Directional Derivative Calculation For F(x, Y, Z) At A Point
Introduction to Directional Derivatives
The directional derivative is a fundamental concept in multivariable calculus that extends the idea of a derivative to higher dimensions. Unlike partial derivatives, which measure the rate of change of a function along the coordinate axes, the directional derivative measures the rate of change of a function along an arbitrary direction. Understanding directional derivatives is crucial for analyzing how a function changes in various directions at a given point, which has applications in physics, engineering, and computer graphics.
In this article, we will delve into the process of finding the directional derivative of the function f(x, y, z) = x^4 + y^2 + z^7 at the point M(-3, 0, -1) in the direction of a unit vector l = 1v - 1w. Here, v represents the direction in which f increases the fastest at M, and w is another specified direction. We will first compute the gradient of f, which points in the direction of the steepest ascent. Then, we will normalize the gradient to obtain the unit vector v. We will also determine the unit vector w and finally calculate the directional derivative using the dot product of the gradient and the unit vector l.
Calculating the Gradient of f(x, y, z)
To find the directional derivative, the first crucial step is to compute the gradient of the function f(x, y, z) = x^4 + y^2 + z^7. The gradient, denoted as ∇f, is a vector-valued function that points in the direction of the greatest rate of increase of the function. It is defined as:
∇f = (∂f/∂x, ∂f/∂y, ∂f/∂z)
Where ∂f/∂x, ∂f/∂y, and ∂f/∂z are the partial derivatives of f with respect to x, y, and z, respectively. Let's calculate these partial derivatives:
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Partial derivative with respect to x:
∂f/∂x = ∂(x^4 + y^2 + z^7)/∂x = 4x^3
This derivative tells us how the function f changes as we vary x while keeping y and z constant.
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Partial derivative with respect to y:
∂f/∂y = ∂(x^4 + y^2 + z^7)/∂y = 2y
Similarly, this derivative shows the rate of change of f with respect to y.
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Partial derivative with respect to z:
∂f/∂z = ∂(x^4 + y^2 + z^7)/∂z = 7z^6
This derivative indicates the rate of change of f with respect to z.
Now, we can write the gradient of f as:
∇f(x, y, z) = (4x^3, 2y, 7z^6)
Next, we need to evaluate the gradient at the point M(-3, 0, -1). This will give us the direction of the steepest ascent at this specific point. Substituting the coordinates of M into the gradient, we get:
∇f(-3, 0, -1) = (4(-3)^3, 2(0), 7(-1)^6) = (-108, 0, 7)
This vector (-108, 0, 7) is crucial because it points in the direction where the function f increases the fastest at the point M. This direction vector will be used to determine the unit vector v.
Determining the Unit Vector v
As discussed in the previous section, the gradient of the function f at point M(-3, 0, -1) is ∇f(-3, 0, -1) = (-108, 0, 7). This vector indicates the direction of the steepest ascent of the function at point M. To find the unit vector v, which represents this direction in a normalized form, we need to divide the gradient vector by its magnitude. This process ensures that the vector has a length of 1, making it a unit vector.
The magnitude of the gradient vector is calculated as follows:
||∇f(-3, 0, -1)|| = √((-108)^2 + 0^2 + 7^2) = √(11664 + 0 + 49) = √11713
Now, we can find the unit vector v by dividing each component of the gradient vector by its magnitude:
v = (∇f(-3, 0, -1)) / ||∇f(-3, 0, -1)|| = (-108/√11713, 0/√11713, 7/√11713)
Simplifying the components, we get:
v = (-108/√11713, 0, 7/√11713)
This unit vector v points in the same direction as the gradient but has a magnitude of 1. It is essential for calculating the directional derivative, as it provides the direction along which we want to measure the rate of change of the function. The components of v represent the direction cosines, indicating the direction's orientation with respect to the coordinate axes.
Defining the Unit Vector w and l
To proceed with calculating the directional derivative, we need to define the unit vector w. The problem statement mentions that l = 1v - 1w, but does not explicitly define w. For the purpose of this discussion, let's assume that w is a unit vector in the direction of the z-axis, which is w = (0, 0, 1). This assumption allows us to demonstrate the process of calculating the directional derivative without additional context for w.
Now that we have v and w, we can determine the vector l using the given relationship:
l = 1v - 1w = v - w
Substituting the values of v and w, we get:
l = (-108/√11713, 0, 7/√11713) - (0, 0, 1)
Performing the subtraction, we obtain:
l = (-108/√11713, 0, 7/√11713 - 1)
This vector l represents the direction in which we want to find the directional derivative. However, to use it in the directional derivative formula, we need to normalize it to a unit vector. The magnitude of l is:
||l|| = √((-108/√11713)^2 + 0^2 + (7/√11713 - 1)^2)
||l|| = √(11664/11713 + (7/√11713 - 1)^2)
||l|| = √(11664/11713 + (49/11713 - 14/√11713 + 1))
||l|| = √(11664/11713 + 49/11713 - 14/√11713 + 1)
||l|| = √(1 + 11713/11713 - 14/√11713)
||l|| = √(2 - 14/√11713)
Therefore, the unit vector in the direction of l, denoted as u, is:
u = l / ||l|| = (-108/(√11713√(2 - 14/√11713)), 0, (7/√11713 - 1)/√(2 - 14/√11713))
This unit vector u is now ready to be used in the directional derivative calculation.
Calculating the Directional Derivative
With the gradient of f at M and the unit vector u in the direction of l determined, we can now calculate the directional derivative. The directional derivative of a function f at a point M in the direction of a unit vector u is given by the dot product of the gradient of f at M and the unit vector u:
D_u f(M) = ∇f(M) · u
We have already found that ∇f(-3, 0, -1) = (-108, 0, 7) and
u = (-108/(√11713√(2 - 14/√11713)), 0, (7/√11713 - 1)/√(2 - 14/√11713))
Now, we compute the dot product:
D_u f(-3, 0, -1) = (-108, 0, 7) · (-108/(√11713√(2 - 14/√11713)), 0, (7/√11713 - 1)/√(2 - 14/√11713))
The dot product is calculated as:
D_u f(-3, 0, -1) = (-108) * (-108/(√11713√(2 - 14/√11713))) + 0 * 0 + 7 * ((7/√11713 - 1)/√(2 - 14/√11713))
Simplifying this expression, we get:
D_u f(-3, 0, -1) = 11664/(√11713√(2 - 14/√11713)) + (49/√11713 - 7)/√(2 - 14/√11713)
This result is the directional derivative of f at M in the direction of l. It represents the rate of change of the function f as we move from the point M in the direction specified by the vector l. This value provides valuable information about the function's behavior along this specific direction.
Conclusion
In this article, we have explored the concept of directional derivatives and demonstrated the process of calculating the directional derivative of the function f(x, y, z) = x^4 + y^2 + z^7 at the point M(-3, 0, -1) in the direction of a unit vector l = 1v - 1w. We began by computing the gradient of f, which gives the direction of the steepest ascent. Then, we determined the unit vector v by normalizing the gradient. We assumed a direction for w and calculated the unit vector u in the direction of l. Finally, we computed the directional derivative using the dot product of the gradient and the unit vector u.
This process highlights the importance of understanding gradients and unit vectors in multivariable calculus. The directional derivative is a powerful tool for analyzing how functions change in different directions, with applications ranging from optimization problems to understanding physical phenomena. By mastering these concepts, one can gain deeper insights into the behavior of functions in higher dimensions.
The directional derivative provides a comprehensive understanding of a function's rate of change along any given direction, making it an indispensable tool in various fields. Whether you are optimizing a function, analyzing fluid flow, or working with computer graphics, the principles and techniques discussed in this article will prove invaluable.
