tangent line to a level curve

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Given the function $f:mathbbR^2tomathbbR$ defined by $$f(x,y)=x^3 + 3x^2y-y^3$$ find the points $(a,b)$ of the plane that satisfy the tangent of the level curve $M=f(a,b)$ in the point $(a,b)$ passes through $(0,1)$.

I tried solving this simply by using the equation of the tangent to a level curve: $$f_x(a,b)(x-a)+f_y(a,b)(y-b)=0$$

Hence, I get $$(3a^2+6ab)(x-a)+(3a^2-3b^2)(y-b) = 0$$

Then should I substitute $x,y$ for $(0,1)$? If so, after working it out, I get $$-a^3 -3a^2b+a^2-b^2+b^3=0$$ But how can I solve the points from here on out?

multivariable-calculus partial-derivative

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edited May 1 '14 at 3:05

symplectomorphic

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asked May 1 '14 at 2:29

user143899

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up vote
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Given the function $f:mathbbR^2tomathbbR$ defined by $$f(x,y)=x^3 + 3x^2y-y^3$$ find the points $(a,b)$ of the plane that satisfy the tangent of the level curve $M=f(a,b)$ in the point $(a,b)$ passes through $(0,1)$.

I tried solving this simply by using the equation of the tangent to a level curve: $$f_x(a,b)(x-a)+f_y(a,b)(y-b)=0$$

Hence, I get $$(3a^2+6ab)(x-a)+(3a^2-3b^2)(y-b) = 0$$

Then should I substitute $x,y$ for $(0,1)$? If so, after working it out, I get $$-a^3 -3a^2b+a^2-b^2+b^3=0$$ But how can I solve the points from here on out?

multivariable-calculus partial-derivative

share|cite|improve this question

edited May 1 '14 at 3:05

symplectomorphic

11.5k21837

asked May 1 '14 at 2:29

user143899

79110

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

Given the function $f:mathbbR^2tomathbbR$ defined by $$f(x,y)=x^3 + 3x^2y-y^3$$ find the points $(a,b)$ of the plane that satisfy the tangent of the level curve $M=f(a,b)$ in the point $(a,b)$ passes through $(0,1)$.

I tried solving this simply by using the equation of the tangent to a level curve: $$f_x(a,b)(x-a)+f_y(a,b)(y-b)=0$$

Hence, I get $$(3a^2+6ab)(x-a)+(3a^2-3b^2)(y-b) = 0$$

Then should I substitute $x,y$ for $(0,1)$? If so, after working it out, I get $$-a^3 -3a^2b+a^2-b^2+b^3=0$$ But how can I solve the points from here on out?

multivariable-calculus partial-derivative

share|cite|improve this question

edited May 1 '14 at 3:05

symplectomorphic

11.5k21837

asked May 1 '14 at 2:29

user143899

79110

Given the function $f:mathbbR^2tomathbbR$ defined by $$f(x,y)=x^3 + 3x^2y-y^3$$ find the points $(a,b)$ of the plane that satisfy the tangent of the level curve $M=f(a,b)$ in the point $(a,b)$ passes through $(0,1)$.

I tried solving this simply by using the equation of the tangent to a level curve: $$f_x(a,b)(x-a)+f_y(a,b)(y-b)=0$$

Hence, I get $$(3a^2+6ab)(x-a)+(3a^2-3b^2)(y-b) = 0$$

Then should I substitute $x,y$ for $(0,1)$? If so, after working it out, I get $$-a^3 -3a^2b+a^2-b^2+b^3=0$$ But how can I solve the points from here on out?

multivariable-calculus partial-derivative

share|cite|improve this question

edited May 1 '14 at 3:05

symplectomorphic

11.5k21837

asked May 1 '14 at 2:29

user143899

79110

share|cite|improve this question

share|cite|improve this question

edited May 1 '14 at 3:05

symplectomorphic

11.5k21837

edited May 1 '14 at 3:05

symplectomorphic

11.5k21837

edited May 1 '14 at 3:05

symplectomorphic

11.5k21837

11.5k21837

asked May 1 '14 at 2:29

user143899

79110

asked May 1 '14 at 2:29

user143899

79110

asked May 1 '14 at 2:29

user143899

79110

79110

add a comment | 

add a comment | 

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You're doing exactly the right thing. So this tells you that a point whose tangent contains $(0,1)$ must satisfy that last equation. It must ALso be a point of the original curve, i.e., satisfy $x^3 - 3x^y - y^3 = M$. Those two equations together may determine just a few points.

share|cite|improve this answer

answered May 1 '14 at 3:49

John Hughes

59.4k23785

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote

You're doing exactly the right thing. So this tells you that a point whose tangent contains $(0,1)$ must satisfy that last equation. It must ALso be a point of the original curve, i.e., satisfy $x^3 - 3x^y - y^3 = M$. Those two equations together may determine just a few points.

share|cite|improve this answer

answered May 1 '14 at 3:49

John Hughes

59.4k23785

add a comment | 

up vote
0
down vote

You're doing exactly the right thing. So this tells you that a point whose tangent contains $(0,1)$ must satisfy that last equation. It must ALso be a point of the original curve, i.e., satisfy $x^3 - 3x^y - y^3 = M$. Those two equations together may determine just a few points.

share|cite|improve this answer

answered May 1 '14 at 3:49

John Hughes

59.4k23785

add a comment | 

up vote
0
down vote

up vote
0
down vote

You're doing exactly the right thing. So this tells you that a point whose tangent contains $(0,1)$ must satisfy that last equation. It must ALso be a point of the original curve, i.e., satisfy $x^3 - 3x^y - y^3 = M$. Those two equations together may determine just a few points.

share|cite|improve this answer

answered May 1 '14 at 3:49

John Hughes

59.4k23785

You're doing exactly the right thing. So this tells you that a point whose tangent contains $(0,1)$ must satisfy that last equation. It must ALso be a point of the original curve, i.e., satisfy $x^3 - 3x^y - y^3 = M$. Those two equations together may determine just a few points.

share|cite|improve this answer

answered May 1 '14 at 3:49

John Hughes

59.4k23785

share|cite|improve this answer

share|cite|improve this answer

answered May 1 '14 at 3:49

John Hughes

59.4k23785

answered May 1 '14 at 3:49

John Hughes

59.4k23785

answered May 1 '14 at 3:49

John Hughes

59.4k23785

59.4k23785

add a comment | 

add a comment | 

 

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