Find a function M The 2019 Stack Overflow Developer Survey Results Are InHow can you prove that a function has no closed form integral?Integral of Vector Valued FunctionHarmonic function and surface integrals.How to solve second order linear partial differential equationIf $f$ is a real valued function, complex differentiable at $z_0$, then $f'(z_0)=0$Find a function s.t. $f(0), f'(0)$ exists but $f''(0)$ doesn'tHow to represent a derivative of a function that has more than 2 vector valued inputsHow to find the dot product inside of a squareDefine in which points function is continuousFind value of $t$ where slope of parametrically-defined curve $=4$ using multivariable calculusTo prove the following result
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Find a function M
The 2019 Stack Overflow Developer Survey Results Are InHow can you prove that a function has no closed form integral?Integral of Vector Valued FunctionHarmonic function and surface integrals.How to solve second order linear partial differential equationIf $f$ is a real valued function, complex differentiable at $z_0$, then $f'(z_0)=0$Find a function s.t. $f(0), f'(0)$ exists but $f''(0)$ doesn'tHow to represent a derivative of a function that has more than 2 vector valued inputsHow to find the dot product inside of a squareDefine in which points function is continuousFind value of $t$ where slope of parametrically-defined curve $=4$ using multivariable calculusTo prove the following result
$begingroup$
Let $x(t)$ be a real valued vector. Can you find a function M such that
$dotM=fractextdMtextdt=dotx^Tdotx$.
I have tried
$M=dotx^Tx,
M=x^Tx$
and many more which don't work. I know that if $x(t)$ is continuous then so will
$dotM=dotx^Tdotx$
be and thus M will exist. But how do i find this M?
multivariable-calculus derivatives vectors
asked Mar 23 at 16:52
holahola
325
$endgroup$
add a comment |
$begingroup$
Let $x(t)$ be a real valued vector. Can you find a function M such that
$dotM=fractextdMtextdt=dotx^Tdotx$.
I have tried
$M=dotx^Tx,
M=x^Tx$
and many more which don't work. I know that if $x(t)$ is continuous then so will
$dotM=dotx^Tdotx$
be and thus M will exist. But how do i find this M?
multivariable-calculus derivatives vectors
asked Mar 23 at 16:52
holahola
325
$endgroup$
$begingroup$
$dotx$ continuous $iff$ $x$ is $C^1$.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 23 at 17:00
add a comment |
$begingroup$
Let $x(t)$ be a real valued vector. Can you find a function M such that
$dotM=fractextdMtextdt=dotx^Tdotx$.
I have tried
$M=dotx^Tx,
M=x^Tx$
and many more which don't work. I know that if $x(t)$ is continuous then so will
$dotM=dotx^Tdotx$
be and thus M will exist. But how do i find this M?
multivariable-calculus derivatives vectors
asked Mar 23 at 16:52
holahola
325
$endgroup$
Let $x(t)$ be a real valued vector. Can you find a function M such that
$dotM=fractextdMtextdt=dotx^Tdotx$.
I have tried
$M=dotx^Tx,
M=x^Tx$
and many more which don't work. I know that if $x(t)$ is continuous then so will
$dotM=dotx^Tdotx$
be and thus M will exist. But how do i find this M?
multivariable-calculus derivatives vectors
multivariable-calculus derivatives vectors
asked Mar 23 at 16:52
holahola
325
asked Mar 23 at 16:52
holahola
325
asked Mar 23 at 16:52
holahola
325
asked Mar 23 at 16:52
holahola
325
asked Mar 23 at 16:52
holahola
325
325
$begingroup$
$dotx$ continuous $iff$ $x$ is $C^1$.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 23 at 17:00
add a comment |
$begingroup$
$dotx$ continuous $iff$ $x$ is $C^1$.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 23 at 17:00
$begingroup$
$dotx$ continuous $iff$ $x$ is $C^1$.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 23 at 17:00
$begingroup$
$dotx$ continuous $iff$ $x$ is $C^1$.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 23 at 17:00
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Impossible a general formula even in dimension 1. In the case, you are asking for
$$int(x'(t))^2,dt$$
and no such formula exists. In concrete cases maybe you can do the integral.
answered Mar 23 at 16:59
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35.4k42972
$endgroup$
$begingroup$
Wouldn't this integral exist if x'(t) is continuous?
$endgroup$
– hola
Mar 23 at 17:01
1
$begingroup$
@hola, it will exist. But you are asking for a formula.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 23 at 17:03
$begingroup$
@hola, see math.stackexchange.com/questions/155/….
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 23 at 17:33
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Impossible a general formula even in dimension 1. In the case, you are asking for
$$int(x'(t))^2,dt$$
and no such formula exists. In concrete cases maybe you can do the integral.
answered Mar 23 at 16:59
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35.4k42972
$endgroup$
$begingroup$
Wouldn't this integral exist if x'(t) is continuous?
$endgroup$
– hola
Mar 23 at 17:01
1
$begingroup$
@hola, it will exist. But you are asking for a formula.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 23 at 17:03
$begingroup$
@hola, see math.stackexchange.com/questions/155/….
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 23 at 17:33
add a comment |
$begingroup$
Impossible a general formula even in dimension 1. In the case, you are asking for
$$int(x'(t))^2,dt$$
and no such formula exists. In concrete cases maybe you can do the integral.
answered Mar 23 at 16:59
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35.4k42972
$endgroup$
$begingroup$
Wouldn't this integral exist if x'(t) is continuous?
$endgroup$
– hola
Mar 23 at 17:01
1
$begingroup$
@hola, it will exist. But you are asking for a formula.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 23 at 17:03
$begingroup$
@hola, see math.stackexchange.com/questions/155/….
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 23 at 17:33
add a comment |
$begingroup$
Impossible a general formula even in dimension 1. In the case, you are asking for
$$int(x'(t))^2,dt$$
and no such formula exists. In concrete cases maybe you can do the integral.
answered Mar 23 at 16:59
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35.4k42972
$endgroup$
Impossible a general formula even in dimension 1. In the case, you are asking for
$$int(x'(t))^2,dt$$
and no such formula exists. In concrete cases maybe you can do the integral.
answered Mar 23 at 16:59
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35.4k42972
answered Mar 23 at 16:59
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35.4k42972
answered Mar 23 at 16:59
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35.4k42972
answered Mar 23 at 16:59
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35.4k42972
35.4k42972
$begingroup$
Wouldn't this integral exist if x'(t) is continuous?
$endgroup$
– hola
Mar 23 at 17:01
1
$begingroup$
@hola, it will exist. But you are asking for a formula.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 23 at 17:03
$begingroup$
@hola, see math.stackexchange.com/questions/155/….
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 23 at 17:33
add a comment |
$begingroup$
Wouldn't this integral exist if x'(t) is continuous?
$endgroup$
– hola
Mar 23 at 17:01
1
$begingroup$
@hola, it will exist. But you are asking for a formula.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 23 at 17:03
$begingroup$
@hola, see math.stackexchange.com/questions/155/….
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 23 at 17:33
$begingroup$
Wouldn't this integral exist if x'(t) is continuous?
$endgroup$
– hola
Mar 23 at 17:01
$begingroup$
Wouldn't this integral exist if x'(t) is continuous?
$endgroup$
– hola
Mar 23 at 17:01
1
1
$begingroup$
@hola, it will exist. But you are asking for a formula.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 23 at 17:03
$begingroup$
@hola, it will exist. But you are asking for a formula.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 23 at 17:03
$begingroup$
@hola, see math.stackexchange.com/questions/155/….
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 23 at 17:33
$begingroup$
@hola, see math.stackexchange.com/questions/155/….
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 23 at 17:33
add a comment |
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$begingroup$
$dotx$ continuous $iff$ $x$ is $C^1$.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 23 at 17:00