Hfrhyu

Raspberry pi 3 B with Ubuntu 18.04 server arm64: what chip

Small nick on power cord from an electric alarm clock, and copper wiring exposed but intact

Is it correct to say moon starry nights?

IC has pull-down resistors on SMBus lines?

Physiological effects of huge anime eyes

How to use ReplaceAll on an expression that contains a rule

Airplane gently rocking its wings during whole flight

Is it convenient to ask the journal's editor for two additional days to complete a review?

Getting Stale Gas Out of a Gas Tank w/out Dropping the Tank

How do I fit a non linear curve?

Film where the government was corrupt with aliens, people sent to kill aliens are given rigged visors not showing the right aliens

Is it ever safe to open a suspicious HTML file (e.g. email attachment)?

What does "shotgun unity" refer to here in this sentence?

Is there a reasonable and studied concept of reduction between regular languages?

Is Nisuin Biblical or Rabbinic?

Is fine stranded wire ok for main supply line?

It is correct to match light sources with the same color temperature?

Yu-Gi-Oh cards in Python 3

Towers in the ocean; How deep can they be built?

What is the process for purifying your home if you believe it may have been previously used for pagan worship?

In the "Harry Potter and the Order of the Phoenix" video game, what potion is used to sabotage Umbridge's speakers?

Plausibility of squid whales

If Nick Fury and Coulson already knew about aliens (Kree and Skrull) why did they wait until Thor's appearance to start making weapons?

What was Carter Burke's job for "the company" in Aliens?

Parametric curve length - calculus [on hold]

2

$begingroup$

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

share|cite|improve this question

edited 2 days ago

Peter Mortensen

565310

asked 2 days ago

McAMcA

264

$endgroup$

put on hold as off-topic by Travis, Jyrki Lahtonen, Leucippus, Eevee Trainer, José Carlos Santos yesterday

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 $$x=5+frac92t^3,y=4+3t^9/2$$?
  $endgroup$
  – Dr. Sonnhard Graubner
  2 days ago
  
  

  
  
  
  

add a comment | 

2

$begingroup$

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

share|cite|improve this question

edited 2 days ago

Peter Mortensen

565310

asked 2 days ago

McAMcA

264

$endgroup$

put on hold as off-topic by Travis, Jyrki Lahtonen, Leucippus, Eevee Trainer, José Carlos Santos yesterday

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 $$x=5+frac92t^3,y=4+3t^9/2$$?
  $endgroup$
  – Dr. Sonnhard Graubner
  2 days ago
  
  

  
  
  
  

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

share|cite|improve this question

edited 2 days ago

Peter Mortensen

565310

asked 2 days ago

McAMcA

264

$endgroup$

share|cite|improve this question

edited 2 days ago

Peter Mortensen

565310

asked 2 days ago

McAMcA

264

share|cite|improve this question

edited 2 days ago

Peter Mortensen

565310

asked 2 days ago

McAMcA

264

edited 2 days ago

Peter Mortensen

565310

edited 2 days ago

Peter Mortensen

565310

asked 2 days ago

McAMcA

264

asked 2 days ago

McAMcA

264

3 Answers
3

active

oldest

votes

2

$begingroup$

You must use the formula $$int_0^2sqrtleft(fracdxdtright)^2+left(fracdydtright)^2dt$$
$$dx=frac923t^2dt$$ and $$dy=3cdot frac92t^7/2dt$$

share|cite|improve this answer

answered 2 days ago

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.4k42867

$endgroup$

add a comment | 

2

$begingroup$

Apply the formula for arc length, we get
$$
int_0^2 frac27t^2,sqrtt^3+12 dt
$$
Then we make the change of variable $v=t^3+1$ to get
$$
int_1^9 frac 9 2 sqrtv dv = 78.
$$

share|cite|improve this answer

answered 2 days ago

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 | 

2

$begingroup$

beginalignedL&=int_0^2 sqrtfrac7294t^4+frac7294t^7dt\&=int_0^2sqrtfrac7294t^4(1+t^3)dt\&=frac272int_0^2t^2(1+t^3)^frac12dt\&=3(1+t^3)^frac32big]_0^2endaligned

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 $f(t)=frac272t^2(1+t^3)^frac12$.

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

Matt A PeltoMatt A Pelto

2,667621

$endgroup$

add a comment | 

2

$begingroup$

You must use the formula $$int_0^2sqrtleft(fracdxdtright)^2+left(fracdydtright)^2dt$$
$$dx=frac923t^2dt$$ and $$dy=3cdot frac92t^7/2dt$$

share|cite|improve this answer

answered 2 days ago

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.4k42867

$endgroup$

add a comment | 

2

$begingroup$

You must use the formula $$int_0^2sqrtleft(fracdxdtright)^2+left(fracdydtright)^2dt$$
$$dx=frac923t^2dt$$ and $$dy=3cdot frac92t^7/2dt$$

share|cite|improve this answer

answered 2 days ago

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.4k42867

$endgroup$

add a comment | 

$begingroup$

You must use the formula $$int_0^2sqrtleft(fracdxdtright)^2+left(fracdydtright)^2dt$$
$$dx=frac923t^2dt$$ and $$dy=3cdot frac92t^7/2dt$$

share|cite|improve this answer

answered 2 days ago

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.4k42867

$endgroup$

share|cite|improve this answer

answered 2 days ago

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.4k42867

answered 2 days ago

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.4k42867

answered 2 days ago

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.4k42867

2

$begingroup$

Apply the formula for arc length, we get
$$
int_0^2 frac27t^2,sqrtt^3+12 dt
$$
Then we make the change of variable $v=t^3+1$ to get
$$
int_1^9 frac 9 2 sqrtv dv = 78.
$$

share|cite|improve this answer

answered 2 days ago

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 | 

2

$begingroup$

Apply the formula for arc length, we get
$$
int_0^2 frac27t^2,sqrtt^3+12 dt
$$
Then we make the change of variable $v=t^3+1$ to get
$$
int_1^9 frac 9 2 sqrtv dv = 78.
$$

share|cite|improve this answer

answered 2 days ago

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
$$
int_0^2 frac27t^2,sqrtt^3+12 dt
$$
Then we make the change of variable $v=t^3+1$ to get
$$
int_1^9 frac 9 2 sqrtv dv = 78.
$$

share|cite|improve this answer

answered 2 days ago

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.

$endgroup$

share|cite|improve this answer

answered 2 days ago

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 2 days ago

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 2 days ago

EagleToLearnEagleToLearn

534

2

$begingroup$

beginalignedL&=int_0^2 sqrtfrac7294t^4+frac7294t^7dt\&=int_0^2sqrtfrac7294t^4(1+t^3)dt\&=frac272int_0^2t^2(1+t^3)^frac12dt\&=3(1+t^3)^frac32big]_0^2endaligned

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 $f(t)=frac272t^2(1+t^3)^frac12$.

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

Matt A PeltoMatt A Pelto

2,667621

$endgroup$

add a comment | 

2

$begingroup$

beginalignedL&=int_0^2 sqrtfrac7294t^4+frac7294t^7dt\&=int_0^2sqrtfrac7294t^4(1+t^3)dt\&=frac272int_0^2t^2(1+t^3)^frac12dt\&=3(1+t^3)^frac32big]_0^2endaligned

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 $f(t)=frac272t^2(1+t^3)^frac12$.

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

Matt A PeltoMatt A Pelto

2,667621

$endgroup$

add a comment | 

$begingroup$

beginalignedL&=int_0^2 sqrtfrac7294t^4+frac7294t^7dt\&=int_0^2sqrtfrac7294t^4(1+t^3)dt\&=frac272int_0^2t^2(1+t^3)^frac12dt\&=3(1+t^3)^frac32big]_0^2endaligned

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 $f(t)=frac272t^2(1+t^3)^frac12$.

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

Matt A PeltoMatt A Pelto

2,667621

$endgroup$

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

Matt A PeltoMatt A Pelto

2,667621

answered 2 days ago

Matt A PeltoMatt A Pelto

2,667621

answered 2 days ago

Matt A PeltoMatt A Pelto

2,667621