es sei das folgende vektorfeld gegeben:
\ f: \mathbb{r}^2 \to \mathbb{r}^2, \ x \mapsto f(x,y) = \begin{pmatrix} f_1(x,y) \\ f_2(x,y) \end{pmatrix} = \begin{pmatrix} 4x^2 + 6y^2 \\ 3x^2 - 4y^2 \end{pmatrix}. \

ordnen sie die folgenden blöcke so an, dass sie einen vollständigen beweis für die totale differenzierbarkeit von $ f $ im gegebenen punkt $ (x_0, y_0) = (3, 1) $ ergeben.

hinweis: nicht alle blöcke werden benötigt.

wählen sie aus diesen bausteinen:
- schreibe $ f(x,y) = f(x_0, y_0) + j_f(x_0, y_0) \begin{pmatrix} x - x_0 \\ y - y_0 \end{pmatrix} + r(x,y) $, also
- für die partiellen ableitungen gilt: $ \partial_x f_1 = 8x $, $ \partial_y f_1 = 12y $, $ \partial_x f_2 = 6x $ und $ \partial_y f_2 = -8y $.
- also ist $ f $ im punkt $ (3, 1) $ total differenzierbar.
- es gilt $ j_f(x,y) = \begin{pmatrix} 8x & 12y \\ 6x & -8y \end{pmatrix} $ und daher
- $ r(x,y) = \begin{pmatrix} 4(x-3)^2 + 6(y-1)^2 \\ 3(x-3)^2 - 4(y-1)^2 \end{pmatrix} $
- es gilt $ j_f(x,y) = \begin{pmatrix} 8x & 6x \\ 12y & -8y \end{pmatrix} $ und daher
- $ j_f(x_0, y_0) = \begin{pmatrix} 24 & 12 \\ 18 & -8 \end{pmatrix} $
- für $ x \to 3, y \to 1 $ ist $ \frac{r_1(x,y)^2}{(x-x_0)^2 + (y-y_0)^2} = \frac{(4(x-3)^2 + 6(y-1)^2)^2}{(x-3)^2 + (y-1)^2} \to 0 $. ($ r_2 $ analog)
- schreibe $ r(x,y) = f(x,y) + j_f(x_0, y_0) \begin{pmatrix} x - x_0 \\ y - y_0 \end{pmatrix} + f(x,y) $, also

Question

es sei das folgende vektorfeld gegeben:
\ f: \mathbb{r}^2 \to \mathbb{r}^2, \ x \mapsto f(x,y) = \begin{pmatrix} f_1(x,y) \\ f_2(x,y) \end{pmatrix} = \begin{pmatrix} 4x^2 + 6y^2 \\ 3x^2 - 4y^2 \end{pmatrix}. \

ordnen sie die folgenden blöcke so an, dass sie einen vollständigen beweis für die totale differenzierbarkeit von $ f $ im gegebenen punkt $ (x_0, y_0) = (3, 1) $ ergeben.

hinweis: nicht alle blöcke werden benötigt.

wählen sie aus diesen bausteinen:
- schreibe $ f(x,y) = f(x_0, y_0) + j_f(x_0, y_0) \begin{pmatrix} x - x_0 \\ y - y_0 \end{pmatrix} + r(x,y) $, also
- für die partiellen ableitungen gilt: $ \partial_x f_1 = 8x $, $ \partial_y f_1 = 12y $, $ \partial_x f_2 = 6x $ und $ \partial_y f_2 = -8y $.
- also ist $ f $ im punkt $ (3, 1) $ total differenzierbar.
- es gilt $ j_f(x,y) = \begin{pmatrix} 8x & 12y \\ 6x & -8y \end{pmatrix} $ und daher
- $ r(x,y) = \begin{pmatrix} 4(x-3)^2 + 6(y-1)^2 \\ 3(x-3)^2 - 4(y-1)^2 \end{pmatrix} $
- es gilt $ j_f(x,y) = \begin{pmatrix} 8x & 6x \\ 12y & -8y \end{pmatrix} $ und daher
- $ j_f(x_0, y_0) = \begin{pmatrix} 24 & 12 \\ 18 & -8 \end{pmatrix} $
- für $ x \to 3, y \to 1 $ ist $ \frac{r_1(x,y)^2}{(x-x_0)^2 + (y-y_0)^2} = \frac{(4(x-3)^2 + 6(y-1)^2)^2}{(x-3)^2 + (y-1)^2} \to 0 $. ($ r_2 $ analog)
- schreibe $ r(x,y) = f(x,y) + j_f(x_0, y_0) \begin{pmatrix} x - x_0 \\ y - y_0 \end{pmatrix} + f(x,y) $, also

Accepted Answer

# Explanation:

🆕 New Concept Discovered: Total Differentiability of Vector Fields
*Showing a function is differentiable using its Jacobian matrix.*

## Step 1: Calculate Partial Derivatives

To prove total differentiability, we first compute the partial derivatives of the component functions $ f_1(x, y) = 4x^2 + 6y^2 $ and $ f_2(x, y) = 3x^2 - 4y^2 $:

$$
\partial_x f_1 = 8x, \quad \partial_y f_1 = 12y, \quad \partial_x f_2 = 6x, \quad \partial_y f_2 = -8y
$$

This corresponds to the block:
* **"Für die partiellen Ableitungen gilt: $\partial_x f_1 = 8x$, $\partial_y f_1 = 12y$, $\partial_x f_2 = 6x$ und $\partial_y f_2 = -8y$."**

---

## Step 2: Construct the Jacobian Matrix

The Jacobian matrix $ J_f(x, y) $ is structured with the partial derivatives of $ f_1 $ in the first row and $ f_2 $ in the second row:

$$
J_f(x, y) = \begin{pmatrix} \partial_x f_1 & \partial_y f_1 \ \partial_x f_2 & \partial_y f_2 \end{pmatrix} = \begin{pmatrix} 8x & 12y \ 6x & -8y \end{pmatrix}
$$

This corresponds to the block:
* **"Es gilt $ J_f(x, y) = \begin{pmatrix} 8x & 12y \ 6x & -8y \end{pmatrix} $ und daher"**

---

## Step 3: Evaluate the Jacobian at the Given Point

We evaluate the Jacobian matrix at the point $(x_0, y_0) = (3, 1)$:

$$
J_f(3, 1) = \begin{pmatrix} 8(3) & 12(1) \ 6(3) & -8(1) \end{pmatrix} = \begin{pmatrix} 24 & 12 \ 18 & -8 \end{pmatrix}
$$

This corresponds to the block:
* **"$ J_f(x_0, y_0) = \begin{pmatrix} 24 & 12 \ 18 & -8 \end{pmatrix} $."**

---

## Step 4: Define the Linear Approximation and Remainder Term

By definition, a function is totally differentiable at $(x_0, y_0)$ if we can write it as a linear approximation plus a remainder term $ r(x, y) $ that vanishes faster than the distance to the point:

$$
f(x, y) = f(x_0, y_0) + J_f(x_0, y_0) \begin{pmatrix} x - x_0 \ y - y_0 \end{pmatrix} + r(x, y)
$$

This corresponds to the block:
* **"Schreibe $ f(x, y) = f(x_0, y_0) + J_f(x_0, y_0) \begin{pmatrix} x - x_0 \ y - y_0 \end{pmatrix} + r(x, y) $, also"**

---

## Step 5: Express the Remainder Term Explicitly

By substituting the functions and their linear approximations, we find the components of the remainder vector $ r(x, y) = \begin{pmatrix} r_1(x, y) \ r_2(x, y) \end{pmatrix} $:

$$
r(x, y) = \begin{pmatrix} 4(x - 3)^2 + 6(y - 1)^2 \ 3(x - 3)^2 - 4(y - 1)^2 \end{pmatrix}
$$

This corresponds to the block:
* **"$ r(x, y) = \begin{pmatrix} 4(x - 3)^2 + 6(y - 1)^2 \ 3(x - 3)^2 - 4(y - 1)^2 \end{pmatrix} $."**

---

## Step 6: Analyze the Limit of the Remainder Term

For total differentiability, the remainder term divided by the Euclidean distance must approach $ 0 $ as $(x, y) \to (3, 1)$:

$$
\lim_{(x, y) \to (3, 1)} \frac{r_i(x, y)}{\sqrt{(x - 3)^2 + (y - 1)^2}} = 0
$$

Note that the block in the exercise uses the squared distance in the denominator, which also correctly shows that the ratio goes to $ 0 $:

This corresponds to the block:
* **"Für $ x \to 3, y \to 1 $ ist $ \frac{r_1(x, y)^2}{(x - x_0)^2 + (y - y_0)^2} = \frac{(4(x - 3)^2 + 6(y - 1)^2)^2}{(x - 3)^2 + (y - 1)^2} \to 0 $. ($ r_2 $ analog)"**

---

## Step 7: Conclude Total Differentiability

Since the limit condition holds, the function is totally differentiable at the given point.

This corresponds to the block:
* **"Also ist $ f $ im Punkt $ (3, 1) $ total differenzierbar."**

---

# Answer:

Arrange the blocks in the following order to complete the proof:

1. **Für die partiellen Ableitungen gilt: $\partial_x f_1 = 8x$, $\partial_y f_1 = 12y$, $\partial_x f_2 = 6x$ und $\partial_y f_2 = -8y$.**
2. **Es gilt $ J_f(x, y) = \begin{pmatrix} 8x & 12y \ 6x & -8y \end{pmatrix} $ und daher**
3. **$ J_f(x_0, y_0) = \begin{pmatrix} 24 & 12 \ 18 & -8 \end{pmatrix} $.**
4. **Schreibe $ f(x, y) = f(x_0, y_0) + J_f(x_0, y_0) \begin{pmatrix} x - x_0 \ y - y_0 \end{pmatrix} + r(x, y) $, also**
5. **$ r(x, y) = \begin{pmatrix} 4(x - 3)^2 + 6(y - 1)^2 \ 3(x - 3)^2 - 4(y - 1)^2 \end{pmatrix} $.**
6. **Für $ x \to 3, y \to 1 $ ist $ \frac{r_1(x, y)^2}{(x - x_0)^2 + (y - y_0)^2} = \frac{(4(x - 3)^2 + 6(y - 1)^2)^2}{(x - 3)^2 + (y - 1)^2} \to 0 $. ($ r_2 $ analog)**
7. **Also ist $ f $ im Punkt $ (3, 1) $ total differenzierbar.**

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Step 1: Calculate Partial Derivatives

Step 2: Construct the Jacobian Matrix

Step 3: Evaluate the Jacobian at the Given Point

Step 4: Define the Linear Approximation and Remainder Term

Step 5: Express the Remainder Term Explicitly

Step 6: Analyze the Limit of the Remainder Term

Step 7: Conclude Total Differentiability

Answer: