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A function \( f \) and a point \( P \) are given. Let \( \theta \) correspond to the direction of the directional derivative. Complete parts (a) through (e). \( f(x, y)=16-3 x^{2}-2 y^{2}, P(2,3) \) 4 (Type any angles in radians between 0 and \( 2 \pi \). Type an exact answer, using \( \pi \) as needed. Use a comma to separate answers as needed.) What angle(s) is/are associated with the direction of maximum decrease? \( \frac{\pi}{4} \) (Type any angles in radians between 0 and \( 2 \pi \). Type an exact answer, using \( \pi \) as needed. Use a comma to separate answers as needed.) What angle(s) is/are associated with the direction of zero change? The angle(s) associated with the directions of zero change is/are \( \frac{3 \pi}{4}, \frac{7 \pi}{4} \). (Type any angles in radians between 0 and \( 2 \pi \). Type an exact answer, using \( \pi \) as needed. Use a comma to separate answers as needed.) c. Write the directional derivative at \( P \) as a function of \( \theta \); call this function \( g(\theta) \). \( g(\theta)=\square \)

Ask by Wheeler Hall. in the United States
Mar 17,2025

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The directional derivative at point \( P(2,3) \) as a function of \( \theta \) is: \[ g(\theta) = -12(\cos\theta + \sin\theta) \]

Solución

We start with the function \[ f(x,y)=16-3x^{2}-2y^{2}, \] and the point \[ P(2,3). \] The gradient of \( f \) is given by \[ \nabla f(x, y) = \left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}\right). \] **Step 1. Compute the Gradient** Find the partial derivatives: \[ \frac{\partial f}{\partial x} = -6x, \qquad \frac{\partial f}{\partial y} = -4y. \] At the point \( P(2,3) \) we have \[ \nabla f(2,3) = \left(-6(2), -4(3)\right) = (-12,-12). \] **Step 2. Write Down the Directional Derivative** The directional derivative of \( f \) at \( P \) in the direction making an angle \( \theta \) with the positive \( x \)-axis is \[ g(\theta)=D_{\mathbf{u}}f(2,3)= \nabla f(2,3) \cdot (\cos\theta,\sin\theta). \] Substitute \(\nabla f(2,3) = (-12,-12)\): \[ g(\theta)= -12\cos\theta -12\sin\theta. \] Thus, \[ g(\theta)= -12(\cos\theta + \sin\theta). \] This is the expression for the directional derivative as a function of \( \theta \).

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Bonus Knowledge

To find the angle associated with the direction of maximum decrease, we first determine the gradient vector of the function, which gives us the direction of maximum increase. The negative of this gradient provides the maximum decrease direction. For \( f(x, y)=16-3x^2-2y^2 \), calculate \( \nabla f \) at the point \( P(2,3) \) to identify the angle. For the directional derivative at \( P \), you can express it in terms of \( \theta \) where \( \theta \) is the angle associated with the direction vector \( ( \cos \theta, \sin \theta ) \). Using the gradient and direction vector, derive the function \( g(\theta) \) to represent how the directional derivative varies with the angle, incorporating cosine and sine components based on the gradient's direction. To dive into the world of directional derivatives, you’d find yourself in a mathematical landscape where the gradient acts like a compass, pointing you toward the steepest slopes! It’s like hiking up a hill; you want to find the most efficient path up, which in calculus terms, is simply following the gradient's direction. Exploring further, perhaps you could look into the practical applications of this in fields like optimization and machine learning, where understanding how to navigate function landscapes can lead to improvements in algorithms. Great methods, such as gradient descent, rely heavily on concepts like these to efficiently find minimum values.

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