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Gegeben sind die Abbildungen $$ \mathbf{v}: \mathbb{R}^{2} \rightarrow \mathbb{R}^{3}, \quad \mathbf{v}(x, y)=\left(\begin{array}{c} x+y^{2} \\ \sin x \cos y \\ e^{x^{2}+y^{2}} \end{array}\right) $$ und $$ \mathbf{w}: \mathbb{R}^{3} \rightarrow \mathbb{R}^{2}, \quad \mathbf{w}(r, s, t)=\left(\begin{array}{c} r+s t \\ \ln \left(1+r^{2}+s^{2}+t^{2}\right) \end{array}\right) $$ (a) Berechnen Sie die Ableitungsmatrizen \(\mathbf{v}^{\prime}(x, y)\) und \(\mathbf{w}^{\prime}(r, s, t)\). (b) Berechnen Sie sämtliche mögliche Verknüpfungen der beiden Abbild gen.

Short Answer

Expert verified
Calculate the derivative matrices for \( \mathbf{v} \) and \( \mathbf{w} \); only the composition \( \mathbf{w}\circ\mathbf{v} \) is possible.

Step by step solution

01

Calculate the Derivative Matrix of v(x, y)

To find the derivative matrix \( \mathbf{v}'(x, y) \), we calculate the partial derivatives of each component of \( \mathbf{v} \) with respect to \( x \) and \( y \):1. The first component is \( x + y^2 \):- Partial derivative with respect to \( x \): 1- Partial derivative with respect to \( y \): \( 2y \)2. The second component is \( \sin x \cos y \):- Partial derivative with respect to \( x \): \( \cos x \cos y \)- Partial derivative with respect to \( y \): \( -\sin x \sin y \)3. The third component is \( e^{x^2 + y^2} \):- Partial derivative with respect to \( x \): \( 2x e^{x^2 + y^2} \)- Partial derivative with respect to \( y \): \( 2y e^{x^2 + y^2} \)Thus, the derivative matrix is:\[\mathbf{v}'(x, y) = \begin{pmatrix} 1 & 2y \cos x \cos y & -\sin x \sin y 2x e^{x^2 + y^2} & 2y e^{x^2 + y^2} \end{pmatrix}\]
02

Calculate the Derivative Matrix of w(r, s, t)

To find the derivative matrix \( \mathbf{w}'(r, s, t) \), calculate the partial derivatives of each component of \( \mathbf{w} \):1. The first component \( r + st \):- Partial derivative with respect to \( r \): 1- Partial derivative with respect to \( s \): \( t \)- Partial derivative with respect to \( t \): \( s \)2. The second component \( \ln(1 + r^2 + s^2 + t^2) \):- Partial derivative with respect to \( r \): \( \frac{2r}{1 + r^2 + s^2 + t^2} \)- Partial derivative with respect to \( s \): \( \frac{2s}{1 + r^2 + s^2 + t^2} \)- Partial derivative with respect to \( t \): \( \frac{2t}{1 + r^2 + s^2 + t^2} \)The derivative matrix \( \mathbf{w}'(r, s, t) \) is:\[\mathbf{w}'(r, s, t) = \begin{pmatrix} 1 & t & s \\frac{2r}{1 + r^2 + s^2 + t^2} & \frac{2s}{1 + r^2 + s^2 + t^2} & \frac{2t}{1 + r^2 + s^2 + t^2} \end{pmatrix}\]
03

Determine Composition Possibilities

Consider the compositions: \( \mathbf{w}(\mathbf{v}(x, y)) \) and \( \mathbf{v}(\mathbf{w}(r,s,t)) \).The function \( \mathbf{v} : \mathbb{R}^2 \to \mathbb{R}^3 \) can only directly compose with a function that maps from \( \mathbb{R}^3 \to \text{something} \). Because \( \mathbf{w} \) maps from \( \mathbb{R}^3 \to \mathbb{R}^2 \), the composition \( \mathbf{v}(\mathbf{w}(r,s,t)) \) makes no sense due to mismatched dimensions.However, \( \mathbf{w} \) mapping from \( \mathbb{R}^3 \to \mathbb{R}^2 \) aligns with the mapping \( \mathbf{v} : \mathbb{R}^2 \to \mathbb{R}^3 \) for the composition \( \mathbf{w}(\mathbf{v}(x, y)) \), creating \( \mathbf{w} \circ \mathbf{v} : \mathbb{R}^2 \to \mathbb{R}^2 \). Only this composition is valid.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Partial Derivatives
Partial derivatives help us understand how a multivariable function changes when we focus on one variable at a time while keeping others constant. If you have a function like \( f(x, y) \), it means that changes in \(x\) or \(y\) can affect the output of \(f\). To find partial derivatives, you calculate the derivative of the function with respect to one variable after fixing all the others.

Take the function component \(x + y^2\):
  • The partial derivative with respect to \(x\) simplifies to 1, since we're treating \(y^2\) as constant.
  • When considering \(y\), it becomes \(2y\), because \(y^2\) differentiates to 2\(y\) while \(x\) is constant.
By examining each variable separately, partial derivatives tell us how each dimension of your input affects the output within a function.
Derivative Matrix
A derivative matrix, or Jacobian matrix, presents all the partial derivatives of a function's components compactly. It organizes these derivatives into a matrix format, which is especially useful when dealing with vector-valued functions. For example, if you have a function \( \mathbf{v}(x, y) \) that outputs a vector, then its derivative matrix \( \mathbf{v}'(x, y) \) shows how each output component changes with each variable.

Consider the vector-valued function \( \mathbf{v}(x, y) \) with three components:
  • The first row in the matrix contains the derivatives of the first component with respect to \(x\) and \(y\).
  • The second row represents the derivatives of the second component, similarly with each variable.
  • The third row continues this pattern for the third component.
This structure allows one to understand how small changes in \(x\) and \(y\) affect each part of the vector output simultaneously, providing a holistic view of the function's behavior.
Function Composition
Function composition is a way of combining two functions to form a new function. It involves using the output of one function as the input of another. When dealing with vector functions like \( \mathbf{v}(x, y) \) and \( \mathbf{w}(r, s, t) \), composability depends heavily on the dimensions.

For instance, the function \( \mathbf{v}: \mathbb{R}^{2} \rightarrow \mathbb{R}^{3} \) can provide the required inputs for a function \( \mathbf{w} \) that starts with a \(\mathbb{R}^{3}\) input. However, reverse composition requiring \( \mathbb{R}^{2} \to \mathbb{R}^{3}\) does not work due to dimension mismatch.
  • Thus, \( \mathbf{w}( \mathbf{v}(x, y)) \) is valid because \( \mathbf{v} \) provides a 3-dimensional vector, matching the input requirements of \( \mathbf{w} \).
  • But \( \mathbf{v}( \mathbf{w}(r, s, t)) \) doesn't succeed mathematically as \( \mathbf{v} \) expects 2-dimensional input.
Function composition emphasizes the importance of compatible input and output dimensions for a smooth, working combination of functions.

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