tangent line to a level curve
$begingroup$
Given the function $f:mathbb{R}^2tomathbb{R}$ 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
edited May 1 '14 at 3:05
symplectomorphic
12.3k22039
asked May 1 '14 at 2:29
user143899user143899
79210
$endgroup$
add a comment |
$begingroup$
Given the function $f:mathbb{R}^2tomathbb{R}$ 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
edited May 1 '14 at 3:05
symplectomorphic
12.3k22039
asked May 1 '14 at 2:29
user143899user143899
79210
$endgroup$
add a comment |
$begingroup$
Given the function $f:mathbb{R}^2tomathbb{R}$ 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
edited May 1 '14 at 3:05
symplectomorphic
12.3k22039
asked May 1 '14 at 2:29
user143899user143899
79210
$endgroup$
Given the function $f:mathbb{R}^2tomathbb{R}$ 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
multivariable-calculus partial-derivative
edited May 1 '14 at 3:05
symplectomorphic
12.3k22039
asked May 1 '14 at 2:29
user143899user143899
79210
edited May 1 '14 at 3:05
symplectomorphic
12.3k22039
asked May 1 '14 at 2:29
user143899user143899
79210
edited May 1 '14 at 3:05
symplectomorphic
12.3k22039
edited May 1 '14 at 3:05
symplectomorphic
12.3k22039
edited May 1 '14 at 3:05
symplectomorphic
12.3k22039
12.3k22039
asked May 1 '14 at 2:29
user143899user143899
79210
asked May 1 '14 at 2:29
user143899user143899
79210
asked May 1 '14 at 2:29
user143899user143899
79210
79210
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
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 $x^3 + 3x^2y - y^3 = M$. Those two equations together may determine just a few points.
edited Dec 20 '18 at 23:49
Tengu
2,63211021
answered May 1 '14 at 3:49
John HughesJohn Hughes
63.3k24090
$endgroup$
add a comment |
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1 Answer
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active
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votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $x^3 + 3x^2y - y^3 = M$. Those two equations together may determine just a few points.
edited Dec 20 '18 at 23:49
Tengu
2,63211021
answered May 1 '14 at 3:49
John HughesJohn Hughes
63.3k24090
$endgroup$
add a comment |
$begingroup$
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 $x^3 + 3x^2y - y^3 = M$. Those two equations together may determine just a few points.
edited Dec 20 '18 at 23:49
Tengu
2,63211021
answered May 1 '14 at 3:49
John HughesJohn Hughes
63.3k24090
$endgroup$
add a comment |
$begingroup$
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 $x^3 + 3x^2y - y^3 = M$. Those two equations together may determine just a few points.
edited Dec 20 '18 at 23:49
Tengu
2,63211021
answered May 1 '14 at 3:49
John HughesJohn Hughes
63.3k24090
$endgroup$
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 $x^3 + 3x^2y - y^3 = M$. Those two equations together may determine just a few points.
edited Dec 20 '18 at 23:49
Tengu
2,63211021
answered May 1 '14 at 3:49
John HughesJohn Hughes
63.3k24090
edited Dec 20 '18 at 23:49
Tengu
2,63211021
edited Dec 20 '18 at 23:49
Tengu
2,63211021
edited Dec 20 '18 at 23:49
Tengu
2,63211021
2,63211021
answered May 1 '14 at 3:49
John HughesJohn Hughes
63.3k24090
answered May 1 '14 at 3:49
John HughesJohn Hughes
63.3k24090
answered May 1 '14 at 3:49
John HughesJohn Hughes
63.3k24090
63.3k24090
add a comment |
add a comment |
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