Question
Consider the system of two equations with five variables
x
,
y
,
z
,
u
,
v
x,y,z,u,v:
x
+
2
u
z
+
y
+
v
z
3
=
4
x+2uz+y+vz
3
=4 and
3
x
+
u
v
−
z
=
2
3x+uv−z=2.
Note that the point
P
P with coordinates
(
x
,
y
,
z
,
u
,
v
)
=
(
1
,
2
,
1
,
0
,
1
)
(x,y,z,u,v)=(1,2,1,0,1) satisfies both equations.
Show that the system can be used to define
z
z and
u
u implicitly as functions of the other three variables
x
x,
y
y and
v
v near the point
P
P.
Find
(
∂
z
∂
x
)
y
,
v
(
∂x
∂z
)
y,v
at the point
(
x
,
y
,
v
)
=
(
1
,
2
,
1
)
(x,y,v)=(1,2,1).
x
,
y
,
z
,
u
,
v
x,y,z,u,v:
x
+
2
u
z
+
y
+
v
z
3
=
4
x+2uz+y+vz
3
=4 and
3
x
+
u
v
−
z
=
2
3x+uv−z=2.
Note that the point
P
P with coordinates
(
x
,
y
,
z
,
u
,
v
)
=
(
1
,
2
,
1
,
0
,
1
)
(x,y,z,u,v)=(1,2,1,0,1) satisfies both equations.
Show that the system can be used to define
z
z and
u
u implicitly as functions of the other three variables
x
x,
y
y and
v
v near the point
P
P.
Find
(
∂
z
∂
x
)
y
,
v
(
∂x
∂z
)
y,v
at the point
(
x
,
y
,
v
)
=
(
1
,
2
,
1
)
(x,y,v)=(1,2,1).
Answers
GPT-4o mini
To show that the system of equations can be used to define \( z \) and \( u \) implicitly as functions of \( x \), \( y \), and \( v \) near the point \( P = (1, 2, 1, 0, 1) \), we need to verify that the implicit function theorem applies.
First, we observe the system of equations:
1. \( f_1(x, y, z, u, v) = x + 2uz + y + vz^3 - 4 \)
2. \( f_2(x, y, z, u, v) = 3x + uv - z - 2 \)
We can express these equations as
\[
\begin{align*}
f_1(x, y, z, u, v) &= 0, \\
f_2(x, y, z, u, v) &= 0.
\end{align*}
\]
### Step 1: Compute the Jacobian Matrix
Next, we will compute the Jacobian matrix of the system with respect to the variables \( z \) and \( u \):
\[
J = \begin{bmatrix}
\frac{\partial f_1}{\partial z} & \frac{\partial f_1}{\partial u} \\
\frac{\partial f_2}{\partial z} & \frac{\partial f_2}{\partial u}
\end{bmatrix}
\]
Let's compute the necessary derivatives at the point \( P = (1, 2, 1, 0, 1) \).
**Derivatives of \( f_1 \)**:
\[
\frac{\partial f_1}{\partial z} = 2u + 3vz^2
\]
At \( P \):
\[
\frac{\partial f_1}{\partial z}(1, 2, 1, 0, 1) = 2(0) + 3(1)(1)^2 = 3
\]
\[
\frac{\partial f_1}{\partial u} = 2z
\]
At \( P \):
\[
\frac{\partial f_1}{\partial u}(1, 2, 1, 0, 1) = 2(1) = 2
\]
**Derivatives of \( f_2 \)**:
\[
\frac{\partial f_2}{\partial z} = -1
\]
At \( P \):
\[
\frac{\partial f_2}{\partial z}(1, 2, 1, 0, 1) = -1
\]
\[
\frac{\partial f_2}{\partial u} = v
\]
At \( P \):
\[
\frac{\partial f_2}{\partial u}(1, 2, 1, 0, 1) = 1
\]
### Step 2: Evaluate the Jacobian \( J \)
Now, we can write the Jacobian matrix evaluated at \( P \):
\[
J(P) = \begin{bmatrix}
3 & 2 \\
-1 & 1
\end{bmatrix}
\]
### Step 3: Check Determinant
Next, we calculate the determinant of \( J \):
\[
\det(J) = (3)(1) - (2)(-1) = 3 + 2 = 5
\]
Since \( \det(J) \neq 0 \), the implicit function theorem guarantees that we can locally express \( z \) and \( u \) as functions of \( x \), \( y \), and \( v \) near \( P \).
### Step 4: Computing \( \frac{\partial z}{\partial x} \)
Now, to find \( \frac{\partial z}{\partial x} \) and \( \frac{\partial u}{\partial x} \), we can use the inverse of the Jacobian matrix \( J \).
Using Cramer's rule or the formula for the inverse of a \( 2 \times 2 \) matrix,
\[
J^{-1} = \frac{1}{\det(J)} \begin{bmatrix}
1 & -2 \\
1 & 3
\end{bmatrix}
\]
So,
\[
J^{-1} = \frac{1}{5} \begin{bmatrix}
1 & -2 \\
1 & 3
\end{bmatrix}
\]
By the chain rule from the implicit function theorem,
\[
\begin{bmatrix}
\frac{\partial z}{\partial x} \\
\frac{\partial u}{\partial x}
\end{bmatrix} = J^{-1} \begin{bmatrix}
\frac{\partial f_1}{\partial x} \\
\frac{\partial f_2}{\partial x}
\end{bmatrix}
\]
### Step 5: Calculate \( \frac{\partial f_1}{\partial x} \) and \( \frac{\partial f_2}{\partial x} \)
We now find:
\[
\frac{\partial f_1}{\partial x} = 1
\]
\[
\frac{\partial f_2}{\partial x} = 3
\]
Putting these derivatives into our equations:
\[
\begin{bmatrix}
\frac{\partial z}{\partial x} \\
\frac{\partial u}{\partial x}
\end{bmatrix} = J^{-1} \begin{bmatrix}
1 \\
3
\end{bmatrix} = \frac{1}{5} \begin{bmatrix}
1 & -2 \\
1 & 3
\end{bmatrix} \begin{bmatrix}
1 \\
3
\end{bmatrix} = \frac{1}{5} \begin{bmatrix}
1 - 6 \\
1 + 9
\end{bmatrix} = \frac{1}{5} \begin{bmatrix}
-5 \\
10
\end{bmatrix} = \begin{bmatrix}
-1 \\
2
\end{bmatrix}
\]
### Final Result
Thus at the point \( (x, y, v) = (1, 2, 1) \):
\[
\frac{\partial z}{\partial x} = -1, \quad \frac{\partial u}{\partial x} = 2
\]
So, the answer is
\[
\frac{\partial z}{\partial x} = -1.
\]
And for \( u \):
\[
\frac{\partial u}{\partial x} = 2.
\]
First, we observe the system of equations:
1. \( f_1(x, y, z, u, v) = x + 2uz + y + vz^3 - 4 \)
2. \( f_2(x, y, z, u, v) = 3x + uv - z - 2 \)
We can express these equations as
\[
\begin{align*}
f_1(x, y, z, u, v) &= 0, \\
f_2(x, y, z, u, v) &= 0.
\end{align*}
\]
### Step 1: Compute the Jacobian Matrix
Next, we will compute the Jacobian matrix of the system with respect to the variables \( z \) and \( u \):
\[
J = \begin{bmatrix}
\frac{\partial f_1}{\partial z} & \frac{\partial f_1}{\partial u} \\
\frac{\partial f_2}{\partial z} & \frac{\partial f_2}{\partial u}
\end{bmatrix}
\]
Let's compute the necessary derivatives at the point \( P = (1, 2, 1, 0, 1) \).
**Derivatives of \( f_1 \)**:
\[
\frac{\partial f_1}{\partial z} = 2u + 3vz^2
\]
At \( P \):
\[
\frac{\partial f_1}{\partial z}(1, 2, 1, 0, 1) = 2(0) + 3(1)(1)^2 = 3
\]
\[
\frac{\partial f_1}{\partial u} = 2z
\]
At \( P \):
\[
\frac{\partial f_1}{\partial u}(1, 2, 1, 0, 1) = 2(1) = 2
\]
**Derivatives of \( f_2 \)**:
\[
\frac{\partial f_2}{\partial z} = -1
\]
At \( P \):
\[
\frac{\partial f_2}{\partial z}(1, 2, 1, 0, 1) = -1
\]
\[
\frac{\partial f_2}{\partial u} = v
\]
At \( P \):
\[
\frac{\partial f_2}{\partial u}(1, 2, 1, 0, 1) = 1
\]
### Step 2: Evaluate the Jacobian \( J \)
Now, we can write the Jacobian matrix evaluated at \( P \):
\[
J(P) = \begin{bmatrix}
3 & 2 \\
-1 & 1
\end{bmatrix}
\]
### Step 3: Check Determinant
Next, we calculate the determinant of \( J \):
\[
\det(J) = (3)(1) - (2)(-1) = 3 + 2 = 5
\]
Since \( \det(J) \neq 0 \), the implicit function theorem guarantees that we can locally express \( z \) and \( u \) as functions of \( x \), \( y \), and \( v \) near \( P \).
### Step 4: Computing \( \frac{\partial z}{\partial x} \)
Now, to find \( \frac{\partial z}{\partial x} \) and \( \frac{\partial u}{\partial x} \), we can use the inverse of the Jacobian matrix \( J \).
Using Cramer's rule or the formula for the inverse of a \( 2 \times 2 \) matrix,
\[
J^{-1} = \frac{1}{\det(J)} \begin{bmatrix}
1 & -2 \\
1 & 3
\end{bmatrix}
\]
So,
\[
J^{-1} = \frac{1}{5} \begin{bmatrix}
1 & -2 \\
1 & 3
\end{bmatrix}
\]
By the chain rule from the implicit function theorem,
\[
\begin{bmatrix}
\frac{\partial z}{\partial x} \\
\frac{\partial u}{\partial x}
\end{bmatrix} = J^{-1} \begin{bmatrix}
\frac{\partial f_1}{\partial x} \\
\frac{\partial f_2}{\partial x}
\end{bmatrix}
\]
### Step 5: Calculate \( \frac{\partial f_1}{\partial x} \) and \( \frac{\partial f_2}{\partial x} \)
We now find:
\[
\frac{\partial f_1}{\partial x} = 1
\]
\[
\frac{\partial f_2}{\partial x} = 3
\]
Putting these derivatives into our equations:
\[
\begin{bmatrix}
\frac{\partial z}{\partial x} \\
\frac{\partial u}{\partial x}
\end{bmatrix} = J^{-1} \begin{bmatrix}
1 \\
3
\end{bmatrix} = \frac{1}{5} \begin{bmatrix}
1 & -2 \\
1 & 3
\end{bmatrix} \begin{bmatrix}
1 \\
3
\end{bmatrix} = \frac{1}{5} \begin{bmatrix}
1 - 6 \\
1 + 9
\end{bmatrix} = \frac{1}{5} \begin{bmatrix}
-5 \\
10
\end{bmatrix} = \begin{bmatrix}
-1 \\
2
\end{bmatrix}
\]
### Final Result
Thus at the point \( (x, y, v) = (1, 2, 1) \):
\[
\frac{\partial z}{\partial x} = -1, \quad \frac{\partial u}{\partial x} = 2
\]
So, the answer is
\[
\frac{\partial z}{\partial x} = -1.
\]
And for \( u \):
\[
\frac{\partial u}{\partial x} = 2.
\]