Chapter 12: Problem 17
Find the directional derivative of \(f(x, y, z)=x y+z^{2}\) a (1.1.1) in the direction toward (5,-3,3)
Short Answer
Expert verified
The directional derivative is \( \frac{2}{3} \).
Step by step solution
01
Understand the Directional Derivative
The directional derivative of a function represents the rate at which the function changes at a point in a particular direction. It is calculated using the gradient of the function and the unit vector in the given direction.
02
Compute the Gradient of f
For the function \[ f(x, y, z) = xy + z^2 \]find the gradient \( abla f \) which is given by the partial derivatives \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) \). \[ \frac{\partial f}{\partial x} = y \]\[ \frac{\partial f}{\partial y} = x \]\[ \frac{\partial f}{\partial z} = 2z \]So, the gradient \( abla f = (y, x, 2z) \).
03
Evaluate the Gradient at the Point (1,1,1)
Now substitute the point (1, 1, 1) into the gradient:\[ abla f(1,1,1) = (1, 1, 2) \]
04
Determine the Direction Vector
The direction given in the exercise is towards the point (5, -3, 3). First, find the direction vector by subtracting (1, 1, 1) from (5, -3, 3):\[ \text{Direction Vector} = (5, -3, 3) - (1, 1, 1) = (4, -4, 2) \]
05
Normalize the Direction Vector
Convert the direction vector into a unit vector by dividing it by its magnitude:\[ \|d\| = \sqrt{4^2 + (-4)^2 + 2^2} = \sqrt{16 + 16 + 4} = \sqrt{36} = 6 \]So, the unit direction vector \( \mathbf{u} = \left( \frac{4}{6}, \frac{-4}{6}, \frac{2}{6} \right) = \left( \frac{2}{3}, \frac{-2}{3}, \frac{1}{3} \right) \).
06
Calculate the Directional Derivative
The directional derivative \( D_u f \) is the dot product of the gradient and the unit vector:\[ D_u f = abla f(1,1,1) \cdot \mathbf{u} \]\[ D_u f = (1, 1, 2) \cdot \left( \frac{2}{3}, \frac{-2}{3}, \frac{1}{3} \right) = 1 \cdot \frac{2}{3} + 1 \cdot \frac{-2}{3} + 2 \cdot \frac{1}{3} \]\[ D_u f = \frac{2}{3} - \frac{2}{3} + \frac{2}{3} = \frac{2}{3} \]
07
Conclusion
The directional derivative of the function \( f(x, y, z) = xy + z^2 \) at the point (1, 1, 1) in the direction towards (5, -3, 3) is \( \frac{2}{3} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Gradient
The gradient is a vector that is crucial in calculating the directional derivative.It is represented by the symbol \( abla f \) and comprises the partial derivatives of a function.For the function \( f(x, y, z) = xy + z^2 \), the gradient is derived from:
These derivatives are combined to form the gradient vector \( abla f = (y, x, 2z) \).This gradient shows the direction of the steepest increase of the function at any point.
Calculating the gradient at a specific point, such as (1, 1, 1), involves substituting these coordinates into the gradient formula, giving \( abla f(1,1,1) = (1, 1, 2) \).This specific vector can then be used to examine how the function changes at this point.
- The partial derivative with respect to \( x \), which is \( \frac{\partial f}{\partial x} = y \).
- The partial derivative with respect to \( y \), which is \( \frac{\partial f}{\partial y} = x \).
- The partial derivative with respect to \( z \), which is \( \frac{\partial f}{\partial z} = 2z \).
These derivatives are combined to form the gradient vector \( abla f = (y, x, 2z) \).This gradient shows the direction of the steepest increase of the function at any point.
Calculating the gradient at a specific point, such as (1, 1, 1), involves substituting these coordinates into the gradient formula, giving \( abla f(1,1,1) = (1, 1, 2) \).This specific vector can then be used to examine how the function changes at this point.
Unit Vector
A unit vector is essential in directional derivatives since it defines the specific direction with magnitude one.To find a unit vector for a given direction, you first need the direction vector.In our exercise, the direction vector points from (1, 1, 1) to (5, -3, 3), calculated as \( (4, -4, 2) \).
Next, you normalize this vector to have a magnitude of one.The magnitude \( \|d\| \) of vector \( (4, -4, 2) \) is calculated as \( \sqrt{4^2 + (-4)^2 + 2^2} = 6 \).Then, each component of the direction vector is divided by this magnitude:
Next, you normalize this vector to have a magnitude of one.The magnitude \( \|d\| \) of vector \( (4, -4, 2) \) is calculated as \( \sqrt{4^2 + (-4)^2 + 2^2} = 6 \).Then, each component of the direction vector is divided by this magnitude:
- \( \frac{4}{6} = \frac{2}{3} \)
- \( \frac{-4}{6} = \frac{-2}{3} \)
- \( \frac{2}{6} = \frac{1}{3} \)
Partial Derivatives
Partial derivatives are vital for understanding how a multivariable function changes when only one variable is varied, while others are held constant.
In this case, the function is \( f(x, y, z) = xy + z^2 \).Partial derivatives with respect to \( x, y, \) and \( z \) are calculated separately to form the gradient.
In this case, the function is \( f(x, y, z) = xy + z^2 \).Partial derivatives with respect to \( x, y, \) and \( z \) are calculated separately to form the gradient.
- The partial derivative with respect to \( x \), \( \frac{\partial f}{\partial x} = y \), shows how the function shifts when \( x \) is altered while \( y \) and \( z \) stay constant.
- Similarly, \( \frac{\partial f}{\partial y} = x \) indicates change when \( y \) moves, holding \( x \) and \( z \) steady.
- Finally, \( \frac{\partial f}{\partial z} = 2z \) describes how \( f \) varies as \( z \) changes, independent of \( x \) and \( y \).