Rigorous deduction of area under parametric curve

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My Calculus book says:

"We know that the area under a curve $y=F(x)$ from $a$ to $b$ is $A=int_a^b F(x) dx$, where $F(x)geqslant0$. If the curve is traced out once by parametric equations $x=f(t)$ and $y=g(t)$, $alpha leqslant t leqslant beta$, then we can calculate an area formula by using the Substitution Rule for Definite Integrals as follows:

$A=int_a^b y dx =int_alpha^beta g(t)f'(t)dt,,,,,,,,,,$[or $int_beta^alpha g(t)f'(t)dt$] ".

Also, it never says that if we have $dx/dt=f'(t)$ we can split the ratio of differentials and obtain $dx=f'(t)dt$.

Could someone deduce the formula of the area under a parametric curve without the use of this mysterious rule of "splitting" the ratio of differentials?

The book deduced the Substituition Rule without doing the split, but said that we could use the split as some non-proved equation to remember how to use this rule. But it is not helpful to me to understand how Substituition Rule was used to obtain the area under parametric curve without using this "split trick".

area parametric parametrization

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edited Jul 28 at 0:19

asked Jul 26 at 22:10

Guilherme Correa Teixeira

216

add a comment | 

up vote
2
down vote

favorite

My Calculus book says:

"We know that the area under a curve $y=F(x)$ from $a$ to $b$ is $A=int_a^b F(x) dx$, where $F(x)geqslant0$. If the curve is traced out once by parametric equations $x=f(t)$ and $y=g(t)$, $alpha leqslant t leqslant beta$, then we can calculate an area formula by using the Substitution Rule for Definite Integrals as follows:

$A=int_a^b y dx =int_alpha^beta g(t)f'(t)dt,,,,,,,,,,$[or $int_beta^alpha g(t)f'(t)dt$] ".

Also, it never says that if we have $dx/dt=f'(t)$ we can split the ratio of differentials and obtain $dx=f'(t)dt$.

Could someone deduce the formula of the area under a parametric curve without the use of this mysterious rule of "splitting" the ratio of differentials?

The book deduced the Substituition Rule without doing the split, but said that we could use the split as some non-proved equation to remember how to use this rule. But it is not helpful to me to understand how Substituition Rule was used to obtain the area under parametric curve without using this "split trick".

area parametric parametrization

share|cite|improve this question

edited Jul 28 at 0:19

asked Jul 26 at 22:10

Guilherme Correa Teixeira

216

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

My Calculus book says:

"We know that the area under a curve $y=F(x)$ from $a$ to $b$ is $A=int_a^b F(x) dx$, where $F(x)geqslant0$. If the curve is traced out once by parametric equations $x=f(t)$ and $y=g(t)$, $alpha leqslant t leqslant beta$, then we can calculate an area formula by using the Substitution Rule for Definite Integrals as follows:

$A=int_a^b y dx =int_alpha^beta g(t)f'(t)dt,,,,,,,,,,$[or $int_beta^alpha g(t)f'(t)dt$] ".

Also, it never says that if we have $dx/dt=f'(t)$ we can split the ratio of differentials and obtain $dx=f'(t)dt$.

Could someone deduce the formula of the area under a parametric curve without the use of this mysterious rule of "splitting" the ratio of differentials?

The book deduced the Substituition Rule without doing the split, but said that we could use the split as some non-proved equation to remember how to use this rule. But it is not helpful to me to understand how Substituition Rule was used to obtain the area under parametric curve without using this "split trick".

area parametric parametrization

share|cite|improve this question

edited Jul 28 at 0:19

asked Jul 26 at 22:10

Guilherme Correa Teixeira

216

My Calculus book says:

"We know that the area under a curve $y=F(x)$ from $a$ to $b$ is $A=int_a^b F(x) dx$, where $F(x)geqslant0$. If the curve is traced out once by parametric equations $x=f(t)$ and $y=g(t)$, $alpha leqslant t leqslant beta$, then we can calculate an area formula by using the Substitution Rule for Definite Integrals as follows:

$A=int_a^b y dx =int_alpha^beta g(t)f'(t)dt,,,,,,,,,,$[or $int_beta^alpha g(t)f'(t)dt$] ".

Also, it never says that if we have $dx/dt=f'(t)$ we can split the ratio of differentials and obtain $dx=f'(t)dt$.

Could someone deduce the formula of the area under a parametric curve without the use of this mysterious rule of "splitting" the ratio of differentials?

The book deduced the Substituition Rule without doing the split, but said that we could use the split as some non-proved equation to remember how to use this rule. But it is not helpful to me to understand how Substituition Rule was used to obtain the area under parametric curve without using this "split trick".

area parametric parametrization

share|cite|improve this question

edited Jul 28 at 0:19

asked Jul 26 at 22:10

Guilherme Correa Teixeira

216

share|cite|improve this question

share|cite|improve this question

edited Jul 28 at 0:19

edited Jul 28 at 0:19

edited Jul 28 at 0:19

asked Jul 26 at 22:10

Guilherme Correa Teixeira

216

asked Jul 26 at 22:10

Guilherme Correa Teixeira

216

asked Jul 26 at 22:10

Guilherme Correa Teixeira

216

216

add a comment | 

add a comment | 

3 Answers
3

active

oldest

votes

up vote
1
down vote



accepted

Ok. thank you gimusi. Your notation gave me an idea, and I think I deduced the formula without the use of equations with differentials like $dx=f'(t)dt$, which is what i was looking for. If someone is reading this and think it is right, please let me know.

So, let me try...

Given a parametric curve with $y=f(t), x=g(t),$ and assuming that $y$ could be writen in terms of x, in other words, $y=F(x)$, we have these equalities:

$y=F(x)=F(g(t))=f(t)$

Now, since the Substitution Rule give us the following equation:

$int f(g(x))g'(x)dx=int f(u)du,,,,,,, with,,,,,,,u=g(x)$

(which can be deduced without the use of equations with differentials like $du=g'(x)dx$), we could rewrite the rule by just changing some letters and have:

$int F(g(t))g'(t)dt=int F(x)dx,,,,,,, with,,,,,,,x=g(t)$

then, using the case of the parametric curve above, that is, $F(g(t))=f(t)$ and $F(x)=y$, we have:

$int f(t)g'(t)dt=int ydx$

That is it. It is enough to satisfy my curiosity, although I think that there is some more steps to prove the relationship between the definite integrals that I firstly asked, which is:

$A=∫^b_aydx=∫^β_αg(t)f′(t)dt,,,,,,,,,[or ∫^α_βg(t)f′(t)dt]$

share|cite|improve this answer

edited Jul 27 at 0:59

answered Jul 27 at 0:45

Guilherme Correa Teixeira

216

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes that’s right, for the latter doubts refer to some calculus book with an explanation of the substitution rule for integrals.
  – gimusi
  Jul 31 at 19:22
  

  
  
  
  

add a comment | 

up vote
0
down vote

It is a simple change of varible for integrals, indeed we have

$$P(x,F(x))$$

and we set

• $x=f(t)implies dx=f'(t)dt$


• $y=F(x(t))=g(t)$

and then

$$A=int_a^b F(x) dx=int_alpha^beta g(t)f'(t) dt$$

If such interpretation don't satisfy your curiosity, for a formal proof you should refer to the theory for the change of variable for integrals.

share|cite|improve this answer

answered Jul 26 at 22:21

gimusi

65k73583

• 
  
  
  

  
  

  
  
  
  
  
  

  
  It should be mandatory that the person who puts a downvote explains their reasons.
  – Piquito
  Jul 31 at 18:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Piquito It should be very useful indeed also for me in order to learn something more.
  – gimusi
  Jul 31 at 19:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Apparently, Gimusi, we can "learn" a lot from certain citizens.
  – Piquito
  Jul 31 at 20:51
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Gimusi, I asked for the proof without the use of "split-trick". But your notation helped me to think how could it be done. Thank you.
  – Guilherme Correa Teixeira
  Aug 1 at 20:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @GuilhermeCorreaTeixeira You are welcome! Bye
  – gimusi
  Aug 1 at 20:10
  

  
  
  
  

add a comment | 

up vote
0
down vote

HINT.- $ydx=ycdotdfracdxdtcdot dt$ and $dfracdxdt=f'(t)$

share|cite|improve this answer

answered Jul 31 at 19:02

Piquito

17.3k31234

add a comment | 

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

Ok. thank you gimusi. Your notation gave me an idea, and I think I deduced the formula without the use of equations with differentials like $dx=f'(t)dt$, which is what i was looking for. If someone is reading this and think it is right, please let me know.

So, let me try...

Given a parametric curve with $y=f(t), x=g(t),$ and assuming that $y$ could be writen in terms of x, in other words, $y=F(x)$, we have these equalities:

$y=F(x)=F(g(t))=f(t)$

Now, since the Substitution Rule give us the following equation:

$int f(g(x))g'(x)dx=int f(u)du,,,,,,, with,,,,,,,u=g(x)$

(which can be deduced without the use of equations with differentials like $du=g'(x)dx$), we could rewrite the rule by just changing some letters and have:

$int F(g(t))g'(t)dt=int F(x)dx,,,,,,, with,,,,,,,x=g(t)$

then, using the case of the parametric curve above, that is, $F(g(t))=f(t)$ and $F(x)=y$, we have:

$int f(t)g'(t)dt=int ydx$

That is it. It is enough to satisfy my curiosity, although I think that there is some more steps to prove the relationship between the definite integrals that I firstly asked, which is:

$A=∫^b_aydx=∫^β_αg(t)f′(t)dt,,,,,,,,,[or ∫^α_βg(t)f′(t)dt]$

share|cite|improve this answer

edited Jul 27 at 0:59

answered Jul 27 at 0:45

Guilherme Correa Teixeira

216

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes that’s right, for the latter doubts refer to some calculus book with an explanation of the substitution rule for integrals.
  – gimusi
  Jul 31 at 19:22
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

Ok. thank you gimusi. Your notation gave me an idea, and I think I deduced the formula without the use of equations with differentials like $dx=f'(t)dt$, which is what i was looking for. If someone is reading this and think it is right, please let me know.

So, let me try...

Given a parametric curve with $y=f(t), x=g(t),$ and assuming that $y$ could be writen in terms of x, in other words, $y=F(x)$, we have these equalities:

$y=F(x)=F(g(t))=f(t)$

Now, since the Substitution Rule give us the following equation:

$int f(g(x))g'(x)dx=int f(u)du,,,,,,, with,,,,,,,u=g(x)$

(which can be deduced without the use of equations with differentials like $du=g'(x)dx$), we could rewrite the rule by just changing some letters and have:

$int F(g(t))g'(t)dt=int F(x)dx,,,,,,, with,,,,,,,x=g(t)$

then, using the case of the parametric curve above, that is, $F(g(t))=f(t)$ and $F(x)=y$, we have:

$int f(t)g'(t)dt=int ydx$

That is it. It is enough to satisfy my curiosity, although I think that there is some more steps to prove the relationship between the definite integrals that I firstly asked, which is:

$A=∫^b_aydx=∫^β_αg(t)f′(t)dt,,,,,,,,,[or ∫^α_βg(t)f′(t)dt]$

share|cite|improve this answer

edited Jul 27 at 0:59

answered Jul 27 at 0:45

Guilherme Correa Teixeira

216

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes that’s right, for the latter doubts refer to some calculus book with an explanation of the substitution rule for integrals.
  – gimusi
  Jul 31 at 19:22
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

Ok. thank you gimusi. Your notation gave me an idea, and I think I deduced the formula without the use of equations with differentials like $dx=f'(t)dt$, which is what i was looking for. If someone is reading this and think it is right, please let me know.

So, let me try...

Given a parametric curve with $y=f(t), x=g(t),$ and assuming that $y$ could be writen in terms of x, in other words, $y=F(x)$, we have these equalities:

$y=F(x)=F(g(t))=f(t)$

Now, since the Substitution Rule give us the following equation:

$int f(g(x))g'(x)dx=int f(u)du,,,,,,, with,,,,,,,u=g(x)$

(which can be deduced without the use of equations with differentials like $du=g'(x)dx$), we could rewrite the rule by just changing some letters and have:

$int F(g(t))g'(t)dt=int F(x)dx,,,,,,, with,,,,,,,x=g(t)$

then, using the case of the parametric curve above, that is, $F(g(t))=f(t)$ and $F(x)=y$, we have:

$int f(t)g'(t)dt=int ydx$

That is it. It is enough to satisfy my curiosity, although I think that there is some more steps to prove the relationship between the definite integrals that I firstly asked, which is:

$A=∫^b_aydx=∫^β_αg(t)f′(t)dt,,,,,,,,,[or ∫^α_βg(t)f′(t)dt]$

share|cite|improve this answer

edited Jul 27 at 0:59

answered Jul 27 at 0:45

Guilherme Correa Teixeira

216

Ok. thank you gimusi. Your notation gave me an idea, and I think I deduced the formula without the use of equations with differentials like $dx=f'(t)dt$, which is what i was looking for. If someone is reading this and think it is right, please let me know.

So, let me try...

Given a parametric curve with $y=f(t), x=g(t),$ and assuming that $y$ could be writen in terms of x, in other words, $y=F(x)$, we have these equalities:

$y=F(x)=F(g(t))=f(t)$

Now, since the Substitution Rule give us the following equation:

$int f(g(x))g'(x)dx=int f(u)du,,,,,,, with,,,,,,,u=g(x)$

(which can be deduced without the use of equations with differentials like $du=g'(x)dx$), we could rewrite the rule by just changing some letters and have:

$int F(g(t))g'(t)dt=int F(x)dx,,,,,,, with,,,,,,,x=g(t)$

then, using the case of the parametric curve above, that is, $F(g(t))=f(t)$ and $F(x)=y$, we have:

$int f(t)g'(t)dt=int ydx$

That is it. It is enough to satisfy my curiosity, although I think that there is some more steps to prove the relationship between the definite integrals that I firstly asked, which is:

$A=∫^b_aydx=∫^β_αg(t)f′(t)dt,,,,,,,,,[or ∫^α_βg(t)f′(t)dt]$

share|cite|improve this answer

edited Jul 27 at 0:59

answered Jul 27 at 0:45

Guilherme Correa Teixeira

216

share|cite|improve this answer

share|cite|improve this answer

edited Jul 27 at 0:59

edited Jul 27 at 0:59

edited Jul 27 at 0:59

answered Jul 27 at 0:45

Guilherme Correa Teixeira

216

answered Jul 27 at 0:45

Guilherme Correa Teixeira

216

answered Jul 27 at 0:45

Guilherme Correa Teixeira

216

216

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes that’s right, for the latter doubts refer to some calculus book with an explanation of the substitution rule for integrals.
  – gimusi
  Jul 31 at 19:22
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes that’s right, for the latter doubts refer to some calculus book with an explanation of the substitution rule for integrals.
  – gimusi
  Jul 31 at 19:22
  

  
  
  
  

Yes that’s right, for the latter doubts refer to some calculus book with an explanation of the substitution rule for integrals.
– gimusi
Jul 31 at 19:22

Yes that’s right, for the latter doubts refer to some calculus book with an explanation of the substitution rule for integrals.
– gimusi
Jul 31 at 19:22

add a comment | 

up vote
0
down vote

It is a simple change of varible for integrals, indeed we have

$$P(x,F(x))$$

and we set

• $x=f(t)implies dx=f'(t)dt$


• $y=F(x(t))=g(t)$

and then

$$A=int_a^b F(x) dx=int_alpha^beta g(t)f'(t) dt$$

If such interpretation don't satisfy your curiosity, for a formal proof you should refer to the theory for the change of variable for integrals.

share|cite|improve this answer

answered Jul 26 at 22:21

gimusi

65k73583

• 
  
  
  

  
  

  
  
  
  
  
  

  
  It should be mandatory that the person who puts a downvote explains their reasons.
  – Piquito
  Jul 31 at 18:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Piquito It should be very useful indeed also for me in order to learn something more.
  – gimusi
  Jul 31 at 19:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Apparently, Gimusi, we can "learn" a lot from certain citizens.
  – Piquito
  Jul 31 at 20:51
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Gimusi, I asked for the proof without the use of "split-trick". But your notation helped me to think how could it be done. Thank you.
  – Guilherme Correa Teixeira
  Aug 1 at 20:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @GuilhermeCorreaTeixeira You are welcome! Bye
  – gimusi
  Aug 1 at 20:10
  

  
  
  
  

add a comment | 

up vote
0
down vote

It is a simple change of varible for integrals, indeed we have

$$P(x,F(x))$$

and we set

• $x=f(t)implies dx=f'(t)dt$


• $y=F(x(t))=g(t)$

and then

$$A=int_a^b F(x) dx=int_alpha^beta g(t)f'(t) dt$$

If such interpretation don't satisfy your curiosity, for a formal proof you should refer to the theory for the change of variable for integrals.

share|cite|improve this answer

answered Jul 26 at 22:21

gimusi

65k73583

• 
  
  
  

  
  

  
  
  
  
  
  

  
  It should be mandatory that the person who puts a downvote explains their reasons.
  – Piquito
  Jul 31 at 18:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Piquito It should be very useful indeed also for me in order to learn something more.
  – gimusi
  Jul 31 at 19:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Apparently, Gimusi, we can "learn" a lot from certain citizens.
  – Piquito
  Jul 31 at 20:51
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Gimusi, I asked for the proof without the use of "split-trick". But your notation helped me to think how could it be done. Thank you.
  – Guilherme Correa Teixeira
  Aug 1 at 20:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @GuilhermeCorreaTeixeira You are welcome! Bye
  – gimusi
  Aug 1 at 20:10
  

  
  
  
  

add a comment | 

up vote
0
down vote

up vote
0
down vote

It is a simple change of varible for integrals, indeed we have

$$P(x,F(x))$$

and we set

• $x=f(t)implies dx=f'(t)dt$


• $y=F(x(t))=g(t)$

and then

$$A=int_a^b F(x) dx=int_alpha^beta g(t)f'(t) dt$$

If such interpretation don't satisfy your curiosity, for a formal proof you should refer to the theory for the change of variable for integrals.

share|cite|improve this answer

answered Jul 26 at 22:21

gimusi

65k73583

It is a simple change of varible for integrals, indeed we have

$$P(x,F(x))$$

and we set

• $x=f(t)implies dx=f'(t)dt$


• $y=F(x(t))=g(t)$

and then

$$A=int_a^b F(x) dx=int_alpha^beta g(t)f'(t) dt$$

If such interpretation don't satisfy your curiosity, for a formal proof you should refer to the theory for the change of variable for integrals.

share|cite|improve this answer

answered Jul 26 at 22:21

gimusi

65k73583

share|cite|improve this answer

share|cite|improve this answer

answered Jul 26 at 22:21

gimusi

65k73583

answered Jul 26 at 22:21

gimusi

65k73583

answered Jul 26 at 22:21

gimusi

65k73583

65k73583

• 
  
  
  

  
  

  
  
  
  
  
  

  
  It should be mandatory that the person who puts a downvote explains their reasons.
  – Piquito
  Jul 31 at 18:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Piquito It should be very useful indeed also for me in order to learn something more.
  – gimusi
  Jul 31 at 19:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Apparently, Gimusi, we can "learn" a lot from certain citizens.
  – Piquito
  Jul 31 at 20:51
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Gimusi, I asked for the proof without the use of "split-trick". But your notation helped me to think how could it be done. Thank you.
  – Guilherme Correa Teixeira
  Aug 1 at 20:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @GuilhermeCorreaTeixeira You are welcome! Bye
  – gimusi
  Aug 1 at 20:10
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  It should be mandatory that the person who puts a downvote explains their reasons.
  – Piquito
  Jul 31 at 18:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Piquito It should be very useful indeed also for me in order to learn something more.
  – gimusi
  Jul 31 at 19:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Apparently, Gimusi, we can "learn" a lot from certain citizens.
  – Piquito
  Jul 31 at 20:51
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Gimusi, I asked for the proof without the use of "split-trick". But your notation helped me to think how could it be done. Thank you.
  – Guilherme Correa Teixeira
  Aug 1 at 20:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @GuilhermeCorreaTeixeira You are welcome! Bye
  – gimusi
  Aug 1 at 20:10
  

  
  
  
  

It should be mandatory that the person who puts a downvote explains their reasons.
– Piquito
Jul 31 at 18:48

It should be mandatory that the person who puts a downvote explains their reasons.
– Piquito
Jul 31 at 18:48

@Piquito It should be very useful indeed also for me in order to learn something more.
– gimusi
Jul 31 at 19:10

@Piquito It should be very useful indeed also for me in order to learn something more.
– gimusi
Jul 31 at 19:10

Apparently, Gimusi, we can "learn" a lot from certain citizens.
– Piquito
Jul 31 at 20:51

Apparently, Gimusi, we can "learn" a lot from certain citizens.
– Piquito
Jul 31 at 20:51

1

1

Gimusi, I asked for the proof without the use of "split-trick". But your notation helped me to think how could it be done. Thank you.
– Guilherme Correa Teixeira
Aug 1 at 20:09

Gimusi, I asked for the proof without the use of "split-trick". But your notation helped me to think how could it be done. Thank you.
– Guilherme Correa Teixeira
Aug 1 at 20:09

@GuilhermeCorreaTeixeira You are welcome! Bye
– gimusi
Aug 1 at 20:10

@GuilhermeCorreaTeixeira You are welcome! Bye
– gimusi
Aug 1 at 20:10

add a comment | 

up vote
0
down vote

HINT.- $ydx=ycdotdfracdxdtcdot dt$ and $dfracdxdt=f'(t)$

share|cite|improve this answer

answered Jul 31 at 19:02

Piquito

17.3k31234

add a comment | 

up vote
0
down vote

HINT.- $ydx=ycdotdfracdxdtcdot dt$ and $dfracdxdt=f'(t)$

share|cite|improve this answer

answered Jul 31 at 19:02

Piquito

17.3k31234

add a comment | 

up vote
0
down vote

up vote
0
down vote

HINT.- $ydx=ycdotdfracdxdtcdot dt$ and $dfracdxdt=f'(t)$

share|cite|improve this answer

answered Jul 31 at 19:02

Piquito

17.3k31234

HINT.- $ydx=ycdotdfracdxdtcdot dt$ and $dfracdxdt=f'(t)$

share|cite|improve this answer

answered Jul 31 at 19:02

Piquito

17.3k31234

share|cite|improve this answer

share|cite|improve this answer

answered Jul 31 at 19:02

Piquito

17.3k31234

answered Jul 31 at 19:02

Piquito

17.3k31234

answered Jul 31 at 19:02

Piquito

17.3k31234

17.3k31234

add a comment | 

add a comment | 

 

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