Parametrisation of a curve(intersection of a circular cone and a plane)

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I am trying to find a parametrisation of the intersection of the graphs of the functions: $f(x, y) = sqrtx^2+y^2$ and $g(x, y) = 20 + x − y$. I used a graphing tool, which gave me the following result:
I tried $x=cost, y=sint, z=20+cost-sint$ but that does not work.
I would appreciate any help.

parametrization

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edited Jul 16 at 1:02

asked Jul 16 at 0:54

Relax295

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I am trying to find a parametrisation of the intersection of the graphs of the functions: $f(x, y) = sqrtx^2+y^2$ and $g(x, y) = 20 + x − y$. I used a graphing tool, which gave me the following result:
I tried $x=cost, y=sint, z=20+cost-sint$ but that does not work.
I would appreciate any help.

parametrization

share|cite|improve this question

edited Jul 16 at 1:02

asked Jul 16 at 0:54

Relax295

849

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

I am trying to find a parametrisation of the intersection of the graphs of the functions: $f(x, y) = sqrtx^2+y^2$ and $g(x, y) = 20 + x − y$. I used a graphing tool, which gave me the following result:
I tried $x=cost, y=sint, z=20+cost-sint$ but that does not work.
I would appreciate any help.

parametrization

share|cite|improve this question

edited Jul 16 at 1:02

asked Jul 16 at 0:54

Relax295

849

I am trying to find a parametrisation of the intersection of the graphs of the functions: $f(x, y) = sqrtx^2+y^2$ and $g(x, y) = 20 + x − y$. I used a graphing tool, which gave me the following result:
I tried $x=cost, y=sint, z=20+cost-sint$ but that does not work.
I would appreciate any help.

parametrization

share|cite|improve this question

edited Jul 16 at 1:02

asked Jul 16 at 0:54

Relax295

849

share|cite|improve this question

share|cite|improve this question

edited Jul 16 at 1:02

edited Jul 16 at 1:02

edited Jul 16 at 1:02

asked Jul 16 at 0:54

Relax295

849

asked Jul 16 at 0:54

Relax295

849

asked Jul 16 at 0:54

Relax295

849

849

add a comment | 

add a comment | 

1 Answer
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$$f(x, y) = sqrtx^2+y^2$$ and $$g(x, y) = 20 + x − y$$ intersect where $$ sqrtx^2+y^2=20 + x − y$$

Square both sides to get $$x^2+y^2=400 +x^2 + y^2 +40x-40y -2xy$$

$$40x-40y -2xy=-400$$

Solve for $y$ to get $$y=frac 40x+40040+2x$$

Parametrize: $$ x=t$$

$$y=frac 40t+40040+2t$$

$$z=sqrt t^2+(frac 40t+40040+2t)^2 $$

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answered Jul 16 at 1:28

Mohammad Riazi-Kermani

27.6k41852

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1 Answer
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active

oldest

votes

1 Answer
1

active

oldest

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active

oldest

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active

oldest

votes

up vote
1
down vote



accepted

$$f(x, y) = sqrtx^2+y^2$$ and $$g(x, y) = 20 + x − y$$ intersect where $$ sqrtx^2+y^2=20 + x − y$$

Square both sides to get $$x^2+y^2=400 +x^2 + y^2 +40x-40y -2xy$$

$$40x-40y -2xy=-400$$

Solve for $y$ to get $$y=frac 40x+40040+2x$$

Parametrize: $$ x=t$$

$$y=frac 40t+40040+2t$$

$$z=sqrt t^2+(frac 40t+40040+2t)^2 $$

share|cite|improve this answer

answered Jul 16 at 1:28

Mohammad Riazi-Kermani

27.6k41852

add a comment | 

up vote
1
down vote



accepted

$$f(x, y) = sqrtx^2+y^2$$ and $$g(x, y) = 20 + x − y$$ intersect where $$ sqrtx^2+y^2=20 + x − y$$

Square both sides to get $$x^2+y^2=400 +x^2 + y^2 +40x-40y -2xy$$

$$40x-40y -2xy=-400$$

Solve for $y$ to get $$y=frac 40x+40040+2x$$

Parametrize: $$ x=t$$

$$y=frac 40t+40040+2t$$

$$z=sqrt t^2+(frac 40t+40040+2t)^2 $$

share|cite|improve this answer

answered Jul 16 at 1:28

Mohammad Riazi-Kermani

27.6k41852

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

$$f(x, y) = sqrtx^2+y^2$$ and $$g(x, y) = 20 + x − y$$ intersect where $$ sqrtx^2+y^2=20 + x − y$$

Square both sides to get $$x^2+y^2=400 +x^2 + y^2 +40x-40y -2xy$$

$$40x-40y -2xy=-400$$

Solve for $y$ to get $$y=frac 40x+40040+2x$$

Parametrize: $$ x=t$$

$$y=frac 40t+40040+2t$$

$$z=sqrt t^2+(frac 40t+40040+2t)^2 $$

share|cite|improve this answer

answered Jul 16 at 1:28

Mohammad Riazi-Kermani

27.6k41852

$$f(x, y) = sqrtx^2+y^2$$ and $$g(x, y) = 20 + x − y$$ intersect where $$ sqrtx^2+y^2=20 + x − y$$

Square both sides to get $$x^2+y^2=400 +x^2 + y^2 +40x-40y -2xy$$

$$40x-40y -2xy=-400$$

Solve for $y$ to get $$y=frac 40x+40040+2x$$

Parametrize: $$ x=t$$

$$y=frac 40t+40040+2t$$

$$z=sqrt t^2+(frac 40t+40040+2t)^2 $$

share|cite|improve this answer

answered Jul 16 at 1:28

Mohammad Riazi-Kermani

27.6k41852

share|cite|improve this answer

share|cite|improve this answer

answered Jul 16 at 1:28

Mohammad Riazi-Kermani

27.6k41852

answered Jul 16 at 1:28

Mohammad Riazi-Kermani

27.6k41852

answered Jul 16 at 1:28

Mohammad Riazi-Kermani

27.6k41852

27.6k41852

add a comment | 

add a comment | 

 

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