Length of curve - CalculusImproper integrals in curve lengthArc length $gamma(t) = (30 sin t, 30 cos t, 50 cos t)$Calculate the length of a curve!Prove that length of a curve is finite / infiniteArc Length parametric curveGiven curve $r(t)=langleunderline acos t,underline bsin trangle$ is the length of the curve $1$?The length of a parametric curveParametrize curve by arc lengthArc length of curve of intersection between cylinder and sphereCould the length of a curve be $0$?
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Length of curve - Calculus
Improper integrals in curve lengthArc length $gamma(t) = (30 sin t, 30 cos t, 50 cos t)$Calculate the length of a curve!Prove that length of a curve is finite / infiniteArc Length parametric curveGiven curve $r(t)=langleunderline acos t,underline bsin trangle$ is the length of the curve $1$?The length of a parametric curveParametrize curve by arc lengthArc length of curve of intersection between cylinder and sphereCould the length of a curve be $0$?
$begingroup$
I need to find the LENGTH of the curve with x,y,z components listed below:
$ x(t) = tsin(2t)$
$ y(t) = tcos(2t)$
$z(t) = (4/3)tsqrt(t) = (4/3)(t^1.5)$
from $t= 0$ to $t=2pi$
can anyone help pls? i have tried but came to a very hard integral which doesnt look right
i get
$L =2 int^2pi_0sqrtt^2 + t$
which looks wrong since the online solution to this is very lengthy.
calculus integration multivariable-calculus curves
edited 2 days ago
Rory Daulton
29.5k63355
asked 2 days ago
ElecLearnElecLearn
11
$endgroup$
add a comment |
$begingroup$
I need to find the LENGTH of the curve with x,y,z components listed below:
$ x(t) = tsin(2t)$
$ y(t) = tcos(2t)$
$z(t) = (4/3)tsqrt(t) = (4/3)(t^1.5)$
from $t= 0$ to $t=2pi$
can anyone help pls? i have tried but came to a very hard integral which doesnt look right
i get
$L =2 int^2pi_0sqrtt^2 + t$
which looks wrong since the online solution to this is very lengthy.
calculus integration multivariable-calculus curves
edited 2 days ago
Rory Daulton
29.5k63355
asked 2 days ago
ElecLearnElecLearn
11
$endgroup$
$begingroup$
SORRY I MEANT LENGTH NOT AREA
$endgroup$
– ElecLearn
2 days ago
$begingroup$
please show your working. The integrand simplifies.
$endgroup$
– David Quinn
2 days ago
$begingroup$
@DavidQuinn i updated it
$endgroup$
– ElecLearn
2 days ago
add a comment |
$begingroup$
I need to find the LENGTH of the curve with x,y,z components listed below:
$ x(t) = tsin(2t)$
$ y(t) = tcos(2t)$
$z(t) = (4/3)tsqrt(t) = (4/3)(t^1.5)$
from $t= 0$ to $t=2pi$
can anyone help pls? i have tried but came to a very hard integral which doesnt look right
i get
$L =2 int^2pi_0sqrtt^2 + t$
which looks wrong since the online solution to this is very lengthy.
calculus integration multivariable-calculus curves
edited 2 days ago
Rory Daulton
29.5k63355
asked 2 days ago
ElecLearnElecLearn
11
$endgroup$
I need to find the LENGTH of the curve with x,y,z components listed below:
$ x(t) = tsin(2t)$
$ y(t) = tcos(2t)$
$z(t) = (4/3)tsqrt(t) = (4/3)(t^1.5)$
from $t= 0$ to $t=2pi$
can anyone help pls? i have tried but came to a very hard integral which doesnt look right
i get
$L =2 int^2pi_0sqrtt^2 + t$
which looks wrong since the online solution to this is very lengthy.
calculus integration multivariable-calculus curves
calculus integration multivariable-calculus curves
edited 2 days ago
Rory Daulton
29.5k63355
asked 2 days ago
ElecLearnElecLearn
11
edited 2 days ago
Rory Daulton
29.5k63355
asked 2 days ago
ElecLearnElecLearn
11
edited 2 days ago
Rory Daulton
29.5k63355
edited 2 days ago
Rory Daulton
29.5k63355
edited 2 days ago
Rory Daulton
29.5k63355
29.5k63355
asked 2 days ago
ElecLearnElecLearn
11
asked 2 days ago
ElecLearnElecLearn
11
asked 2 days ago
ElecLearnElecLearn
11
11
$begingroup$
SORRY I MEANT LENGTH NOT AREA
$endgroup$
– ElecLearn
2 days ago
$begingroup$
please show your working. The integrand simplifies.
$endgroup$
– David Quinn
2 days ago
$begingroup$
@DavidQuinn i updated it
$endgroup$
– ElecLearn
2 days ago
add a comment |
$begingroup$
SORRY I MEANT LENGTH NOT AREA
$endgroup$
– ElecLearn
2 days ago
$begingroup$
please show your working. The integrand simplifies.
$endgroup$
– David Quinn
2 days ago
$begingroup$
@DavidQuinn i updated it
$endgroup$
– ElecLearn
2 days ago
$begingroup$
SORRY I MEANT LENGTH NOT AREA
$endgroup$
– ElecLearn
2 days ago
$begingroup$
SORRY I MEANT LENGTH NOT AREA
$endgroup$
– ElecLearn
2 days ago
$begingroup$
please show your working. The integrand simplifies.
$endgroup$
– David Quinn
2 days ago
$begingroup$
please show your working. The integrand simplifies.
$endgroup$
– David Quinn
2 days ago
$begingroup$
@DavidQuinn i updated it
$endgroup$
– ElecLearn
2 days ago
$begingroup$
@DavidQuinn i updated it
$endgroup$
– ElecLearn
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You have to use the formula of the length of a curve: $int ||varphi|| dt$ from $0$ to $2pi$.
in this case the derivative of the components are:
$$x'=sin(2t) + 2tcos(2t)$$
$$y'=cos(2t)- 2tsin(2t)$$
$$z'=2t^0.5$$
follows that the function as an integral is $2t+1$ because $2t+1$ is positive in $[0,2pi]$. And now this integral is very easy.
answered 2 days ago
Antonio SanniaAntonio Sannia
212
New contributor
Antonio Sannia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
did you use $int sqrtx'^2 +y'^2 +z'^2$?
$endgroup$
– ElecLearn
2 days ago
$begingroup$
because i obtain $2int sqrt t^2 +t $
$endgroup$
– ElecLearn
2 days ago
$begingroup$
yes I used it. Under the root I obtain (2t+1)^2
$endgroup$
– Antonio Sannia
2 days ago
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You have to use the formula of the length of a curve: $int ||varphi|| dt$ from $0$ to $2pi$.
in this case the derivative of the components are:
$$x'=sin(2t) + 2tcos(2t)$$
$$y'=cos(2t)- 2tsin(2t)$$
$$z'=2t^0.5$$
follows that the function as an integral is $2t+1$ because $2t+1$ is positive in $[0,2pi]$. And now this integral is very easy.
answered 2 days ago
Antonio SanniaAntonio Sannia
212
New contributor
Antonio Sannia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
did you use $int sqrtx'^2 +y'^2 +z'^2$?
$endgroup$
– ElecLearn
2 days ago
$begingroup$
because i obtain $2int sqrt t^2 +t $
$endgroup$
– ElecLearn
2 days ago
$begingroup$
yes I used it. Under the root I obtain (2t+1)^2
$endgroup$
– Antonio Sannia
2 days ago
add a comment |
$begingroup$
You have to use the formula of the length of a curve: $int ||varphi|| dt$ from $0$ to $2pi$.
in this case the derivative of the components are:
$$x'=sin(2t) + 2tcos(2t)$$
$$y'=cos(2t)- 2tsin(2t)$$
$$z'=2t^0.5$$
follows that the function as an integral is $2t+1$ because $2t+1$ is positive in $[0,2pi]$. And now this integral is very easy.
answered 2 days ago
Antonio SanniaAntonio Sannia
212
New contributor
Antonio Sannia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
did you use $int sqrtx'^2 +y'^2 +z'^2$?
$endgroup$
– ElecLearn
2 days ago
$begingroup$
because i obtain $2int sqrt t^2 +t $
$endgroup$
– ElecLearn
2 days ago
$begingroup$
yes I used it. Under the root I obtain (2t+1)^2
$endgroup$
– Antonio Sannia
2 days ago
add a comment |
$begingroup$
You have to use the formula of the length of a curve: $int ||varphi|| dt$ from $0$ to $2pi$.
in this case the derivative of the components are:
$$x'=sin(2t) + 2tcos(2t)$$
$$y'=cos(2t)- 2tsin(2t)$$
$$z'=2t^0.5$$
follows that the function as an integral is $2t+1$ because $2t+1$ is positive in $[0,2pi]$. And now this integral is very easy.
answered 2 days ago
Antonio SanniaAntonio Sannia
212
New contributor
Antonio Sannia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
You have to use the formula of the length of a curve: $int ||varphi|| dt$ from $0$ to $2pi$.
in this case the derivative of the components are:
$$x'=sin(2t) + 2tcos(2t)$$
$$y'=cos(2t)- 2tsin(2t)$$
$$z'=2t^0.5$$
follows that the function as an integral is $2t+1$ because $2t+1$ is positive in $[0,2pi]$. And now this integral is very easy.
answered 2 days ago
Antonio SanniaAntonio Sannia
212
New contributor
Antonio Sannia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
answered 2 days ago
Antonio SanniaAntonio Sannia
212
New contributor
Antonio Sannia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 days ago
Antonio SanniaAntonio Sannia
212
answered 2 days ago
Antonio SanniaAntonio Sannia
212
212
New contributor
Antonio Sannia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Antonio Sannia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Antonio Sannia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
did you use $int sqrtx'^2 +y'^2 +z'^2$?
$endgroup$
– ElecLearn
2 days ago
$begingroup$
because i obtain $2int sqrt t^2 +t $
$endgroup$
– ElecLearn
2 days ago
$begingroup$
yes I used it. Under the root I obtain (2t+1)^2
$endgroup$
– Antonio Sannia
2 days ago
add a comment |
$begingroup$
did you use $int sqrtx'^2 +y'^2 +z'^2$?
$endgroup$
– ElecLearn
2 days ago
$begingroup$
because i obtain $2int sqrt t^2 +t $
$endgroup$
– ElecLearn
2 days ago
$begingroup$
yes I used it. Under the root I obtain (2t+1)^2
$endgroup$
– Antonio Sannia
2 days ago
$begingroup$
did you use $int sqrtx'^2 +y'^2 +z'^2$?
$endgroup$
– ElecLearn
2 days ago
$begingroup$
did you use $int sqrtx'^2 +y'^2 +z'^2$?
$endgroup$
– ElecLearn
2 days ago
$begingroup$
because i obtain $2int sqrt t^2 +t $
$endgroup$
– ElecLearn
2 days ago
$begingroup$
because i obtain $2int sqrt t^2 +t $
$endgroup$
– ElecLearn
2 days ago
$begingroup$
yes I used it. Under the root I obtain (2t+1)^2
$endgroup$
– Antonio Sannia
2 days ago
$begingroup$
yes I used it. Under the root I obtain (2t+1)^2
$endgroup$
– Antonio Sannia
2 days ago
add a comment |
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$begingroup$
SORRY I MEANT LENGTH NOT AREA
$endgroup$
– ElecLearn
2 days ago
$begingroup$
please show your working. The integrand simplifies.
$endgroup$
– David Quinn
2 days ago
$begingroup$
@DavidQuinn i updated it
$endgroup$
– ElecLearn
2 days ago