Chapter 4: Problem 276
For the following exercises, find the directional derivative of the function in the direction of the unit vector \(\mathbf{u}=\cos \theta \mathbf{i}+\sin \theta \mathbf{j}\). $$f(x, y)=\cos (3 x+y), \theta=\frac{\pi}{4}$$
Short Answer
Expert verified
The directional derivative is \(-2\sqrt{2}\sin(3x+y)\).
Step by step solution
01
Understand the Components
The directional derivative of a function, denoted as \( D_{\mathbf{u}} f(x, y) \), represents how \( f(x, y) \) changes as you move in the direction of a unit vector \( \mathbf{u} \). We are given the function \( f(x, y)=\cos(3x+y) \) and the direction vector given by the angle \( \theta = \frac{\pi}{4} \). The unit vector \( \mathbf{u} \) is \( \cos \theta \mathbf{i} + \sin \theta \mathbf{j} \).
02
Calculate the Gradient Vector
Calculate the gradient of \( f(x, y) \), denoted as \( abla f(x, y) \). It is obtained by taking the partial derivatives of \( f \) with respect to \( x \) and \( y \). So, \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \).
03
Partial Derivative with Respect to x
Find \( \frac{\partial f}{\partial x} \). Since \( f(x, y) = \cos(3x+y) \), apply the chain rule: \( \frac{\partial f}{\partial x} = -3\sin(3x+y) \).
04
Partial Derivative with Respect to y
Find \( \frac{\partial f}{\partial y} \). Again, apply the chain rule to get \( \frac{\partial f}{\partial y} = -\sin(3x+y) \).
05
Form the Gradient Vector
The gradient vector \( abla f \) is \( (-3\sin(3x+y), -\sin(3x+y)) \).
06
Evaluate the Unit Vector
The unit vector \( \mathbf{u} \) in terms of the angle \( \theta = \frac{\pi}{4} \) is \( \left( \cos\left(\frac{\pi}{4}\right), \sin\left(\frac{\pi}{4}\right) \right) = \left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) \).
07
Compute the Dot Product
Calculate the dot product of \( abla f(x, y) \) and \( \mathbf{u} \): \[ D_{\mathbf{u}} f(x, y) = abla f \cdot \mathbf{u} = (-3\sin(3x+y), -\sin(3x+y)) \cdot \left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right) \] \[ = -3\sin(3x+y) \times \frac{\sqrt{2}}{2} + (-\sin(3x+y)) \times \frac{\sqrt{2}}{2} \] \[ = -2\sqrt{2}\sin(3x+y) \].
08
Finalize the Directional Derivative
The directional derivative of the function \( f(x, y) \) in the direction of \( \mathbf{u} \) when \( \theta = \frac{\pi}{4} \) is \( -2\sqrt{2}\sin(3x+y) \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Gradient Vector
The gradient vector is a crucial concept in multivariable calculus. It's like a compass pointing in the direction of the steepest ascent of a function. For functions of two variables, the gradient vector is expressed as \(abla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \).This vector gives us a multi-dimensional rate of change, showing us how much the function increases with small changes in both the \(x\) and \(y\) directions.
- The first component is the partial derivative with respect to \(x\), showing how sensitive the function is to \(x\).
- The second component is the partial derivative with respect to \(y\), indicating how sensitive it is to changes in \(y\).
Partial Derivatives
Partial derivatives offer insight into how a multivariable function changes when altering one independent variable, keeping the others constant. In essence, they isolate the sensitivity of the function to a single variable.
To find the partial derivative of a function, say \(f(x, y)\) with respect to \(x\), the notation \(\frac{\partial f}{\partial x}\) is used.
To find the partial derivative of a function, say \(f(x, y)\) with respect to \(x\), the notation \(\frac{\partial f}{\partial x}\) is used.
- For our given function \(f(x, y) = \cos(3x + y)\), the partial derivative with respect to \(x\) was found using the chain rule, yielding \(-3\sin(3x+y)\).
- Similarly, the partial derivative with respect to \(y\) is \(-\sin(3x+y)\).
Chain Rule
The chain rule is a fundamental technique in calculus used to differentiate composite functions. It enables us to tackle complex functions by breaking them down into compositions of simpler functions. This rule states that if a variable \(z\) depends on a variable \(u\), which in turn depends on another variable \(x\), then the rate of change of \(z\) with respect to \(x\) can be found by multiplying the derivatives:\[\frac{dz}{dx} = \frac{dz}{du} \cdot \frac{du}{dx} \].
In the context of multivariable functions like \(f(x, y) = \cos(3x+y)\), the chain rule is essential:
In the context of multivariable functions like \(f(x, y) = \cos(3x+y)\), the chain rule is essential:
- We used the chain rule to find partial derivatives for both \(x\) and \(y\).
- For \(\frac{\partial f}{\partial x}\), this involved differentiating the inner function \((3x+y)\) and multiplying by the derivative of the outer function, \(-\sin\).