroup$
And I found your solution very clear, thanks for the answer, it helped a lot
$endgroup$
– Kaan Izmir
Jan 12 at 19:53
|
show 2 more comments
$begingroup$
Give a man a fish, ....
$endgroup$
– irchans
Jan 12 at 19:33
$begingroup$
Is that it? When the question says the gradient vector I thought it would be different. I got confused by the word vector.
$endgroup$
– Kaan Izmir
Jan 12 at 19:36
$begingroup$
@irchans Please concern yourself with own methods of instruction. I will concern myself with mine.
$endgroup$
– NicNic8
Jan 12 at 19:36
1
$begingroup$
@NicNic8 Sorry. For what it's worth, I found your explanation to be very clear.
$endgroup$
– irchans
Jan 12 at 19:46
1
$begingroup$
And I found your solution very clear, thanks for the answer, it helped a lot
$endgroup$
– Kaan Izmir
Jan 12 at 19:53
$begingroup$
Give a man a fish, ....
$endgroup$
– irchans
Jan 12 at 19:33
$begingroup$
Give a man a fish, ....
$endgroup$
– irchans
Jan 12 at 19:33
$begingroup$
Is that it? When the question says the gradient vector I thought it would be different. I got confused by the word vector.
$endgroup$
– Kaan Izmir
Jan 12 at 19:36
$begingroup$
Is that it? When the question says the gradient vector I thought it would be different. I got confused by the word vector.
$endgroup$
– Kaan Izmir
Jan 12 at 19:36
$begingroup$
@irchans Please concern yourself with own methods of instruction. I will concern myself with mine.
$endgroup$
– NicNic8
Jan 12 at 19:36
$begingroup$
@irchans Please concern yourself with own methods of instruction. I will concern myself with mine.
$endgroup$
– NicNic8
Jan 12 at 19:36
1
1
$begingroup$
@NicNic8 Sorry. For what it's worth, I found your explanation to be very clear.
$endgroup$
– irchans
Jan 12 at 19:46
$begingroup$
@NicNic8 Sorry. For what it's worth, I found your explanation to be very clear.
$endgroup$
– irchans
Jan 12 at 19:46
1
1
$begingroup$
And I found your solution very clear, thanks for the answer, it helped a lot
$endgroup$
– Kaan Izmir
Jan 12 at 19:53
$begingroup$
And I found your solution very clear, thanks for the answer, it helped a lot
$endgroup$
– Kaan Izmir
Jan 12 at 19:53
|
show 2 more comments
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1
$begingroup$
Do you know the relation between the directional derivative and the gradient?
$endgroup$
– Ian
Jan 12 at 19:32
$begingroup$
Yes, I know the relation between them.
$endgroup$
– Kaan Izmir
Jan 12 at 19:34
1
$begingroup$
Knowing that and the bound $-| x | | y | leq x cdot y leq | x | | y |$ (called the Cauchy-Schwarz inequality), you can finish the problem.
$endgroup$
– Ian
Jan 12 at 19:36
$begingroup$
Yeah, thank you for that
$endgroup$
– Kaan Izmir
Jan 12 at 19:44