function from $mathbbR^2 to mathbbR^3$ Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Validating a mathematical model (Lagrange formulation and geometry)Concave function propertiesTangent lines of a smooth curve $C subseteq mathbbP^2$Finding a curve from its evoluteWhat are the points in $mathbbR^2$ equidistant to the curves $y=sin(x)$ and $y=cos(x)$?Proving that the function is a distance in $mathbbR^2$Calculate time using velocity as a function of distanceProjection of Rectangle from $mathbbR^3$ into $mathbbR^2$Trying to calculate the normal derivative of a function on the sphere but getting an 'inverted' representation?How can I develop an appropriate circular arc connecting two points?
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function from $mathbbR^2 to mathbbR^3$
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Validating a mathematical model (Lagrange formulation and geometry)Concave function propertiesTangent lines of a smooth curve $C subseteq mathbbP^2$Finding a curve from its evoluteWhat are the points in $mathbbR^2$ equidistant to the curves $y=sin(x)$ and $y=cos(x)$?Proving that the function is a distance in $mathbbR^2$Calculate time using velocity as a function of distanceProjection of Rectangle from $mathbbR^3$ into $mathbbR^2$Trying to calculate the normal derivative of a function on the sphere but getting an 'inverted' representation?How can I develop an appropriate circular arc connecting two points?
$begingroup$
Suppose $phi:mathbbR^2tomathbbR^3$ by $phi(x_1,x_2)= (u(x_1,x_2),v(x_1,x_2),w(x_1,x_2))$ is a smooth function. Furthermore, suppose that $alpha:mathbbRtomathbbR^2$ is a smooth curve given by $alpha(t) = (f(t),g(t))$. Using the chain rule, find an expression for $fracddt(phi(alpha(t))$, in terms of $u,v,w,f$ and $g$ (and/or their derivatives).
geometry
edited Apr 2 at 14:21
Winther
20.9k33156
asked Sep 4 '17 at 19:46
user477465user477465
152114
$endgroup$
add a comment |
$begingroup$
Suppose $phi:mathbbR^2tomathbbR^3$ by $phi(x_1,x_2)= (u(x_1,x_2),v(x_1,x_2),w(x_1,x_2))$ is a smooth function. Furthermore, suppose that $alpha:mathbbRtomathbbR^2$ is a smooth curve given by $alpha(t) = (f(t),g(t))$. Using the chain rule, find an expression for $fracddt(phi(alpha(t))$, in terms of $u,v,w,f$ and $g$ (and/or their derivatives).
geometry
edited Apr 2 at 14:21
Winther
20.9k33156
asked Sep 4 '17 at 19:46
user477465user477465
152114
$endgroup$
2
$begingroup$
And what aspect of this problem are you having trouble with?
$endgroup$
– eyeballfrog
Sep 4 '17 at 19:49
$begingroup$
Sorry if I didn't explain properly, I'm confused about nearly every aspect. I know what the chain rule is but I don't know how to find the given expression. And I'm not sure what the relationship between a smooth function and a smooth curve is and why its relevant.
$endgroup$
– user477465
Sep 4 '17 at 20:46
$begingroup$
A smooth function, curve, or whatever is one that has continuous derivatives of all orders at all of its points. Note that this condition does not imply the function can be expanded as a power series.
$endgroup$
– eyeballfrog
Sep 4 '17 at 21:25
add a comment |
$begingroup$
Suppose $phi:mathbbR^2tomathbbR^3$ by $phi(x_1,x_2)= (u(x_1,x_2),v(x_1,x_2),w(x_1,x_2))$ is a smooth function. Furthermore, suppose that $alpha:mathbbRtomathbbR^2$ is a smooth curve given by $alpha(t) = (f(t),g(t))$. Using the chain rule, find an expression for $fracddt(phi(alpha(t))$, in terms of $u,v,w,f$ and $g$ (and/or their derivatives).
geometry
edited Apr 2 at 14:21
Winther
20.9k33156
asked Sep 4 '17 at 19:46
user477465user477465
152114
$endgroup$
Suppose $phi:mathbbR^2tomathbbR^3$ by $phi(x_1,x_2)= (u(x_1,x_2),v(x_1,x_2),w(x_1,x_2))$ is a smooth function. Furthermore, suppose that $alpha:mathbbRtomathbbR^2$ is a smooth curve given by $alpha(t) = (f(t),g(t))$. Using the chain rule, find an expression for $fracddt(phi(alpha(t))$, in terms of $u,v,w,f$ and $g$ (and/or their derivatives).
geometry
geometry
edited Apr 2 at 14:21
Winther
20.9k33156
asked Sep 4 '17 at 19:46
user477465user477465
152114
edited Apr 2 at 14:21
Winther
20.9k33156
asked Sep 4 '17 at 19:46
user477465user477465
152114
edited Apr 2 at 14:21
Winther
20.9k33156
edited Apr 2 at 14:21
Winther
20.9k33156
edited Apr 2 at 14:21
Winther
20.9k33156
20.9k33156
asked Sep 4 '17 at 19:46
user477465user477465
152114
asked Sep 4 '17 at 19:46
user477465user477465
152114
asked Sep 4 '17 at 19:46
user477465user477465
152114
152114
2
$begingroup$
And what aspect of this problem are you having trouble with?
$endgroup$
– eyeballfrog
Sep 4 '17 at 19:49
$begingroup$
Sorry if I didn't explain properly, I'm confused about nearly every aspect. I know what the chain rule is but I don't know how to find the given expression. And I'm not sure what the relationship between a smooth function and a smooth curve is and why its relevant.
$endgroup$
– user477465
Sep 4 '17 at 20:46
$begingroup$
A smooth function, curve, or whatever is one that has continuous derivatives of all orders at all of its points. Note that this condition does not imply the function can be expanded as a power series.
$endgroup$
– eyeballfrog
Sep 4 '17 at 21:25
add a comment |
2
$begingroup$
And what aspect of this problem are you having trouble with?
$endgroup$
– eyeballfrog
Sep 4 '17 at 19:49
$begingroup$
Sorry if I didn't explain properly, I'm confused about nearly every aspect. I know what the chain rule is but I don't know how to find the given expression. And I'm not sure what the relationship between a smooth function and a smooth curve is and why its relevant.
$endgroup$
– user477465
Sep 4 '17 at 20:46
$begingroup$
A smooth function, curve, or whatever is one that has continuous derivatives of all orders at all of its points. Note that this condition does not imply the function can be expanded as a power series.
$endgroup$
– eyeballfrog
Sep 4 '17 at 21:25
2
2
$begingroup$
And what aspect of this problem are you having trouble with?
$endgroup$
– eyeballfrog
Sep 4 '17 at 19:49
$begingroup$
And what aspect of this problem are you having trouble with?
$endgroup$
– eyeballfrog
Sep 4 '17 at 19:49
$begingroup$
Sorry if I didn't explain properly, I'm confused about nearly every aspect. I know what the chain rule is but I don't know how to find the given expression. And I'm not sure what the relationship between a smooth function and a smooth curve is and why its relevant.
$endgroup$
– user477465
Sep 4 '17 at 20:46
$begingroup$
Sorry if I didn't explain properly, I'm confused about nearly every aspect. I know what the chain rule is but I don't know how to find the given expression. And I'm not sure what the relationship between a smooth function and a smooth curve is and why its relevant.
$endgroup$
– user477465
Sep 4 '17 at 20:46
$begingroup$
A smooth function, curve, or whatever is one that has continuous derivatives of all orders at all of its points. Note that this condition does not imply the function can be expanded as a power series.
$endgroup$
– eyeballfrog
Sep 4 '17 at 21:25
$begingroup$
A smooth function, curve, or whatever is one that has continuous derivatives of all orders at all of its points. Note that this condition does not imply the function can be expanded as a power series.
$endgroup$
– eyeballfrog
Sep 4 '17 at 21:25
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Immediately from the question:
$fracddt(phi(alpha(t))) = fracddtbig( u(f(t),g(t)),v(f(t),g(t)),w(f(t),g(t))big)$
Now you can use the chain rule on each component yourself.
$ = big( fracddtu(f(t),g(t)),fracddtv(f(t),g(t)),fracddtw(f(t),g(t))big)$
The functions are only said to be smooth to justify differentiation.
answered Sep 4 '17 at 21:33
Christian FieldhouseChristian Fieldhouse
464114
$endgroup$
$begingroup$
Thanks for answering. So does that mean that for example taking $fracddt u(f(t),g(t)) = fracpartialalphapartialu = fracpartialalphapartialf(t)fracpartialf(t)partialu+fracpartialalphapartialg(t)fracpartialg(t)partialu$
$endgroup$
– user477465
Sep 5 '17 at 23:02
$begingroup$
And once i have that for $fracddt u(f(t),g(t))$ then what do I do? Do I do the same thing for the other components and once i get the different expressions, how do I present my answer?
$endgroup$
– user477465
Sep 5 '17 at 23:03
add a comment |
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$begingroup$
Immediately from the question:
$fracddt(phi(alpha(t))) = fracddtbig( u(f(t),g(t)),v(f(t),g(t)),w(f(t),g(t))big)$
Now you can use the chain rule on each component yourself.
$ = big( fracddtu(f(t),g(t)),fracddtv(f(t),g(t)),fracddtw(f(t),g(t))big)$
The functions are only said to be smooth to justify differentiation.
answered Sep 4 '17 at 21:33
Christian FieldhouseChristian Fieldhouse
464114
$endgroup$
$begingroup$
Thanks for answering. So does that mean that for example taking $fracddt u(f(t),g(t)) = fracpartialalphapartialu = fracpartialalphapartialf(t)fracpartialf(t)partialu+fracpartialalphapartialg(t)fracpartialg(t)partialu$
$endgroup$
– user477465
Sep 5 '17 at 23:02
$begingroup$
And once i have that for $fracddt u(f(t),g(t))$ then what do I do? Do I do the same thing for the other components and once i get the different expressions, how do I present my answer?
$endgroup$
– user477465
Sep 5 '17 at 23:03
add a comment |
$begingroup$
Immediately from the question:
$fracddt(phi(alpha(t))) = fracddtbig( u(f(t),g(t)),v(f(t),g(t)),w(f(t),g(t))big)$
Now you can use the chain rule on each component yourself.
$ = big( fracddtu(f(t),g(t)),fracddtv(f(t),g(t)),fracddtw(f(t),g(t))big)$
The functions are only said to be smooth to justify differentiation.
answered Sep 4 '17 at 21:33
Christian FieldhouseChristian Fieldhouse
464114
$endgroup$
$begingroup$
Thanks for answering. So does that mean that for example taking $fracddt u(f(t),g(t)) = fracpartialalphapartialu = fracpartialalphapartialf(t)fracpartialf(t)partialu+fracpartialalphapartialg(t)fracpartialg(t)partialu$
$endgroup$
– user477465
Sep 5 '17 at 23:02
$begingroup$
And once i have that for $fracddt u(f(t),g(t))$ then what do I do? Do I do the same thing for the other components and once i get the different expressions, how do I present my answer?
$endgroup$
– user477465
Sep 5 '17 at 23:03
add a comment |
$begingroup$
Immediately from the question:
$fracddt(phi(alpha(t))) = fracddtbig( u(f(t),g(t)),v(f(t),g(t)),w(f(t),g(t))big)$
Now you can use the chain rule on each component yourself.
$ = big( fracddtu(f(t),g(t)),fracddtv(f(t),g(t)),fracddtw(f(t),g(t))big)$
The functions are only said to be smooth to justify differentiation.
answered Sep 4 '17 at 21:33
Christian FieldhouseChristian Fieldhouse
464114
$endgroup$
Immediately from the question:
$fracddt(phi(alpha(t))) = fracddtbig( u(f(t),g(t)),v(f(t),g(t)),w(f(t),g(t))big)$
Now you can use the chain rule on each component yourself.
$ = big( fracddtu(f(t),g(t)),fracddtv(f(t),g(t)),fracddtw(f(t),g(t))big)$
The functions are only said to be smooth to justify differentiation.
answered Sep 4 '17 at 21:33
Christian FieldhouseChristian Fieldhouse
464114
answered Sep 4 '17 at 21:33
Christian FieldhouseChristian Fieldhouse
464114
answered Sep 4 '17 at 21:33
Christian FieldhouseChristian Fieldhouse
464114
answered Sep 4 '17 at 21:33
Christian FieldhouseChristian Fieldhouse
464114
464114
$begingroup$
Thanks for answering. So does that mean that for example taking $fracddt u(f(t),g(t)) = fracpartialalphapartialu = fracpartialalphapartialf(t)fracpartialf(t)partialu+fracpartialalphapartialg(t)fracpartialg(t)partialu$
$endgroup$
– user477465
Sep 5 '17 at 23:02
$begingroup$
And once i have that for $fracddt u(f(t),g(t))$ then what do I do? Do I do the same thing for the other components and once i get the different expressions, how do I present my answer?
$endgroup$
– user477465
Sep 5 '17 at 23:03
add a comment |
$begingroup$
Thanks for answering. So does that mean that for example taking $fracddt u(f(t),g(t)) = fracpartialalphapartialu = fracpartialalphapartialf(t)fracpartialf(t)partialu+fracpartialalphapartialg(t)fracpartialg(t)partialu$
$endgroup$
– user477465
Sep 5 '17 at 23:02
$begingroup$
And once i have that for $fracddt u(f(t),g(t))$ then what do I do? Do I do the same thing for the other components and once i get the different expressions, how do I present my answer?
$endgroup$
– user477465
Sep 5 '17 at 23:03
$begingroup$
Thanks for answering. So does that mean that for example taking $fracddt u(f(t),g(t)) = fracpartialalphapartialu = fracpartialalphapartialf(t)fracpartialf(t)partialu+fracpartialalphapartialg(t)fracpartialg(t)partialu$
$endgroup$
– user477465
Sep 5 '17 at 23:02
$begingroup$
Thanks for answering. So does that mean that for example taking $fracddt u(f(t),g(t)) = fracpartialalphapartialu = fracpartialalphapartialf(t)fracpartialf(t)partialu+fracpartialalphapartialg(t)fracpartialg(t)partialu$
$endgroup$
– user477465
Sep 5 '17 at 23:02
$begingroup$
And once i have that for $fracddt u(f(t),g(t))$ then what do I do? Do I do the same thing for the other components and once i get the different expressions, how do I present my answer?
$endgroup$
– user477465
Sep 5 '17 at 23:03
$begingroup$
And once i have that for $fracddt u(f(t),g(t))$ then what do I do? Do I do the same thing for the other components and once i get the different expressions, how do I present my answer?
$endgroup$
– user477465
Sep 5 '17 at 23:03
add a comment |
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2
$begingroup$
And what aspect of this problem are you having trouble with?
$endgroup$
– eyeballfrog
Sep 4 '17 at 19:49
$begingroup$
Sorry if I didn't explain properly, I'm confused about nearly every aspect. I know what the chain rule is but I don't know how to find the given expression. And I'm not sure what the relationship between a smooth function and a smooth curve is and why its relevant.
$endgroup$
– user477465
Sep 4 '17 at 20:46
$begingroup$
A smooth function, curve, or whatever is one that has continuous derivatives of all orders at all of its points. Note that this condition does not imply the function can be expanded as a power series.
$endgroup$
– eyeballfrog
Sep 4 '17 at 21:25