If $f:mathbb R^2to mathbb R$ is defined by $f(x,y)=x^2+xy$. What is the meaning of the derivative/linear transformation, at the point $(1,2)$?
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Suppose $f:mathbb R^2to mathbb R$ is defined by $f(x,y)=x^2+xy$.
The gradient is $(2x+y, x)$ and so the Jacobian is $beginbmatrix 2x+y& xendbmatrix$.
At the point $(1,2)$, the derivative is the linear transformation $beginbmatrix 4& 1endbmatrix$, which is the matrix of the transformation $T(x,y)=4x+y$.
For the sake of an example, the directional derivative in the direction of $(1,1)$ is $beginbmatrix 4& 1endbmatrixbeginbmatrix 1/sqrt2\ 1/sqrt2endbmatrix=5/sqrt2$. Which I understand is the slope of the tangent plane in the direction of $(1,1)$.
However, I am not quite understanding what the linear transformation $T(x,y)=4x+y$ represented by the matrix $beginbmatrix 4& 1endbmatrix$ is supposed to represent.
The tangent plane lives tangent to the graph of $f$ at the point $f(1,2)=3$; i.e., at the point $(1,2,3) in mathbb R^3$.
So, what exactly does the linear transformation $beginbmatrix 4& 1endbmatrix$ represent?
What is being transformed?
calculus linear-algebra geometry derivatives linear-transformations
asked Aug 26 at 20:38
Al Jebr
4,01943071
add a comment |Â
up vote
3
down vote
favorite
Suppose $f:mathbb R^2to mathbb R$ is defined by $f(x,y)=x^2+xy$.
The gradient is $(2x+y, x)$ and so the Jacobian is $beginbmatrix 2x+y& xendbmatrix$.
At the point $(1,2)$, the derivative is the linear transformation $beginbmatrix 4& 1endbmatrix$, which is the matrix of the transformation $T(x,y)=4x+y$.
For the sake of an example, the directional derivative in the direction of $(1,1)$ is $beginbmatrix 4& 1endbmatrixbeginbmatrix 1/sqrt2\ 1/sqrt2endbmatrix=5/sqrt2$. Which I understand is the slope of the tangent plane in the direction of $(1,1)$.
However, I am not quite understanding what the linear transformation $T(x,y)=4x+y$ represented by the matrix $beginbmatrix 4& 1endbmatrix$ is supposed to represent.
The tangent plane lives tangent to the graph of $f$ at the point $f(1,2)=3$; i.e., at the point $(1,2,3) in mathbb R^3$.
So, what exactly does the linear transformation $beginbmatrix 4& 1endbmatrix$ represent?
What is being transformed?
calculus linear-algebra geometry derivatives linear-transformations
asked Aug 26 at 20:38
Al Jebr
4,01943071
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Suppose $f:mathbb R^2to mathbb R$ is defined by $f(x,y)=x^2+xy$.
The gradient is $(2x+y, x)$ and so the Jacobian is $beginbmatrix 2x+y& xendbmatrix$.
At the point $(1,2)$, the derivative is the linear transformation $beginbmatrix 4& 1endbmatrix$, which is the matrix of the transformation $T(x,y)=4x+y$.
For the sake of an example, the directional derivative in the direction of $(1,1)$ is $beginbmatrix 4& 1endbmatrixbeginbmatrix 1/sqrt2\ 1/sqrt2endbmatrix=5/sqrt2$. Which I understand is the slope of the tangent plane in the direction of $(1,1)$.
However, I am not quite understanding what the linear transformation $T(x,y)=4x+y$ represented by the matrix $beginbmatrix 4& 1endbmatrix$ is supposed to represent.
The tangent plane lives tangent to the graph of $f$ at the point $f(1,2)=3$; i.e., at the point $(1,2,3) in mathbb R^3$.
So, what exactly does the linear transformation $beginbmatrix 4& 1endbmatrix$ represent?
What is being transformed?
calculus linear-algebra geometry derivatives linear-transformations
asked Aug 26 at 20:38
Al Jebr
4,01943071
Suppose $f:mathbb R^2to mathbb R$ is defined by $f(x,y)=x^2+xy$.
The gradient is $(2x+y, x)$ and so the Jacobian is $beginbmatrix 2x+y& xendbmatrix$.
At the point $(1,2)$, the derivative is the linear transformation $beginbmatrix 4& 1endbmatrix$, which is the matrix of the transformation $T(x,y)=4x+y$.
For the sake of an example, the directional derivative in the direction of $(1,1)$ is $beginbmatrix 4& 1endbmatrixbeginbmatrix 1/sqrt2\ 1/sqrt2endbmatrix=5/sqrt2$. Which I understand is the slope of the tangent plane in the direction of $(1,1)$.
However, I am not quite understanding what the linear transformation $T(x,y)=4x+y$ represented by the matrix $beginbmatrix 4& 1endbmatrix$ is supposed to represent.
The tangent plane lives tangent to the graph of $f$ at the point $f(1,2)=3$; i.e., at the point $(1,2,3) in mathbb R^3$.
So, what exactly does the linear transformation $beginbmatrix 4& 1endbmatrix$ represent?
What is being transformed?
calculus linear-algebra geometry derivatives linear-transformations
asked Aug 26 at 20:38
Al Jebr
4,01943071
edited Aug 26 at 21:53
asked Aug 26 at 20:38
Al Jebr
4,01943071
asked Aug 26 at 20:38
Al Jebr
4,01943071
asked Aug 26 at 20:38
Al Jebr
4,01943071
4,01943071
add a comment |Â
add a comment |Â
2 Answers
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In two dimensions, you can move in a circles's worth of directions. These directions are taken in a tangent space around the point at which the derivative is calculated. In your example that is the point $(1,2)$.
The derivative $[4,1]$ means that if I move by an amount $(dx_1, dx_2)$ from $(1,2)$ then I will move $dy = 4dx_1+dx_2$ from $3$ which is $f(1,2)$.
In general, if $f:mathbb R^mtomathbb R^n$ is differentiable at $(x_1,...,x_m)$ then $Df(x_1,...,x_m)$ is a linear map that maps $(dx_1,...,dx_m)$ to $(dy_1,...,dy_n)$ where $(dx_1,...,dx_m)$ is in the tangent space of $(x_1,...,x_m)$ and $(dy_1,...,dy_n)$ is in the tangent space of $f(x_1,...,x_m)$.
answered Aug 26 at 20:51
John Douma
4,64111217
So, is the transformation acting on the tangent plane to the point $f(1,2)$?
– Al Jebr
Aug 26 at 21:49
@AlJebr The transformation is acting on vectors in the plane whose origin is $(1,2)$. Those vectors represent changes which is why we use differentials. The transformed vectors are changes on the real line whose origin is $f(1,2)=3$.
– John Douma
Aug 26 at 22:03
I see. Can you elaborate on your last paragraph. What happens if we have a map from $mathbb R^2 to mathbb R^2$? The derivative at the point $(x,y)$ is a transformation from $mathbb R^2 to mathbb R^2$. In this case, we don't get a real number for slope anymore. What happens in this case?
– Al Jebr
Aug 26 at 22:28
@AlJebr In that case our derivative would be a two by two matrix that maps $(dx_1,dx_2)$ to $(dy_1,dy_2)$.
– John Douma
Aug 26 at 22:29
In the above example with derivative $beginbmatrix 4& 1endbmatrix$, is it correct to say that $4$ is the slope in the $x$ direction and $1$ is the slope in the $y$ direction? I ask because the tangent plane at $(1,2)$ has equation $f(1,2)+4(x-1)+1(y-2)$.
– Al Jebr
Aug 26 at 23:14
 |Â
show 1 more comment
up vote
0
down vote
The equation of the tangent plane at $(1,2)$ is $$z=3+4(x-1)+1(y-2)$$
Thus the transformation [4,1] linearly approximates your function near the point $(1,2,3)$ and we have $$Delta z approx 4Delta x + 1Delta y$$
answered Aug 26 at 21:21
Mohammad Riazi-Kermani
30.5k41852
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
In two dimensions, you can move in a circles's worth of directions. These directions are taken in a tangent space around the point at which the derivative is calculated. In your example that is the point $(1,2)$.
The derivative $[4,1]$ means that if I move by an amount $(dx_1, dx_2)$ from $(1,2)$ then I will move $dy = 4dx_1+dx_2$ from $3$ which is $f(1,2)$.
In general, if $f:mathbb R^mtomathbb R^n$ is differentiable at $(x_1,...,x_m)$ then $Df(x_1,...,x_m)$ is a linear map that maps $(dx_1,...,dx_m)$ to $(dy_1,...,dy_n)$ where $(dx_1,...,dx_m)$ is in the tangent space of $(x_1,...,x_m)$ and $(dy_1,...,dy_n)$ is in the tangent space of $f(x_1,...,x_m)$.
answered Aug 26 at 20:51
John Douma
4,64111217
So, is the transformation acting on the tangent plane to the point $f(1,2)$?
– Al Jebr
Aug 26 at 21:49
@AlJebr The transformation is acting on vectors in the plane whose origin is $(1,2)$. Those vectors represent changes which is why we use differentials. The transformed vectors are changes on the real line whose origin is $f(1,2)=3$.
– John Douma
Aug 26 at 22:03
I see. Can you elaborate on your last paragraph. What happens if we have a map from $mathbb R^2 to mathbb R^2$? The derivative at the point $(x,y)$ is a transformation from $mathbb R^2 to mathbb R^2$. In this case, we don't get a real number for slope anymore. What happens in this case?
– Al Jebr
Aug 26 at 22:28
@AlJebr In that case our derivative would be a two by two matrix that maps $(dx_1,dx_2)$ to $(dy_1,dy_2)$.
– John Douma
Aug 26 at 22:29
In the above example with derivative $beginbmatrix 4& 1endbmatrix$, is it correct to say that $4$ is the slope in the $x$ direction and $1$ is the slope in the $y$ direction? I ask because the tangent plane at $(1,2)$ has equation $f(1,2)+4(x-1)+1(y-2)$.
– Al Jebr
Aug 26 at 23:14
 |Â
show 1 more comment
up vote
1
down vote
accepted
In two dimensions, you can move in a circles's worth of directions. These directions are taken in a tangent space around the point at which the derivative is calculated. In your example that is the point $(1,2)$.
The derivative $[4,1]$ means that if I move by an amount $(dx_1, dx_2)$ from $(1,2)$ then I will move $dy = 4dx_1+dx_2$ from $3$ which is $f(1,2)$.
In general, if $f:mathbb R^mtomathbb R^n$ is differentiable at $(x_1,...,x_m)$ then $Df(x_1,...,x_m)$ is a linear map that maps $(dx_1,...,dx_m)$ to $(dy_1,...,dy_n)$ where $(dx_1,...,dx_m)$ is in the tangent space of $(x_1,...,x_m)$ and $(dy_1,...,dy_n)$ is in the tangent space of $f(x_1,...,x_m)$.
answered Aug 26 at 20:51
John Douma
4,64111217
So, is the transformation acting on the tangent plane to the point $f(1,2)$?
– Al Jebr
Aug 26 at 21:49
@AlJebr The transformation is acting on vectors in the plane whose origin is $(1,2)$. Those vectors represent changes which is why we use differentials. The transformed vectors are changes on the real line whose origin is $f(1,2)=3$.
– John Douma
Aug 26 at 22:03
I see. Can you elaborate on your last paragraph. What happens if we have a map from $mathbb R^2 to mathbb R^2$? The derivative at the point $(x,y)$ is a transformation from $mathbb R^2 to mathbb R^2$. In this case, we don't get a real number for slope anymore. What happens in this case?
– Al Jebr
Aug 26 at 22:28
@AlJebr In that case our derivative would be a two by two matrix that maps $(dx_1,dx_2)$ to $(dy_1,dy_2)$.
– John Douma
Aug 26 at 22:29
In the above example with derivative $beginbmatrix 4& 1endbmatrix$, is it correct to say that $4$ is the slope in the $x$ direction and $1$ is the slope in the $y$ direction? I ask because the tangent plane at $(1,2)$ has equation $f(1,2)+4(x-1)+1(y-2)$.
– Al Jebr
Aug 26 at 23:14
 |Â
show 1 more comment
up vote
1
down vote
accepted
up vote
1
down vote
accepted
In two dimensions, you can move in a circles's worth of directions. These directions are taken in a tangent space around the point at which the derivative is calculated. In your example that is the point $(1,2)$.
The derivative $[4,1]$ means that if I move by an amount $(dx_1, dx_2)$ from $(1,2)$ then I will move $dy = 4dx_1+dx_2$ from $3$ which is $f(1,2)$.
In general, if $f:mathbb R^mtomathbb R^n$ is differentiable at $(x_1,...,x_m)$ then $Df(x_1,...,x_m)$ is a linear map that maps $(dx_1,...,dx_m)$ to $(dy_1,...,dy_n)$ where $(dx_1,...,dx_m)$ is in the tangent space of $(x_1,...,x_m)$ and $(dy_1,...,dy_n)$ is in the tangent space of $f(x_1,...,x_m)$.
answered Aug 26 at 20:51
John Douma
4,64111217
In two dimensions, you can move in a circles's worth of directions. These directions are taken in a tangent space around the point at which the derivative is calculated. In your example that is the point $(1,2)$.
The derivative $[4,1]$ means that if I move by an amount $(dx_1, dx_2)$ from $(1,2)$ then I will move $dy = 4dx_1+dx_2$ from $3$ which is $f(1,2)$.
In general, if $f:mathbb R^mtomathbb R^n$ is differentiable at $(x_1,...,x_m)$ then $Df(x_1,...,x_m)$ is a linear map that maps $(dx_1,...,dx_m)$ to $(dy_1,...,dy_n)$ where $(dx_1,...,dx_m)$ is in the tangent space of $(x_1,...,x_m)$ and $(dy_1,...,dy_n)$ is in the tangent space of $f(x_1,...,x_m)$.
answered Aug 26 at 20:51
John Douma
4,64111217
answered Aug 26 at 20:51
John Douma
4,64111217
answered Aug 26 at 20:51
John Douma
4,64111217
answered Aug 26 at 20:51
John Douma
4,64111217
4,64111217
So, is the transformation acting on the tangent plane to the point $f(1,2)$?
– Al Jebr
Aug 26 at 21:49
@AlJebr The transformation is acting on vectors in the plane whose origin is $(1,2)$. Those vectors represent changes which is why we use differentials. The transformed vectors are changes on the real line whose origin is $f(1,2)=3$.
– John Douma
Aug 26 at 22:03
I see. Can you elaborate on your last paragraph. What happens if we have a map from $mathbb R^2 to mathbb R^2$? The derivative at the point $(x,y)$ is a transformation from $mathbb R^2 to mathbb R^2$. In this case, we don't get a real number for slope anymore. What happens in this case?
– Al Jebr
Aug 26 at 22:28
@AlJebr In that case our derivative would be a two by two matrix that maps $(dx_1,dx_2)$ to $(dy_1,dy_2)$.
– John Douma
Aug 26 at 22:29
In the above example with derivative $beginbmatrix 4& 1endbmatrix$, is it correct to say that $4$ is the slope in the $x$ direction and $1$ is the slope in the $y$ direction? I ask because the tangent plane at $(1,2)$ has equation $f(1,2)+4(x-1)+1(y-2)$.
– Al Jebr
Aug 26 at 23:14
 |Â
show 1 more comment
So, is the transformation acting on the tangent plane to the point $f(1,2)$?
– Al Jebr
Aug 26 at 21:49
@AlJebr The transformation is acting on vectors in the plane whose origin is $(1,2)$. Those vectors represent changes which is why we use differentials. The transformed vectors are changes on the real line whose origin is $f(1,2)=3$.
– John Douma
Aug 26 at 22:03
I see. Can you elaborate on your last paragraph. What happens if we have a map from $mathbb R^2 to mathbb R^2$? The derivative at the point $(x,y)$ is a transformation from $mathbb R^2 to mathbb R^2$. In this case, we don't get a real number for slope anymore. What happens in this case?
– Al Jebr
Aug 26 at 22:28
@AlJebr In that case our derivative would be a two by two matrix that maps $(dx_1,dx_2)$ to $(dy_1,dy_2)$.
– John Douma
Aug 26 at 22:29
In the above example with derivative $beginbmatrix 4& 1endbmatrix$, is it correct to say that $4$ is the slope in the $x$ direction and $1$ is the slope in the $y$ direction? I ask because the tangent plane at $(1,2)$ has equation $f(1,2)+4(x-1)+1(y-2)$.
– Al Jebr
Aug 26 at 23:14
So, is the transformation acting on the tangent plane to the point $f(1,2)$?
– Al Jebr
Aug 26 at 21:49
So, is the transformation acting on the tangent plane to the point $f(1,2)$?
– Al Jebr
Aug 26 at 21:49
@AlJebr The transformation is acting on vectors in the plane whose origin is $(1,2)$. Those vectors represent changes which is why we use differentials. The transformed vectors are changes on the real line whose origin is $f(1,2)=3$.
– John Douma
Aug 26 at 22:03
@AlJebr The transformation is acting on vectors in the plane whose origin is $(1,2)$. Those vectors represent changes which is why we use differentials. The transformed vectors are changes on the real line whose origin is $f(1,2)=3$.
– John Douma
Aug 26 at 22:03
I see. Can you elaborate on your last paragraph. What happens if we have a map from $mathbb R^2 to mathbb R^2$? The derivative at the point $(x,y)$ is a transformation from $mathbb R^2 to mathbb R^2$. In this case, we don't get a real number for slope anymore. What happens in this case?
– Al Jebr
Aug 26 at 22:28
I see. Can you elaborate on your last paragraph. What happens if we have a map from $mathbb R^2 to mathbb R^2$? The derivative at the point $(x,y)$ is a transformation from $mathbb R^2 to mathbb R^2$. In this case, we don't get a real number for slope anymore. What happens in this case?
– Al Jebr
Aug 26 at 22:28
@AlJebr In that case our derivative would be a two by two matrix that maps $(dx_1,dx_2)$ to $(dy_1,dy_2)$.
– John Douma
Aug 26 at 22:29
@AlJebr In that case our derivative would be a two by two matrix that maps $(dx_1,dx_2)$ to $(dy_1,dy_2)$.
– John Douma
Aug 26 at 22:29
In the above example with derivative $beginbmatrix 4& 1endbmatrix$, is it correct to say that $4$ is the slope in the $x$ direction and $1$ is the slope in the $y$ direction? I ask because the tangent plane at $(1,2)$ has equation $f(1,2)+4(x-1)+1(y-2)$.
– Al Jebr
Aug 26 at 23:14
In the above example with derivative $beginbmatrix 4& 1endbmatrix$, is it correct to say that $4$ is the slope in the $x$ direction and $1$ is the slope in the $y$ direction? I ask because the tangent plane at $(1,2)$ has equation $f(1,2)+4(x-1)+1(y-2)$.
– Al Jebr
Aug 26 at 23:14
 |Â
show 1 more comment
up vote
0
down vote
The equation of the tangent plane at $(1,2)$ is $$z=3+4(x-1)+1(y-2)$$
Thus the transformation [4,1] linearly approximates your function near the point $(1,2,3)$ and we have $$Delta z approx 4Delta x + 1Delta y$$
answered Aug 26 at 21:21
Mohammad Riazi-Kermani
30.5k41852
add a comment |Â
up vote
0
down vote
The equation of the tangent plane at $(1,2)$ is $$z=3+4(x-1)+1(y-2)$$
Thus the transformation [4,1] linearly approximates your function near the point $(1,2,3)$ and we have $$Delta z approx 4Delta x + 1Delta y$$
answered Aug 26 at 21:21
Mohammad Riazi-Kermani
30.5k41852
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The equation of the tangent plane at $(1,2)$ is $$z=3+4(x-1)+1(y-2)$$
Thus the transformation [4,1] linearly approximates your function near the point $(1,2,3)$ and we have $$Delta z approx 4Delta x + 1Delta y$$
answered Aug 26 at 21:21
Mohammad Riazi-Kermani
30.5k41852
The equation of the tangent plane at $(1,2)$ is $$z=3+4(x-1)+1(y-2)$$
Thus the transformation [4,1] linearly approximates your function near the point $(1,2,3)$ and we have $$Delta z approx 4Delta x + 1Delta y$$
answered Aug 26 at 21:21
Mohammad Riazi-Kermani
30.5k41852
answered Aug 26 at 21:21
Mohammad Riazi-Kermani
30.5k41852
answered Aug 26 at 21:21
Mohammad Riazi-Kermani
30.5k41852
answered Aug 26 at 21:21
Mohammad Riazi-Kermani
30.5k41852
30.5k41852
add a comment |Â
add a comment |Â
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