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Magazine Chapter 13: Problem 12
Find the directional derivative of \(f\) at \(P\) in the direction of \(\mathbf{a} .\) $$ f(x, y)=e^{x} \cos y ; P(0, \pi / 4) ; \mathbf{a}=5 \mathbf{i}-2 \mathbf{j} $$
Short Answer
Expert verified
The directional derivative is \( \frac{7\sqrt{58}}{58} \).
Step by step solution
01
Compute the Gradient of the Function
The gradient of the function, denoted as \( abla f \), is a vector of the partial derivatives of \( f \). Calculate: \[ abla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \] Find \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \): \[ \frac{\partial f}{\partial x} = e^{x} \cos y \] \[ \frac{\partial f}{\partial y} = -e^{x} \sin y \] Thus, \( abla f(x, y) = \left( e^{x} \cos y, -e^{x} \sin y \right) \).
02
Evaluate the Gradient at Point P
Substitute the coordinates of point \( P(0, \pi/4) \) into the gradient to find \( abla f \) at \( P \): \[ abla f(0, \pi/4) = \left( e^{0} \cos \frac{\pi}{4}, -e^{0} \sin \frac{\pi}{4} \right) = \left( \frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right) \].
03
Normalize the Direction Vector
The direction vector \( \mathbf{a} = 5\mathbf{i} - 2\mathbf{j} \) needs to be normalized. The magnitude of \( \mathbf{a} \) is given by: \[ \|\mathbf{a}\| = \sqrt{5^2 + (-2)^2} = \sqrt{29} \] Thus, the unit vector in the direction of \( \mathbf{a} \) is: \[ \mathbf{u} = \left( \frac{5}{\sqrt{29}}, \frac{-2}{\sqrt{29}} \right) \].
04
Calculate the Directional Derivative
The directional derivative of \( f \) at \( P \) in the direction of \( \mathbf{a} \) is given by: \[ D_{\mathbf{u}} f = abla f(0, \pi/4) \cdot \mathbf{u} \] Substitute the values: \[ D_{\mathbf{u}} f = \left( \frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right) \cdot \left( \frac{5}{\sqrt{29}}, \frac{-2}{\sqrt{29}} \right) \] \[ = \frac{\sqrt{2}}{2} \times \frac{5}{\sqrt{29}} + \left(-\frac{\sqrt{2}}{2}\right) \times \frac{-2}{\sqrt{29}} \] \[ = \frac{5\sqrt{2}}{2\sqrt{29}} + \frac{2\sqrt{2}}{2\sqrt{29}} \] \[ = \frac{7\sqrt{2}}{2\sqrt{29}} \].
05
Simplify the Expression
Further simplification gives: \[ D_{\mathbf{u}} f = \frac{7\sqrt{2}}{2\sqrt{29}} = \frac{7\sqrt{58}}{58} \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Gradient
The gradient of a function is like a compass that points in the direction of the steepest ascent. Mathematically, the gradient is represented as a vector, consisting of all partial derivatives of a function. For our example with the function \( f(x, y) = e^x \cos y \), the gradient is calculated as:
- First, take the partial derivative with respect to \( x \), which yields \( \frac{\partial f}{\partial x} = e^x \cos y \).
- Second, find the partial derivative with respect to \( y \), resulting in \( \frac{\partial f}{\partial y} = -e^x \sin y \).
Partial Derivatives
Partial derivatives are the building blocks of a gradient. They measure how a function changes as you tweak one variable, keeping others constant. For example, in the function \( f(x, y) = e^x \cos y \), each variable, \( x \) and \( y \), has a role in determining the function's behavior:
- \( \frac{\partial f}{\partial x} = e^x \cos y \) examines changes in \( f \) as \( x \) changes.
- \( \frac{\partial f}{\partial y} = -e^x \sin y \) checks how \( f \) evolves as \( y \) varies.
Unit Vector
In vector mathematics, a unit vector provides direction without affecting magnitude. To find a unit vector, you normalize the given vector by dividing it by its magnitude. Given the direction vector \( \mathbf{a} = 5\mathbf{i} - 2\mathbf{j} \), its magnitude can be computed as:\[ \| \mathbf{a} \| = \sqrt{5^2 + (-2)^2} = \sqrt{29} \]The unit vector, \( \mathbf{u} \), is then:
- \( \mathbf{u} = \left( \frac{5}{\sqrt{29}}, \frac{-2}{\sqrt{29}} \right) \)
Function Evaluation
Evaluating a function lets you understand its value at specific points of interest. For instance, consider evaluating \( abla f(x, y) \) at point \( P(0, \pi/4) \):\[ abla f(0, \pi/4) = \left( e^{0} \cos \frac{\pi}{4}, -e^{0} \sin \frac{\pi}{4} \right) = \left( \frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right) \]Through such evaluations, we extract precise vector or scalar values showing the behavior or trajectory of the function at pinpointed coordinates. This practice is essential in understanding how a function behaves not just theoretically, but also in real-world contexts or at crucial points, such as those used for calculating the directional derivative.
