Chapter 13: Problem 60
Find a unit vector in the direction in which \(f\) decreases most rapidly at \(P,\) and find the rate of change of \(f\) at \(P\) in that direction. $$ f(x, y, z)=4 e^{x y} \cos z ; P(0,1, \pi / 4) $$
Short Answer
Expert verified
The unit vector is \((-\frac{\sqrt{2}}{2}, 0, \frac{\sqrt{2}}{2})\) and the rate of change is \(-4\).
Step by step solution
01
Calculate the Gradient of f
The gradient of a function \( f(x, y, z) \) is given by \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) \). Compute each partial derivative separately. \( \frac{\partial f}{\partial x} = 4y e^{xy} \cos z \), \( \frac{\partial f}{\partial y} = 4x e^{xy} \cos z \), and \( \frac{\partial f}{\partial z} = -4 e^{xy} \sin z \). Thus, \( abla f = \left( 4y e^{xy} \cos z, 4x e^{xy} \cos z, -4 e^{xy} \sin z \right) \).
02
Evaluate the Gradient at Point P
Substitute \( P(0,1,\frac{\pi}{4}) \) into the gradient. We find the gradient components: \( \frac{\partial f}{\partial x} = 4 \times 1 \times e^{0} \times \cos(\frac{\pi}{4}) = 2\sqrt{2} \), \( \frac{\partial f}{\partial y} = 0 \times e^{0} \times \cos(\frac{\pi}{4}) = 0 \), \( \frac{\partial f}{\partial z} = -4 \times e^{0} \times \sin(\frac{\pi}{4}) = -2\sqrt{2} \). So, \( abla f(0, 1, \frac{\pi}{4}) = \left( 2\sqrt{2}, 0, -2\sqrt{2} \right) \).
03
Find the Unit Vector in the Direction of Maximum Decrease
The direction in which \( f \) decreases most rapidly is the negative gradient vector. The unit vector in this direction is \( -\frac{abla f}{\|abla f\|} \). First, calculate the magnitude of the gradient vector: \( \|abla f\| = \sqrt{(2\sqrt{2})^2 + 0^2 + (-2\sqrt{2})^2} = \sqrt{8 + 8} = \sqrt{16} = 4 \). The unit vector is \( -\frac{1}{4} \left( 2\sqrt{2}, 0, -2\sqrt{2} \right) = \left( -\frac{\sqrt{2}}{2}, 0, \frac{\sqrt{2}}{2} \right) \).
04
Determine the Rate of Change of f in the Direction of Maximum Decrease
The rate of change of \( f \) at \( P \) in the direction of the unit vector \( \mathbf{u} \) is given by the dot product \( abla f \cdot \mathbf{u} \). Thus, \( abla f \cdot \left( -\frac{\sqrt{2}}{2}, 0, \frac{\sqrt{2}}{2} \right) = 2\sqrt{2} \times -\frac{\sqrt{2}}{2} + 0 \times 0 + (-2\sqrt{2}) \times \frac{\sqrt{2}}{2} = -2 - 2 = -4 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Derivatives
Partial derivatives are essential tools in multivariable calculus. They measure how a function changes as each variable changes, one at a time. If you have a function of several variables, like \( f(x, y, z) = 4e^{xy} \cos z \), partial derivatives help us see how \( f \) changes with respect to each variable: \( x \), \( y \), and \( z \).
- The partial derivative with respect to \( x \) is \( \frac{\partial f}{\partial x} = 4y e^{xy} \cos z \). This shows how \( f \) changes as \( x \) varies, keeping \( y \) and \( z \) constant.
- The partial derivative with respect to \( y \) is \( \frac{\partial f}{\partial y} = 4x e^{xy} \cos z \). It represents the rate of change with \( y \), while \( x \) and \( z \) are held constant.
- The partial derivative with respect to \( z \) is \( \frac{\partial f}{\partial z} = -4 e^{xy} \sin z \). Here we see the sensitivity of the function to changes in \( z \).
Directional Derivative
The directional derivative extends the idea of partial derivatives, but it considers changes in any direction instead of just along the axes. When you want to know how a function changes in a specific direction, that's when directional derivatives come in handy. The directional derivative of a function \( f(x, y, z) \) at a point \( P \) is the rate at which the function changes as you move from \( P \) in the direction of a vector \( \mathbf{v} \).
To calculate it, the key step is to use the gradient of the function, \( abla f \), which we will discuss shortly. Then, you find the unit vector \( \mathbf{u} \) in the direction you're interested in, and the directional derivative is the dot product \( abla f \cdot \mathbf{u} \).
To calculate it, the key step is to use the gradient of the function, \( abla f \), which we will discuss shortly. Then, you find the unit vector \( \mathbf{u} \) in the direction you're interested in, and the directional derivative is the dot product \( abla f \cdot \mathbf{u} \).
- This process shows us how quickly \( f \) changes as you move in the direction of \( \mathbf{u} \).
- It is particularly useful in optimization problems and for understanding contour maps within multivariable calculus.
- The directional derivative is maximal along the gradient vector and minimal in the opposite direction, which gives insight into the function's behavior on a surface.
Calculating Gradients
The gradient itself is a vector that carries all the partial derivatives of a function at a point. It's like the compass that leads us in the direction of steepest ascent of the function. For a function \( f(x, y, z) \), the gradient \( abla f \) is represented as \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) \).
For example, when we worked out the gradient of \( f(x, y, z) = 4e^{xy} \cos z \) at the point \( P(0, 1, \pi/4) \), it was \( abla f(0, 1, \pi/4) = \left( 2\sqrt{2}, 0, -2\sqrt{2} \right) \).
For example, when we worked out the gradient of \( f(x, y, z) = 4e^{xy} \cos z \) at the point \( P(0, 1, \pi/4) \), it was \( abla f(0, 1, \pi/4) = \left( 2\sqrt{2}, 0, -2\sqrt{2} \right) \).
- This vector points in the direction where the function increases most rapidly.
- The magnitude of the gradient gives the rate of increase in that direction.
- If you want to find the direction of steepest descent, you simply take the negative of the gradient.