Chapter 5: Problem 14
Gegeben sei eine Funktion $$ f: \mathbb{R}^{2} \rightarrow \mathbb{R} \text { mit } \quad f^{\prime}(x, y)=\left(\sin \left(x^{2}\right) y\right) $$ sowie die Abbildung $$ \left.\mathbf{g}: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2} \quad \text { mit } \quad \mathbf{g}(x, y)=(x y, x+2 y)^{T}\right) $$ Ermitteln Sie für die Funktion $$ h: \mathbb{R}^{2} \rightarrow \mathbb{R} \text { mit } h=f \circ \mathbf{g} $$ im Punkt \((1,0)\) die Richtung des stärksten Anstiegs.
Short Answer
Step by step solution
Define the Composite Function
Find Expression for \( f' \)
Compute the Jacobian of \( \mathbf{g} \)
Compute the Gradient of \( f \)
Use Chain Rule to Find Gradient of \( h \)
Determine Direction of Steepest Ascent
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Composite Function
- Here, \( \mathbf{g}(x, y) = (xy, x + 2y)^T \), a transformation that combines and changes the input pair.
- Then, these transformed values are fed into the function \( f \), making the composite function \( h(x, y) \), which essentially alters inputs through two layers.
Gradient
In the exercise, the gradient of \( f \) is calculated in terms of the transformed variables \( (u, v) \) from the function \( g(x, y) \). It is expressed as \( [\frac{\partial f}{\partial u}, \frac{\partial f}{\partial v}]^T \), and in this case becomes \([\sin(u^2)v, \sin(u^2)]^T\) after substituting the expressions for \( f' \).
- For a practical approach, evaluating the gradient entails substituting specific values of \( (u, v) \) to assess how \( f \) changes around that point.
- The direction and magnitude indicated by the gradient are pivotal in finding local maximums and minimums in real-world applications.
Jacobian Matrix
For the mapping \( \mathbf{g}(x, y) = (xy, x + 2y)^T \), the Jacobian matrix \( J_g \) is calculated as follows:
- The first row includes the partial derivatives of the first component \( xy \) with respect to \( x \) and \( y \): \( [y, x] \).
- The second row includes the partial derivatives of the second component \( x + 2y \): \( [1, 2] \).
Chain Rule
The exercise uses the chain rule to find the gradient of the composite function \( h(x, y) = f(\mathbf{g}(x, y)) \). The chain rule here states that:
- First, compute the gradient of \( f \) with respect to the variables it directly depends on, \( (u, v) \).
- Then, multiply this gradient by the transpose of the Jacobian matrix \( J_g \) of the transformation \( g \).