Finding the direction in which the derivative is exactly $2$ Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Is there a difference between “in the direction of (1.1)” and “in the direction toward (1.1)?”Directional derivative for differentiable functionTo find $a,b,c$ as the directional derivative of $f(x,y,z)=axy^2+byz+cz^2x^3$ , at $(1,2-1)$ , is atmost $64$ in a direction parallel to $z$-axis?Directional Derivatives - Geometric intuitionCan every total differential of a function w at some point also a directional derivative and the gradient at the level curve for that pint?In which direction directional derivative should be maximumFind the direction in which the directional derivative at a point has the value 2A name for the directional derivative with respect to a unit vector?Find the directional derivative of the function $f(x,y)=fracxx^2+y^2$ at the point $(1,2)$ in the direction of the path $textbfr(t)= (t^3,2t)$How does a gradient allow the calculation of the directional derivative
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Finding the direction in which the derivative is exactly $2$
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Is there a difference between “in the direction of (1.1)” and “in the direction toward (1.1)?”Directional derivative for differentiable functionTo find $a,b,c$ as the directional derivative of $f(x,y,z)=axy^2+byz+cz^2x^3$ , at $(1,2-1)$ , is atmost $64$ in a direction parallel to $z$-axis?Directional Derivatives - Geometric intuitionCan every total differential of a function w at some point also a directional derivative and the gradient at the level curve for that pint?In which direction directional derivative should be maximumFind the direction in which the directional derivative at a point has the value 2A name for the directional derivative with respect to a unit vector?Find the directional derivative of the function $f(x,y)=fracxx^2+y^2$ at the point $(1,2)$ in the direction of the path $textbfr(t)= (t^3,2t)$How does a gradient allow the calculation of the directional derivative
$begingroup$
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
edited Mar 27 at 9:19
Ernie060
2,940719
asked Mar 27 at 4:29
krauser126krauser126
636
$endgroup$
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
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
edited Mar 27 at 9:19
Ernie060
2,940719
asked Mar 27 at 4:29
krauser126krauser126
636
$endgroup$
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
multivariable-calculus partial-derivative
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
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
edited Mar 27 at 9:19
Ernie060
2,940719
2,940719
asked Mar 27 at 4:29
krauser126krauser126
636
asked Mar 27 at 4:29
krauser126krauser126
636
asked Mar 27 at 4:29
krauser126krauser126
636
636
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 |
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
3
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
$begingroup$
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 |
2 Answers
2
active
oldest
votes
$begingroup$
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)
answered Mar 27 at 5:13
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40.6k33678
$endgroup$
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)$$
answered Mar 27 at 8:10
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$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)
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)
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)
answered Mar 27 at 5:13
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40.6k33678
$endgroup$
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)
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
answered Mar 27 at 5:13
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40.6k33678
40.6k33678
add a comment |
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)$$
answered Mar 27 at 8:10
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
add a comment |
$begingroup$
You are right all along. Now, it is a bit of time solving the following system of variables:$$begincasesx+2y=2\x^2+y^2=1endcases$$which has two answers $$(x,y)=(0,1)\(x,y)=(0.8,0.6)$$
answered Mar 27 at 8:10
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
add a comment |
$begingroup$
You are right all along. Now, it is a bit of time solving the following system of variables:$$begincasesx+2y=2\x^2+y^2=1endcases$$which has two answers $$(x,y)=(0,1)\(x,y)=(0.8,0.6)$$
answered Mar 27 at 8:10
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
You are right all along. Now, it is a bit of time solving the following system of variables:$$begincasesx+2y=2\x^2+y^2=1endcases$$which has two answers $$(x,y)=(0,1)\(x,y)=(0.8,0.6)$$
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
answered Mar 27 at 8:10
Mostafa AyazMostafa Ayaz
18.1k31040
18.1k31040
add a comment |
add a comment |
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$begingroup$
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