- #1
Benny
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Hi, can someone check/fix my errors in the following question?
Q. Consider the equations
u = x^3 + 2xy + y^2, v = x^2 + y
(i) Show that these equations can be solved locally for x and y as C^1 functions of u and v near the point where x = 1, y = 1, u = 4, v = 2.
(ii) Find [tex]\frac{{\partial x}}{{\partial u}}[/tex] and [tex]\frac{{\partial x}}{{\partial v}}[/tex] at this point.
(i) u = x^3 + 2xy + y^2, v = x^2 + y. I rewrite these equations as
[tex]f\left( {u,v,x,y} \right) = \left( {x^3 + 2xy + y^2 - u,x^2 + y - v} \right) = \left( {0,0} \right)[/tex] where [itex]f:R^4 \to R^2 [/itex] is a C^1 function.
I verified that f(4,2,1,1) = (0,0).
[tex]
\det \left[ {\begin{array}{*{20}c}
{\frac{{\partial f_1 }}{{\partial x}}} & {\frac{{\partial f_1 }}{{\partial y}}} \\
{\frac{{\partial f_2 }}{{\partial x}}} & {\frac{{\partial f_2 }}{{\partial y}}} \\
\end{array}} \right]
[/tex]
[tex]
= \left| {\begin{array}{*{20}c}
{3x^2 + 2y} & {2x + 2y} \\
{2x} & 1 \\
\end{array}} \right|
[/tex]
[tex]
= \left| {\begin{array}{*{20}c}
5 & 4 \\
2 & 1 \\
\end{array}} \right| = - 3 \ne 0
[/tex]
at (u,v,x,y) = (4,2,1,1).
So is that enough (i)?
(ii) I think I use the general formula for the derivative in problems of this type to obtain the derivative of (x,y) = h(u,v) in matrix form and then take the relevant parts.
By (i) there is a function of the form [itex]\left( {x,y} \right) = h\left( {u,v} \right)[/itex] near (u,v,x,y) = (4,2,1,1).
[tex]
h'\left( {u,v} \right) = \left[ {\begin{array}{*{20}c}
{\frac{{\partial f_1 }}{{\partial x}}} & {\frac{{\partial f_1 }}{{\partial y}}} \\
{\frac{{\partial f_2 }}{{\partial x}}} & {\frac{{\partial f_2 }}{{\partial y}}} \\
\end{array}} \right]^{ - 1} \left[ {\begin{array}{*{20}c}
{\frac{{\partial f_1 }}{{\partial u}}} & {\frac{{\partial f_1 }}{{\partial v}}} \\
{\frac{{\partial f_2 }}{{\partial u}}} & {\frac{{\partial f_2 }}{{\partial v}}} \\
\end{array}} \right]
[/tex]
[tex]
= \left[ {\begin{array}{*{20}c}
{\frac{{\partial x}}{{\partial u}}} & {\frac{{\partial x}}{{\partial v}}} \\
{\frac{{\partial y}}{{\partial u}}} & {\frac{{\partial y}}{{\partial v}}} \\
\end{array}} \right]
[/tex] where the last equality follows from the definition of (x,y) = h(u,v).
Do I plug the values of u,v,x,y into the above? I've been having some problems using the implicit function theorem so I don't really know how to do (i) and (ii). Any comments would be good thanks.
Q. Consider the equations
u = x^3 + 2xy + y^2, v = x^2 + y
(i) Show that these equations can be solved locally for x and y as C^1 functions of u and v near the point where x = 1, y = 1, u = 4, v = 2.
(ii) Find [tex]\frac{{\partial x}}{{\partial u}}[/tex] and [tex]\frac{{\partial x}}{{\partial v}}[/tex] at this point.
(i) u = x^3 + 2xy + y^2, v = x^2 + y. I rewrite these equations as
[tex]f\left( {u,v,x,y} \right) = \left( {x^3 + 2xy + y^2 - u,x^2 + y - v} \right) = \left( {0,0} \right)[/tex] where [itex]f:R^4 \to R^2 [/itex] is a C^1 function.
I verified that f(4,2,1,1) = (0,0).
[tex]
\det \left[ {\begin{array}{*{20}c}
{\frac{{\partial f_1 }}{{\partial x}}} & {\frac{{\partial f_1 }}{{\partial y}}} \\
{\frac{{\partial f_2 }}{{\partial x}}} & {\frac{{\partial f_2 }}{{\partial y}}} \\
\end{array}} \right]
[/tex]
[tex]
= \left| {\begin{array}{*{20}c}
{3x^2 + 2y} & {2x + 2y} \\
{2x} & 1 \\
\end{array}} \right|
[/tex]
[tex]
= \left| {\begin{array}{*{20}c}
5 & 4 \\
2 & 1 \\
\end{array}} \right| = - 3 \ne 0
[/tex]
at (u,v,x,y) = (4,2,1,1).
So is that enough (i)?
(ii) I think I use the general formula for the derivative in problems of this type to obtain the derivative of (x,y) = h(u,v) in matrix form and then take the relevant parts.
By (i) there is a function of the form [itex]\left( {x,y} \right) = h\left( {u,v} \right)[/itex] near (u,v,x,y) = (4,2,1,1).
[tex]
h'\left( {u,v} \right) = \left[ {\begin{array}{*{20}c}
{\frac{{\partial f_1 }}{{\partial x}}} & {\frac{{\partial f_1 }}{{\partial y}}} \\
{\frac{{\partial f_2 }}{{\partial x}}} & {\frac{{\partial f_2 }}{{\partial y}}} \\
\end{array}} \right]^{ - 1} \left[ {\begin{array}{*{20}c}
{\frac{{\partial f_1 }}{{\partial u}}} & {\frac{{\partial f_1 }}{{\partial v}}} \\
{\frac{{\partial f_2 }}{{\partial u}}} & {\frac{{\partial f_2 }}{{\partial v}}} \\
\end{array}} \right]
[/tex]
[tex]
= \left[ {\begin{array}{*{20}c}
{\frac{{\partial x}}{{\partial u}}} & {\frac{{\partial x}}{{\partial v}}} \\
{\frac{{\partial y}}{{\partial u}}} & {\frac{{\partial y}}{{\partial v}}} \\
\end{array}} \right]
[/tex] where the last equality follows from the definition of (x,y) = h(u,v).
Do I plug the values of u,v,x,y into the above? I've been having some problems using the implicit function theorem so I don't really know how to do (i) and (ii). Any comments would be good thanks.
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