Solving Equations: Check/Fix Errors & Find Partial Derivatives

In summary, the conversation was about solving a system of equations for x and y as C^1 functions of u and v near a specific point. The method used was the implicit function theorem, and the values of the partial derivatives were calculated using the Jacobian matrix. The conversation also included a discussion about using Mathematica to verify the results.
  • #1
Benny
584
0
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.
 
Last edited:
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  • #2
(i) looks fine. For (ii), look at the proof you're given for the implicit function theorem. There must be some sort of construction for h. Use this construction to find the partials. As it stands, I don't know where you're getting the first equality from.
 
  • #3
I'll do a simple one first:

[tex]F:\mathbb{R}^2\rightarrow\mathbb{R}^2,\quad F(x,y)=(2x+y,x-y)=(u,v)[/tex]

and apply the Implicit Function Theorem near the point (x,y)=(1,2).

The Jacobian of F is:

[tex]\mathcal{J}\left\{F\right\}=

\left|\begin{array}{cc}2 & 1 \\
1 & -1 \end{array}\right|\neq 0[/tex]

Therefore by the Inverse Function Theorem:

[tex]F^{-1}:\mathbb{R}^2\rightarrow\mathbb{R}^2,\quad F^{-1}(u,v)=(g_1(u,v),g_2(u,v))=(x,y)[/tex]

near the point (1,2). Thus:

[tex]\left(F^{-1}\right)^{'}(4,-1)=\left[F^{'}\right]^{-1}(1,2)[/tex]

or:

[tex]\left(F^{-1}\right)^{'}(4,-1)=\left[\begin{array}{cc}1/3 & 1/3 \\
1/3 & -2/3 \end{array}\right][/tex]

That is:

[tex]x_u=1/3[/tex]

[tex]x_v=1/3[/tex]

[tex]y_u=1/3[/tex]

[tex]y_v=-2/3[/tex]
 
  • #4
Thanks for the help guys.

I appreciate the help and I hope I don't sound arrogant but I don't think my set up is incorrect at this stage. Although I did leave out a negative sign in the first equality (the part with the product of two matrices). I used the same method on Saltydog's example (this time with the negative sign included as it should be according to my notes) and I obtained the same answer as him. I'll go over the theorem again to see if I've missed something.
 
  • #5
Benny said:
Thanks for the help guys.

I appreciate the help and I hope I don't sound arrogant but I don't think my set up is incorrect at this stage. Although I did leave out a negative sign in the first equality (the part with the product of two matrices). I used the same method on Saltydog's example (this time with the negative sign included as it should be according to my notes) and I obtained the same answer as him. I'll go over the theorem again to see if I've missed something.

Good for you Benny. Didn't wish to imply you were doing anything wrong. Just wanted to work a simple problem. Now, can you verify the values of the derivatives for you problem directly without relying on the Implicit Function Theorem? What for? Whatever.

Oh yea, it could happen. Got Mathematica? Here's part of the code:

Code:
[tex]\text{array}=\left\{x,y\right\}/.Solve[\left\{u==x^3+2xy+y^2,v==
x^2+y\right\},\left\{x,y\right\}][/tex]

Note that quintics are involved so necessarilly Mathematica returns 4 sets of (very messy) solutions, 3 of which are extraneous. Find the good one, calculate the partials and verify:

[tex]x_u(4,2)=-1/3\quad x_v(4,2)=4/3[/tex]

[tex]y_u(4,2)=2/3\quad y_v(4,2)=-5/3[/tex]
 
Last edited:
  • #6
Thanks for assisting me further Saltydog. When I get around to finishing off that question I'll see if I obtain the same values as the ones you have posted.
 

1. What are the common errors when solving equations?

Some common errors when solving equations include forgetting to distribute terms when simplifying, making sign errors when combining like terms, and forgetting to apply the order of operations.

2. How can I check if my solution to an equation is correct?

You can check your solution by plugging it back into the original equation and seeing if it satisfies the equation. You can also use a graphing calculator to graph both sides of the equation and see if they intersect at the same point.

3. Why is it important to find and fix errors when solving equations?

Finding and fixing errors when solving equations is important because even small mistakes can completely change the solution to the equation. It ensures that the solution is accurate and reliable.

4. What are partial derivatives and how do you find them?

Partial derivatives are a way of finding the rate of change of a function with respect to one of its variables, while holding all other variables constant. To find a partial derivative, you take the derivative of the function with respect to the variable you are interested in, treating all other variables as constants.

5. How do partial derivatives relate to solving equations?

Partial derivatives can be used to solve equations involving multiple variables. By finding the partial derivatives of each variable and setting them equal to each other, you can find the values of the variables that satisfy the equation.

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