Parametric curve length - calculus [on hold]
$begingroup$
Find the length of the following parametric curve.
$x = 5 + frac92 t^3$, $y = 4 + 3 t^{frac92}$, $0 leq t leq 2$.
I used integration and after some point I got lost :( What are the steps?
calculus parametric
edited Mar 30 at 1:16
Peter Mortensen
563310
asked Mar 29 at 19:19
McAMcA
264
$endgroup$
put on hold as off-topic by Travis, Jyrki Lahtonen, Leucippus, Eevee Trainer, José Carlos Santos 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Travis, Jyrki Lahtonen, Leucippus, Eevee Trainer, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Find the length of the following parametric curve.
$x = 5 + frac92 t^3$, $y = 4 + 3 t^{frac92}$, $0 leq t leq 2$.
I used integration and after some point I got lost :( What are the steps?
calculus parametric
edited Mar 30 at 1:16
Peter Mortensen
563310
asked Mar 29 at 19:19
McAMcA
264
$endgroup$
put on hold as off-topic by Travis, Jyrki Lahtonen, Leucippus, Eevee Trainer, José Carlos Santos 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Travis, Jyrki Lahtonen, Leucippus, Eevee Trainer, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Is this $$x=5+frac{9}{2}t^3,y=4+3t^{9/2}$$?
$endgroup$
– Dr. Sonnhard Graubner
Mar 29 at 19:28
add a comment |
$begingroup$
Find the length of the following parametric curve.
$x = 5 + frac92 t^3$, $y = 4 + 3 t^{frac92}$, $0 leq t leq 2$.
I used integration and after some point I got lost :( What are the steps?
calculus parametric
edited Mar 30 at 1:16
Peter Mortensen
563310
asked Mar 29 at 19:19
McAMcA
264
$endgroup$
Find the length of the following parametric curve.
$x = 5 + frac92 t^3$, $y = 4 + 3 t^{frac92}$, $0 leq t leq 2$.
I used integration and after some point I got lost :( What are the steps?
calculus parametric
calculus parametric
edited Mar 30 at 1:16
Peter Mortensen
563310
asked Mar 29 at 19:19
McAMcA
264
edited Mar 30 at 1:16
Peter Mortensen
563310
asked Mar 29 at 19:19
McAMcA
264
edited Mar 30 at 1:16
Peter Mortensen
563310
edited Mar 30 at 1:16
Peter Mortensen
563310
edited Mar 30 at 1:16
Peter Mortensen
563310
563310
asked Mar 29 at 19:19
McAMcA
264
asked Mar 29 at 19:19
McAMcA
264
asked Mar 29 at 19:19
McAMcA
264
264
put on hold as off-topic by Travis, Jyrki Lahtonen, Leucippus, Eevee Trainer, José Carlos Santos 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Travis, Jyrki Lahtonen, Leucippus, Eevee Trainer, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by Travis, Jyrki Lahtonen, Leucippus, Eevee Trainer, José Carlos Santos 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Travis, Jyrki Lahtonen, Leucippus, Eevee Trainer, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Is this $$x=5+frac{9}{2}t^3,y=4+3t^{9/2}$$?
$endgroup$
– Dr. Sonnhard Graubner
Mar 29 at 19:28
add a comment |
$begingroup$
Is this $$x=5+frac{9}{2}t^3,y=4+3t^{9/2}$$?
$endgroup$
– Dr. Sonnhard Graubner
Mar 29 at 19:28
$begingroup$
Is this $$x=5+frac{9}{2}t^3,y=4+3t^{9/2}$$?
$endgroup$
– Dr. Sonnhard Graubner
Mar 29 at 19:28
$begingroup$
Is this $$x=5+frac{9}{2}t^3,y=4+3t^{9/2}$$?
$endgroup$
– Dr. Sonnhard Graubner
Mar 29 at 19:28
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
You must use the formula $$int_{0}^{2}sqrt{left(frac{dx}{dt}right)^2+left(frac{dy}{dt}right)^2}dt$$
$$dx=frac{9}{2}3t^2dt$$ and $$dy=3cdot frac{9}{2}t^{7/2}dt$$
answered Mar 29 at 19:32
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
$endgroup$
add a comment |
$begingroup$
Apply the formula for arc length, we get
$$
int_0^2 frac{27{{t}^{2}},sqrt{{{t}^{3}}+1}}{2} dt
$$
Then we make the change of variable $v=t^3+1$ to get
$$
int_1^9 frac 9 2 sqrt{v} dv = 78.
$$
answered Mar 29 at 19:38
EagleToLearnEagleToLearn
534
New contributor
EagleToLearn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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add a comment |
$begingroup$
begin{aligned}L&=int_0^2 sqrt{frac{729}4t^4+frac{729}4t^7}dt\&=int_0^2sqrt{frac{729}4t^4(1+t^3)}dt\&=frac{27}2int_0^2t^2(1+t^3)^{frac12}dt\&=3(1+t^3)^{frac32}big]_0^2end{aligned}
Made the leap from the third line to the fourth line by recognizing that $F(t)=3(1+t^3)^{frac32}$ is an antiderivative of $f(t)=frac{27}2t^2(1+t^3)^{frac12}$.
answered Mar 29 at 19:48
Matt A PeltoMatt A Pelto
2,677621
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add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You must use the formula $$int_{0}^{2}sqrt{left(frac{dx}{dt}right)^2+left(frac{dy}{dt}right)^2}dt$$
$$dx=frac{9}{2}3t^2dt$$ and $$dy=3cdot frac{9}{2}t^{7/2}dt$$
answered Mar 29 at 19:32
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
$endgroup$
add a comment |
$begingroup$
You must use the formula $$int_{0}^{2}sqrt{left(frac{dx}{dt}right)^2+left(frac{dy}{dt}right)^2}dt$$
$$dx=frac{9}{2}3t^2dt$$ and $$dy=3cdot frac{9}{2}t^{7/2}dt$$
answered Mar 29 at 19:32
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
$endgroup$
add a comment |
$begingroup$
You must use the formula $$int_{0}^{2}sqrt{left(frac{dx}{dt}right)^2+left(frac{dy}{dt}right)^2}dt$$
$$dx=frac{9}{2}3t^2dt$$ and $$dy=3cdot frac{9}{2}t^{7/2}dt$$
answered Mar 29 at 19:32
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
$endgroup$
You must use the formula $$int_{0}^{2}sqrt{left(frac{dx}{dt}right)^2+left(frac{dy}{dt}right)^2}dt$$
$$dx=frac{9}{2}3t^2dt$$ and $$dy=3cdot frac{9}{2}t^{7/2}dt$$
answered Mar 29 at 19:32
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
answered Mar 29 at 19:32
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
answered Mar 29 at 19:32
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
answered Mar 29 at 19:32
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
78.4k42867
add a comment |
add a comment |
$begingroup$
Apply the formula for arc length, we get
$$
int_0^2 frac{27{{t}^{2}},sqrt{{{t}^{3}}+1}}{2} dt
$$
Then we make the change of variable $v=t^3+1$ to get
$$
int_1^9 frac 9 2 sqrt{v} dv = 78.
$$
answered Mar 29 at 19:38
EagleToLearnEagleToLearn
534
New contributor
EagleToLearn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Apply the formula for arc length, we get
$$
int_0^2 frac{27{{t}^{2}},sqrt{{{t}^{3}}+1}}{2} dt
$$
Then we make the change of variable $v=t^3+1$ to get
$$
int_1^9 frac 9 2 sqrt{v} dv = 78.
$$
answered Mar 29 at 19:38
EagleToLearnEagleToLearn
534
New contributor
EagleToLearn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Apply the formula for arc length, we get
$$
int_0^2 frac{27{{t}^{2}},sqrt{{{t}^{3}}+1}}{2} dt
$$
Then we make the change of variable $v=t^3+1$ to get
$$
int_1^9 frac 9 2 sqrt{v} dv = 78.
$$
answered Mar 29 at 19:38
EagleToLearnEagleToLearn
534
New contributor
EagleToLearn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Apply the formula for arc length, we get
$$
int_0^2 frac{27{{t}^{2}},sqrt{{{t}^{3}}+1}}{2} dt
$$
Then we make the change of variable $v=t^3+1$ to get
$$
int_1^9 frac 9 2 sqrt{v} dv = 78.
$$
answered Mar 29 at 19:38
EagleToLearnEagleToLearn
534
New contributor
EagleToLearn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Mar 29 at 19:38
EagleToLearnEagleToLearn
534
New contributor
EagleToLearn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Mar 29 at 19:38
EagleToLearnEagleToLearn
534
answered Mar 29 at 19:38
EagleToLearnEagleToLearn
534
534
New contributor
EagleToLearn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
EagleToLearn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
EagleToLearn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
$begingroup$
begin{aligned}L&=int_0^2 sqrt{frac{729}4t^4+frac{729}4t^7}dt\&=int_0^2sqrt{frac{729}4t^4(1+t^3)}dt\&=frac{27}2int_0^2t^2(1+t^3)^{frac12}dt\&=3(1+t^3)^{frac32}big]_0^2end{aligned}
Made the leap from the third line to the fourth line by recognizing that $F(t)=3(1+t^3)^{frac32}$ is an antiderivative of $f(t)=frac{27}2t^2(1+t^3)^{frac12}$.
answered Mar 29 at 19:48
Matt A PeltoMatt A Pelto
2,677621
$endgroup$
add a comment |
$begingroup$
begin{aligned}L&=int_0^2 sqrt{frac{729}4t^4+frac{729}4t^7}dt\&=int_0^2sqrt{frac{729}4t^4(1+t^3)}dt\&=frac{27}2int_0^2t^2(1+t^3)^{frac12}dt\&=3(1+t^3)^{frac32}big]_0^2end{aligned}
Made the leap from the third line to the fourth line by recognizing that $F(t)=3(1+t^3)^{frac32}$ is an antiderivative of $f(t)=frac{27}2t^2(1+t^3)^{frac12}$.
answered Mar 29 at 19:48
Matt A PeltoMatt A Pelto
2,677621
$endgroup$
add a comment |
$begingroup$
begin{aligned}L&=int_0^2 sqrt{frac{729}4t^4+frac{729}4t^7}dt\&=int_0^2sqrt{frac{729}4t^4(1+t^3)}dt\&=frac{27}2int_0^2t^2(1+t^3)^{frac12}dt\&=3(1+t^3)^{frac32}big]_0^2end{aligned}
Made the leap from the third line to the fourth line by recognizing that $F(t)=3(1+t^3)^{frac32}$ is an antiderivative of $f(t)=frac{27}2t^2(1+t^3)^{frac12}$.
answered Mar 29 at 19:48
Matt A PeltoMatt A Pelto
2,677621
$endgroup$
begin{aligned}L&=int_0^2 sqrt{frac{729}4t^4+frac{729}4t^7}dt\&=int_0^2sqrt{frac{729}4t^4(1+t^3)}dt\&=frac{27}2int_0^2t^2(1+t^3)^{frac12}dt\&=3(1+t^3)^{frac32}big]_0^2end{aligned}
Made the leap from the third line to the fourth line by recognizing that $F(t)=3(1+t^3)^{frac32}$ is an antiderivative of $f(t)=frac{27}2t^2(1+t^3)^{frac12}$.
answered Mar 29 at 19:48
Matt A PeltoMatt A Pelto
2,677621
edited Mar 29 at 20:20
answered Mar 29 at 19:48
Matt A PeltoMatt A Pelto
2,677621
answered Mar 29 at 19:48
Matt A PeltoMatt A Pelto
2,677621
answered Mar 29 at 19:48
Matt A PeltoMatt A Pelto
2,677621
2,677621
add a comment |
add a comment |
$begingroup$
Is this $$x=5+frac{9}{2}t^3,y=4+3t^{9/2}$$?
$endgroup$
– Dr. Sonnhard Graubner
Mar 29 at 19:28