Transform $x^3+y^3-3xy=0$
I was given the following task:
Calculate the area B, described by the curve: $$tag{x,y > 0}x^3+y^3-3xy=0$$
There are many ways to do this but my university told me to transform the curve into $$r(t); r=sqrt{x^2+y^2}; t = frac{y}{x}$$
I've tried many things but did not get a closed form of $r(t)$:
$$x^3+y^3-3xy=0$$
$$x^3+y^3=3xy$$
$$x^2cdot x+y^2cdot y=3xy$$
$$x^2+y^2cdot t=3y$$
$$dfrac{y^2cdot t}{y^2cdot t}x^2+y^2cdot t=3y$$
$$...$$
This keeps on going but I did not find a good result. I am very happy if someone could help me with transforming the curve into $r(t)$
real-analysis
asked Nov 25 at 20:55
Finn Eggers
364213
add a comment |
I was given the following task:
Calculate the area B, described by the curve: $$tag{x,y > 0}x^3+y^3-3xy=0$$
There are many ways to do this but my university told me to transform the curve into $$r(t); r=sqrt{x^2+y^2}; t = frac{y}{x}$$
I've tried many things but did not get a closed form of $r(t)$:
$$x^3+y^3-3xy=0$$
$$x^3+y^3=3xy$$
$$x^2cdot x+y^2cdot y=3xy$$
$$x^2+y^2cdot t=3y$$
$$dfrac{y^2cdot t}{y^2cdot t}x^2+y^2cdot t=3y$$
$$...$$
This keeps on going but I did not find a good result. I am very happy if someone could help me with transforming the curve into $r(t)$
real-analysis
asked Nov 25 at 20:55
Finn Eggers
364213
add a comment |
I was given the following task:
Calculate the area B, described by the curve: $$tag{x,y > 0}x^3+y^3-3xy=0$$
There are many ways to do this but my university told me to transform the curve into $$r(t); r=sqrt{x^2+y^2}; t = frac{y}{x}$$
I've tried many things but did not get a closed form of $r(t)$:
$$x^3+y^3-3xy=0$$
$$x^3+y^3=3xy$$
$$x^2cdot x+y^2cdot y=3xy$$
$$x^2+y^2cdot t=3y$$
$$dfrac{y^2cdot t}{y^2cdot t}x^2+y^2cdot t=3y$$
$$...$$
This keeps on going but I did not find a good result. I am very happy if someone could help me with transforming the curve into $r(t)$
real-analysis
asked Nov 25 at 20:55
Finn Eggers
364213
I was given the following task:
Calculate the area B, described by the curve: $$tag{x,y > 0}x^3+y^3-3xy=0$$
There are many ways to do this but my university told me to transform the curve into $$r(t); r=sqrt{x^2+y^2}; t = frac{y}{x}$$
I've tried many things but did not get a closed form of $r(t)$:
$$x^3+y^3-3xy=0$$
$$x^3+y^3=3xy$$
$$x^2cdot x+y^2cdot y=3xy$$
$$x^2+y^2cdot t=3y$$
$$dfrac{y^2cdot t}{y^2cdot t}x^2+y^2cdot t=3y$$
$$...$$
This keeps on going but I did not find a good result. I am very happy if someone could help me with transforming the curve into $r(t)$
real-analysis
real-analysis
asked Nov 25 at 20:55
Finn Eggers
364213
asked Nov 25 at 20:55
Finn Eggers
364213
asked Nov 25 at 20:55
Finn Eggers
364213
asked Nov 25 at 20:55
Finn Eggers
364213
asked Nov 25 at 20:55
Finn Eggers
364213
364213
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
Tip:
Replace $y$ with $tx$. You obtain
$$x^3(1+t^3)-3x^2t=0iff begin{cases}x=0qquadtext{ or } \x(1+t^3)=3t
end{cases}$$
Can you proceed?
answered Nov 25 at 21:07
Bernard
118k638111
I didn't think of that. Thank you! I will try to proceed
– Finn Eggers
Nov 25 at 21:09
add a comment |
I think you need to square the equation at some stage, because $r$ is expressed in terms of even powers of $x$ and $y$ and $t$ is homogeneous, while the original equation is not, and includes odd powers.
So I would go $$y=tx$$
$$r^2=x^2+y^2=x^2(1+t^2)$$
Substitute the first of these in the original equation $$(1+t^3)x^3=3tx^2$$
Now, either $x=0$ or $(1+t^3)x=3t$. Now square this to eliminate $x$ and get an expression in terms of $r^2$ and then you can take the square root confident that $r$ is positive.
answered Nov 25 at 21:07
Mark Bennet
80.1k981179
I am stuck at "square this to eliminate x". Could you clarify?
– Finn Eggers
Nov 25 at 21:18
@FinnEggers well you get $x^2$ in terms of $t$ and $r$ from the second equation - or you could just take the square root of the second equation to get $x$ directly.
– Mark Bennet
Nov 25 at 21:22
add a comment |
The transformation is close to polar coordinates, with $t=tantheta$.
Hence plugging in the given equation,
$$r^3(cos^3theta+sin^3theta)-3r^2costhetasintheta=0$$
or
$$r=frac{3costhetasintheta}{cos^3theta+sin^3theta}=frac{3dfrac1{sqrt{t^2+1}}dfrac t{sqrt{t^2+1}}}{dfrac1{(sqrt{t^2+1})^3}+dfrac{t^3}{(sqrt{t^2+1})^3}}=3tfrac{sqrt{t^2+1}}{t^3+1}.$$
answered Nov 25 at 21:16
Yves Daoust
124k671221
add a comment |
Polar coordinates with $$x=rcos theta, y=rsin theta , t=tan theta$$
Results in $$ r(1+t^3)-3tsqrt {1+t^2}=0$$
You can solve for $r$ as a function of $t$
answered Nov 25 at 21:09
Mohammad Riazi-Kermani
40.6k42058
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
Tip:
Replace $y$ with $tx$. You obtain
$$x^3(1+t^3)-3x^2t=0iff begin{cases}x=0qquadtext{ or } \x(1+t^3)=3t
end{cases}$$
Can you proceed?
answered Nov 25 at 21:07
Bernard
118k638111
I didn't think of that. Thank you! I will try to proceed
– Finn Eggers
Nov 25 at 21:09
add a comment |
Tip:
Replace $y$ with $tx$. You obtain
$$x^3(1+t^3)-3x^2t=0iff begin{cases}x=0qquadtext{ or } \x(1+t^3)=3t
end{cases}$$
Can you proceed?
answered Nov 25 at 21:07
Bernard
118k638111
I didn't think of that. Thank you! I will try to proceed
– Finn Eggers
Nov 25 at 21:09
add a comment |
Tip:
Replace $y$ with $tx$. You obtain
$$x^3(1+t^3)-3x^2t=0iff begin{cases}x=0qquadtext{ or } \x(1+t^3)=3t
end{cases}$$
Can you proceed?
answered Nov 25 at 21:07
Bernard
118k638111
Tip:
Replace $y$ with $tx$. You obtain
$$x^3(1+t^3)-3x^2t=0iff begin{cases}x=0qquadtext{ or } \x(1+t^3)=3t
end{cases}$$
Can you proceed?
answered Nov 25 at 21:07
Bernard
118k638111
answered Nov 25 at 21:07
Bernard
118k638111
answered Nov 25 at 21:07
Bernard
118k638111
answered Nov 25 at 21:07
Bernard
118k638111
118k638111
I didn't think of that. Thank you! I will try to proceed
– Finn Eggers
Nov 25 at 21:09
add a comment |
I didn't think of that. Thank you! I will try to proceed
– Finn Eggers
Nov 25 at 21:09
I didn't think of that. Thank you! I will try to proceed
– Finn Eggers
Nov 25 at 21:09
I didn't think of that. Thank you! I will try to proceed
– Finn Eggers
Nov 25 at 21:09
add a comment |
I think you need to square the equation at some stage, because $r$ is expressed in terms of even powers of $x$ and $y$ and $t$ is homogeneous, while the original equation is not, and includes odd powers.
So I would go $$y=tx$$
$$r^2=x^2+y^2=x^2(1+t^2)$$
Substitute the first of these in the original equation $$(1+t^3)x^3=3tx^2$$
Now, either $x=0$ or $(1+t^3)x=3t$. Now square this to eliminate $x$ and get an expression in terms of $r^2$ and then you can take the square root confident that $r$ is positive.
answered Nov 25 at 21:07
Mark Bennet
80.1k981179
I am stuck at "square this to eliminate x". Could you clarify?
– Finn Eggers
Nov 25 at 21:18
@FinnEggers well you get $x^2$ in terms of $t$ and $r$ from the second equation - or you could just take the square root of the second equation to get $x$ directly.
– Mark Bennet
Nov 25 at 21:22
add a comment |
I think you need to square the equation at some stage, because $r$ is expressed in terms of even powers of $x$ and $y$ and $t$ is homogeneous, while the original equation is not, and includes odd powers.
So I would go $$y=tx$$
$$r^2=x^2+y^2=x^2(1+t^2)$$
Substitute the first of these in the original equation $$(1+t^3)x^3=3tx^2$$
Now, either $x=0$ or $(1+t^3)x=3t$. Now square this to eliminate $x$ and get an expression in terms of $r^2$ and then you can take the square root confident that $r$ is positive.
answered Nov 25 at 21:07
Mark Bennet
80.1k981179
I am stuck at "square this to eliminate x". Could you clarify?
– Finn Eggers
Nov 25 at 21:18
@FinnEggers well you get $x^2$ in terms of $t$ and $r$ from the second equation - or you could just take the square root of the second equation to get $x$ directly.
– Mark Bennet
Nov 25 at 21:22
add a comment |
I think you need to square the equation at some stage, because $r$ is expressed in terms of even powers of $x$ and $y$ and $t$ is homogeneous, while the original equation is not, and includes odd powers.
So I would go $$y=tx$$
$$r^2=x^2+y^2=x^2(1+t^2)$$
Substitute the first of these in the original equation $$(1+t^3)x^3=3tx^2$$
Now, either $x=0$ or $(1+t^3)x=3t$. Now square this to eliminate $x$ and get an expression in terms of $r^2$ and then you can take the square root confident that $r$ is positive.
answered Nov 25 at 21:07
Mark Bennet
80.1k981179
I think you need to square the equation at some stage, because $r$ is expressed in terms of even powers of $x$ and $y$ and $t$ is homogeneous, while the original equation is not, and includes odd powers.
So I would go $$y=tx$$
$$r^2=x^2+y^2=x^2(1+t^2)$$
Substitute the first of these in the original equation $$(1+t^3)x^3=3tx^2$$
Now, either $x=0$ or $(1+t^3)x=3t$. Now square this to eliminate $x$ and get an expression in terms of $r^2$ and then you can take the square root confident that $r$ is positive.
answered Nov 25 at 21:07
Mark Bennet
80.1k981179
answered Nov 25 at 21:07
Mark Bennet
80.1k981179
answered Nov 25 at 21:07
Mark Bennet
80.1k981179
answered Nov 25 at 21:07
Mark Bennet
80.1k981179
80.1k981179
I am stuck at "square this to eliminate x". Could you clarify?
– Finn Eggers
Nov 25 at 21:18
@FinnEggers well you get $x^2$ in terms of $t$ and $r$ from the second equation - or you could just take the square root of the second equation to get $x$ directly.
– Mark Bennet
Nov 25 at 21:22
add a comment |
I am stuck at "square this to eliminate x". Could you clarify?
– Finn Eggers
Nov 25 at 21:18
@FinnEggers well you get $x^2$ in terms of $t$ and $r$ from the second equation - or you could just take the square root of the second equation to get $x$ directly.
– Mark Bennet
Nov 25 at 21:22
I am stuck at "square this to eliminate x". Could you clarify?
– Finn Eggers
Nov 25 at 21:18
I am stuck at "square this to eliminate x". Could you clarify?
– Finn Eggers
Nov 25 at 21:18
@FinnEggers well you get $x^2$ in terms of $t$ and $r$ from the second equation - or you could just take the square root of the second equation to get $x$ directly.
– Mark Bennet
Nov 25 at 21:22
@FinnEggers well you get $x^2$ in terms of $t$ and $r$ from the second equation - or you could just take the square root of the second equation to get $x$ directly.
– Mark Bennet
Nov 25 at 21:22
add a comment |
The transformation is close to polar coordinates, with $t=tantheta$.
Hence plugging in the given equation,
$$r^3(cos^3theta+sin^3theta)-3r^2costhetasintheta=0$$
or
$$r=frac{3costhetasintheta}{cos^3theta+sin^3theta}=frac{3dfrac1{sqrt{t^2+1}}dfrac t{sqrt{t^2+1}}}{dfrac1{(sqrt{t^2+1})^3}+dfrac{t^3}{(sqrt{t^2+1})^3}}=3tfrac{sqrt{t^2+1}}{t^3+1}.$$
answered Nov 25 at 21:16
Yves Daoust
124k671221
add a comment |
The transformation is close to polar coordinates, with $t=tantheta$.
Hence plugging in the given equation,
$$r^3(cos^3theta+sin^3theta)-3r^2costhetasintheta=0$$
or
$$r=frac{3costhetasintheta}{cos^3theta+sin^3theta}=frac{3dfrac1{sqrt{t^2+1}}dfrac t{sqrt{t^2+1}}}{dfrac1{(sqrt{t^2+1})^3}+dfrac{t^3}{(sqrt{t^2+1})^3}}=3tfrac{sqrt{t^2+1}}{t^3+1}.$$
answered Nov 25 at 21:16
Yves Daoust
124k671221
add a comment |
The transformation is close to polar coordinates, with $t=tantheta$.
Hence plugging in the given equation,
$$r^3(cos^3theta+sin^3theta)-3r^2costhetasintheta=0$$
or
$$r=frac{3costhetasintheta}{cos^3theta+sin^3theta}=frac{3dfrac1{sqrt{t^2+1}}dfrac t{sqrt{t^2+1}}}{dfrac1{(sqrt{t^2+1})^3}+dfrac{t^3}{(sqrt{t^2+1})^3}}=3tfrac{sqrt{t^2+1}}{t^3+1}.$$
answered Nov 25 at 21:16
Yves Daoust
124k671221
The transformation is close to polar coordinates, with $t=tantheta$.
Hence plugging in the given equation,
$$r^3(cos^3theta+sin^3theta)-3r^2costhetasintheta=0$$
or
$$r=frac{3costhetasintheta}{cos^3theta+sin^3theta}=frac{3dfrac1{sqrt{t^2+1}}dfrac t{sqrt{t^2+1}}}{dfrac1{(sqrt{t^2+1})^3}+dfrac{t^3}{(sqrt{t^2+1})^3}}=3tfrac{sqrt{t^2+1}}{t^3+1}.$$
answered Nov 25 at 21:16
Yves Daoust
124k671221
edited Nov 25 at 22:35
answered Nov 25 at 21:16
Yves Daoust
124k671221
answered Nov 25 at 21:16
Yves Daoust
124k671221
answered Nov 25 at 21:16
Yves Daoust
124k671221
124k671221
add a comment |
add a comment |
Polar coordinates with $$x=rcos theta, y=rsin theta , t=tan theta$$
Results in $$ r(1+t^3)-3tsqrt {1+t^2}=0$$
You can solve for $r$ as a function of $t$
answered Nov 25 at 21:09
Mohammad Riazi-Kermani
40.6k42058
add a comment |
Polar coordinates with $$x=rcos theta, y=rsin theta , t=tan theta$$
Results in $$ r(1+t^3)-3tsqrt {1+t^2}=0$$
You can solve for $r$ as a function of $t$
answered Nov 25 at 21:09
Mohammad Riazi-Kermani
40.6k42058
add a comment |
Polar coordinates with $$x=rcos theta, y=rsin theta , t=tan theta$$
Results in $$ r(1+t^3)-3tsqrt {1+t^2}=0$$
You can solve for $r$ as a function of $t$
answered Nov 25 at 21:09
Mohammad Riazi-Kermani
40.6k42058
Polar coordinates with $$x=rcos theta, y=rsin theta , t=tan theta$$
Results in $$ r(1+t^3)-3tsqrt {1+t^2}=0$$
You can solve for $r$ as a function of $t$
answered Nov 25 at 21:09
Mohammad Riazi-Kermani
40.6k42058
answered Nov 25 at 21:09
Mohammad Riazi-Kermani
40.6k42058
answered Nov 25 at 21:09
Mohammad Riazi-Kermani
40.6k42058
answered Nov 25 at 21:09
Mohammad Riazi-Kermani
40.6k42058
40.6k42058
add a comment |
add a comment |
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