Mixing
![Probability of attaining same prime factor [on hold]](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhZCTB4-bb-s-32SAm6ytjzFOU06cH9o8q-a_wq_UnH6lcd8hvws23CmBWHM6g256LzTX9yK1EiCCzTC1NSgtxlNk_J9hdr9bA3tD0Nqntbl521j4bLAm_uXPANuWLBhcCKTKzns2gaUodx/s72-c/1.jpg)
Find area given two curves which belong to the orthogonal family to $x^2+2y^2=K$
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
Calculate the region area of the $xy$-plane limited by curves $C_1$ and $C_2$ with $|x|leq4$, knowing that $C_1$ passes through $(1, 1)$, $C_2$ passes through $(1, -1)$ and both curves belong to the orthogonal family a $x^2+2y^2=K$, with $K$ arbitrary constant.
I did this:
$$beginmatrix
mathfrak F:&2x+4yy'=0&(1)\\
mathfrak F^perp:&2x-4dfrac yy'=0&(2)\\
mathfrak F^perp:&x=2dfrac yy'&(3)\\
mathfrak F^perp:&dfracy'y=dfrac2x&(4)\\
mathfrak F^perp:&displaystyleintdfractext dyy=2displaystyleintdfractext dxx&(5)\\
mathfrak F^perp:&ln|y|=ln(x^2)+C&(6)\\
mathfrak F^perp:&y=kx^2.&(7)
endmatrix$$
As we have $y(1)=1$ for $C_1$ and $y(1)=-1$ for $C_2$ then $$beginmatrix
C_1:&y=x^2\
C_2:&y=-x^2,\
endmatrix$$ so the integral for $|x|leq4$ becomes $$textArea=dfrac2563,$$ according to WolframAlpha.
Is this correct? By the way I am not sure if the first steps are correct, because when I derive in the first equation $K$ dissapears (I know this is a constant, but sometimes we need to replace $x^2+2y^2=K$ inside the $mathfrak F$, and this time that did not happen).
Any help?
Thank you!
integration differential-equations area
asked Aug 3 at 1:12
manooooh
415211
 |Â
show 1 more comment
up vote
2
down vote
favorite
Calculate the region area of the $xy$-plane limited by curves $C_1$ and $C_2$ with $|x|leq4$, knowing that $C_1$ passes through $(1, 1)$, $C_2$ passes through $(1, -1)$ and both curves belong to the orthogonal family a $x^2+2y^2=K$, with $K$ arbitrary constant.
I did this:
$$beginmatrix
mathfrak F:&2x+4yy'=0&(1)\\
mathfrak F^perp:&2x-4dfrac yy'=0&(2)\\
mathfrak F^perp:&x=2dfrac yy'&(3)\\
mathfrak F^perp:&dfracy'y=dfrac2x&(4)\\
mathfrak F^perp:&displaystyleintdfractext dyy=2displaystyleintdfractext dxx&(5)\\
mathfrak F^perp:&ln|y|=ln(x^2)+C&(6)\\
mathfrak F^perp:&y=kx^2.&(7)
endmatrix$$
As we have $y(1)=1$ for $C_1$ and $y(1)=-1$ for $C_2$ then $$beginmatrix
C_1:&y=x^2\
C_2:&y=-x^2,\
endmatrix$$ so the integral for $|x|leq4$ becomes $$textArea=dfrac2563,$$ according to WolframAlpha.
Is this correct? By the way I am not sure if the first steps are correct, because when I derive in the first equation $K$ dissapears (I know this is a constant, but sometimes we need to replace $x^2+2y^2=K$ inside the $mathfrak F$, and this time that did not happen).
Any help?
Thank you!
integration differential-equations area
asked Aug 3 at 1:12
manooooh
415211
@amd when I plot $x^2+2y^2=K$ with the condition $y(1)=1,-1$ then $K=3$ so the ellipse becomes $x^2+2y^2=3$, and graphing this curve with $y=x^2$ and $y=-x^2$ I noticed that, in $(1,1)$ and $(1,-1)$, the curves are orthogonal. Please see this image. What am I doing wrong? How would you solve it?
– manooooh
Aug 3 at 2:09
@amd ok. But if a family of curves with $y'$ then for find the orthogonal family we need to replace $y'$ with $-1/y'$, so that is what I did between steps $(1)$ and $(2)$.
– manooooh
Aug 3 at 2:17
1
Sorry. I made a mistake of my own. The orthogonal curves are indeed parabolas. I’ll note that the $y$-axis is also an orthogonal curve to the family of ellipses, but can’t be expressed in the form $y=kx^2$. Consolidating the two constants of integration eliminates a solution to the differential equation, fortunately not an issue for this particular problem.
– amd
Aug 3 at 3:17
@amd no problem, the experts also have mistakes:). So, how is my work?
– manooooh
Aug 3 at 3:24
1
Seems fine to me. Did you really have to turn to W|A to integrate $2x^2$?
– amd
Aug 3 at 3:25
 |Â
show 1 more comment
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Calculate the region area of the $xy$-plane limited by curves $C_1$ and $C_2$ with $|x|leq4$, knowing that $C_1$ passes through $(1, 1)$, $C_2$ passes through $(1, -1)$ and both curves belong to the orthogonal family a $x^2+2y^2=K$, with $K$ arbitrary constant.
I did this:
$$beginmatrix
mathfrak F:&2x+4yy'=0&(1)\\
mathfrak F^perp:&2x-4dfrac yy'=0&(2)\\
mathfrak F^perp:&x=2dfrac yy'&(3)\\
mathfrak F^perp:&dfracy'y=dfrac2x&(4)\\
mathfrak F^perp:&displaystyleintdfractext dyy=2displaystyleintdfractext dxx&(5)\\
mathfrak F^perp:&ln|y|=ln(x^2)+C&(6)\\
mathfrak F^perp:&y=kx^2.&(7)
endmatrix$$
As we have $y(1)=1$ for $C_1$ and $y(1)=-1$ for $C_2$ then $$beginmatrix
C_1:&y=x^2\
C_2:&y=-x^2,\
endmatrix$$ so the integral for $|x|leq4$ becomes $$textArea=dfrac2563,$$ according to WolframAlpha.
Is this correct? By the way I am not sure if the first steps are correct, because when I derive in the first equation $K$ dissapears (I know this is a constant, but sometimes we need to replace $x^2+2y^2=K$ inside the $mathfrak F$, and this time that did not happen).
Any help?
Thank you!
integration differential-equations area
asked Aug 3 at 1:12
manooooh
415211
Calculate the region area of the $xy$-plane limited by curves $C_1$ and $C_2$ with $|x|leq4$, knowing that $C_1$ passes through $(1, 1)$, $C_2$ passes through $(1, -1)$ and both curves belong to the orthogonal family a $x^2+2y^2=K$, with $K$ arbitrary constant.
I did this:
$$beginmatrix
mathfrak F:&2x+4yy'=0&(1)\\
mathfrak F^perp:&2x-4dfrac yy'=0&(2)\\
mathfrak F^perp:&x=2dfrac yy'&(3)\\
mathfrak F^perp:&dfracy'y=dfrac2x&(4)\\
mathfrak F^perp:&displaystyleintdfractext dyy=2displaystyleintdfractext dxx&(5)\\
mathfrak F^perp:&ln|y|=ln(x^2)+C&(6)\\
mathfrak F^perp:&y=kx^2.&(7)
endmatrix$$
As we have $y(1)=1$ for $C_1$ and $y(1)=-1$ for $C_2$ then $$beginmatrix
C_1:&y=x^2\
C_2:&y=-x^2,\
endmatrix$$ so the integral for $|x|leq4$ becomes $$textArea=dfrac2563,$$ according to WolframAlpha.
Is this correct? By the way I am not sure if the first steps are correct, because when I derive in the first equation $K$ dissapears (I know this is a constant, but sometimes we need to replace $x^2+2y^2=K$ inside the $mathfrak F$, and this time that did not happen).
Any help?
Thank you!
integration differential-equations area
asked Aug 3 at 1:12
manooooh
415211
asked Aug 3 at 1:12
manooooh
415211
asked Aug 3 at 1:12
manooooh
415211
asked Aug 3 at 1:12
manooooh
415211
415211
@amd when I plot $x^2+2y^2=K$ with the condition $y(1)=1,-1$ then $K=3$ so the ellipse becomes $x^2+2y^2=3$, and graphing this curve with $y=x^2$ and $y=-x^2$ I noticed that, in $(1,1)$ and $(1,-1)$, the curves are orthogonal. Please see this image. What am I doing wrong? How would you solve it?
– manooooh
Aug 3 at 2:09
@amd ok. But if a family of curves with $y'$ then for find the orthogonal family we need to replace $y'$ with $-1/y'$, so that is what I did between steps $(1)$ and $(2)$.
– manooooh
Aug 3 at 2:17
1
Sorry. I made a mistake of my own. The orthogonal curves are indeed parabolas. I’ll note that the $y$-axis is also an orthogonal curve to the family of ellipses, but can’t be expressed in the form $y=kx^2$. Consolidating the two constants of integration eliminates a solution to the differential equation, fortunately not an issue for this particular problem.
– amd
Aug 3 at 3:17
@amd no problem, the experts also have mistakes:). So, how is my work?
– manooooh
Aug 3 at 3:24
1
Seems fine to me. Did you really have to turn to W|A to integrate $2x^2$?
– amd
Aug 3 at 3:25
 |Â
show 1 more comment
@amd when I plot $x^2+2y^2=K$ with the condition $y(1)=1,-1$ then $K=3$ so the ellipse becomes $x^2+2y^2=3$, and graphing this curve with $y=x^2$ and $y=-x^2$ I noticed that, in $(1,1)$ and $(1,-1)$, the curves are orthogonal. Please see this image. What am I doing wrong? How would you solve it?
– manooooh
Aug 3 at 2:09
@amd ok. But if a family of curves with $y'$ then for find the orthogonal family we need to replace $y'$ with $-1/y'$, so that is what I did between steps $(1)$ and $(2)$.
– manooooh
Aug 3 at 2:17
1
Sorry. I made a mistake of my own. The orthogonal curves are indeed parabolas. I’ll note that the $y$-axis is also an orthogonal curve to the family of ellipses, but can’t be expressed in the form $y=kx^2$. Consolidating the two constants of integration eliminates a solution to the differential equation, fortunately not an issue for this particular problem.
– amd
Aug 3 at 3:17
@amd no problem, the experts also have mistakes:). So, how is my work?
– manooooh
Aug 3 at 3:24
1
Seems fine to me. Did you really have to turn to W|A to integrate $2x^2$?
– amd
Aug 3 at 3:25
@amd when I plot $x^2+2y^2=K$ with the condition $y(1)=1,-1$ then $K=3$ so the ellipse becomes $x^2+2y^2=3$, and graphing this curve with $y=x^2$ and $y=-x^2$ I noticed that, in $(1,1)$ and $(1,-1)$, the curves are orthogonal. Please see this image. What am I doing wrong? How would you solve it?
– manooooh
Aug 3 at 2:09
@amd when I plot $x^2+2y^2=K$ with the condition $y(1)=1,-1$ then $K=3$ so the ellipse becomes $x^2+2y^2=3$, and graphing this curve with $y=x^2$ and $y=-x^2$ I noticed that, in $(1,1)$ and $(1,-1)$, the curves are orthogonal. Please see this image. What am I doing wrong? How would you solve it?
– manooooh
Aug 3 at 2:09
@amd ok. But if a family of curves with $y'$ then for find the orthogonal family we need to replace $y'$ with $-1/y'$, so that is what I did between steps $(1)$ and $(2)$.
– manooooh
Aug 3 at 2:17
@amd ok. But if a family of curves with $y'$ then for find the orthogonal family we need to replace $y'$ with $-1/y'$, so that is what I did between steps $(1)$ and $(2)$.
– manooooh
Aug 3 at 2:17
1
1
Sorry. I made a mistake of my own. The orthogonal curves are indeed parabolas. I’ll note that the $y$-axis is also an orthogonal curve to the family of ellipses, but can’t be expressed in the form $y=kx^2$. Consolidating the two constants of integration eliminates a solution to the differential equation, fortunately not an issue for this particular problem.
– amd
Aug 3 at 3:17
Sorry. I made a mistake of my own. The orthogonal curves are indeed parabolas. I’ll note that the $y$-axis is also an orthogonal curve to the family of ellipses, but can’t be expressed in the form $y=kx^2$. Consolidating the two constants of integration eliminates a solution to the differential equation, fortunately not an issue for this particular problem.
– amd
Aug 3 at 3:17
@amd no problem, the experts also have mistakes
:). So, how is my work?– manooooh
Aug 3 at 3:24
@amd no problem, the experts also have mistakes
:). So, how is my work?– manooooh
Aug 3 at 3:24
1
1
Seems fine to me. Did you really have to turn to W|A to integrate $2x^2$?
– amd
Aug 3 at 3:25
Seems fine to me. Did you really have to turn to W|A to integrate $2x^2$?
– amd
Aug 3 at 3:25
 |Â
show 1 more comment
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2870641%2ffind-area-given-two-curves-which-belong-to-the-orthogonal-family-to-x22y2-k%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
@amd when I plot $x^2+2y^2=K$ with the condition $y(1)=1,-1$ then $K=3$ so the ellipse becomes $x^2+2y^2=3$, and graphing this curve with $y=x^2$ and $y=-x^2$ I noticed that, in $(1,1)$ and $(1,-1)$, the curves are orthogonal. Please see this image. What am I doing wrong? How would you solve it?
– manooooh
Aug 3 at 2:09
@amd ok. But if a family of curves with $y'$ then for find the orthogonal family we need to replace $y'$ with $-1/y'$, so that is what I did between steps $(1)$ and $(2)$.
– manooooh
Aug 3 at 2:17
1
Sorry. I made a mistake of my own. The orthogonal curves are indeed parabolas. I’ll note that the $y$-axis is also an orthogonal curve to the family of ellipses, but can’t be expressed in the form $y=kx^2$. Consolidating the two constants of integration eliminates a solution to the differential equation, fortunately not an issue for this particular problem.
– amd
Aug 3 at 3:17
@amd no problem, the experts also have mistakes
:). So, how is my work?– manooooh
Aug 3 at 3:24
1
Seems fine to me. Did you really have to turn to W|A to integrate $2x^2$?
– amd
Aug 3 at 3:25