Question

Find the gradient of the function and the maximum value of the directional derivative at the given point. $$ \frac{\text { Function }}{f(x, y, z)=x e^{y z}} \frac{\text { Point }}{(2,0,-4)} $$

Short answer

The gradient of the function at the point (2,0,-4) is <1, -8, 0> and the maximum value of the directional derivative at this point is \(\sqrt{65}\)

Step-by-step solution

Compute the partial derivatives

First, we compute the partial derivatives, \(\frac{\partial f}{\partial x}\), \(\frac{\partial f}{\partial y}\), \(\frac{\partial f}{\partial z}\), of the function \(f(x, y, z) = xe^{yz}\). Using the product rule of differentiation, we find: \(\frac{\partial f}{\partial x} = e^{yz}\), \(\frac{\partial f}{\partial y} = xze^{yz}\), and \(\frac{\partial f}{\partial z} = xye^{yz}\)

Evaluate the gradients at the given point

Second, we evaluate these partial derivatives at the point (2,0,-4). Plugging x = 2, y = 0, and z = -4 into our expressions gives us: \(\frac{\partial f}{\partial x} = e^{0} = 1\), \(\frac{\partial f}{\partial y} = 2*-4*e^{0} = -8\), and \(\frac{\partial f}{\partial z} = 2*0*e^{0} = 0\)

Calculate the gradient vector

Third, with all the values for the partial derivatives computed, we can find the gradient vector of the function at the given point. The gradient vector \(\nabla f\) is given by \(\nabla f = <1, -8, 0>\)

Find the maximum value of the directional derivative

Finally, in order to determine the maximum value of the directional derivative, one should note that this is achieved in the direction of the gradient vector \(\nabla f\). The magnitude of the gradient vector, |\(\nabla f\)| is the maximum rate of increase of the function, which also represents the maximum value of the directional derivative. The magnitude of \(\nabla f = <1, -8, 0>\) is given by \(\sqrt{(1)^2 + (-8)^2 + (0)^2} = \sqrt{65}\)

Key concepts

Partial Derivatives

Understanding partial derivatives is crucial for many applications in calculus, especially when dealing with functions of multiple variables. In simple terms, a partial derivative measures how a function changes as one of the variables is varied while the other variables are held constant.

For instance, consider the function from the exercise, \(f(x, y, z) = xe^{yz}\). Here, we have a function of three variables: \(x\), \(y\), and \(z\). When computing \(\frac{\partial f}{\partial x}\), we treat \(y\) and \(z\) as constants and only differentiate with respect to \(x\). This results in \(e^{yz}\), since the derivative of \(x\) with respect to itself is 1, and \(e^{yz}\) acts as a constant multiplier.

Similarly, \(\frac{\partial f}{\partial y} = xze^{yz}\) and \(\frac{\partial f}{\partial z} = xye^{yz}\) are calculated by focusing on the respective variables \(y\) and \(z\), whilst keeping the other variables static. These computations are the foundation for finding the rate at which the function changes in any direction, which brings us to the concept of directional derivatives.

Directional Derivative

The directional derivative, in vector calculus, indicates how a function \(f\) changes at a point in a given direction. It is a generalization of the concept of a derivative in one dimension to higher dimensions. To find the directional derivative of a function at a particular point along a specified vector \(\mathbf{v}\), one must project the gradient vector of \(f\) onto \(\mathbf{v}\).

The basic formula for the directional derivative along a unit vector \(\mathbf{u}\) is given by \(D_{\mathbf{u}}f = abla f \cdot \mathbf{u}\), where \(abla f\) is the gradient vector of the function \(f\), and \(\cdot\) denotes the dot product. The outcome represents how much the function \(f\) is increasing or decreasing at the point in the direction of the vector \(\mathbf{u}\).

In the exercise, we're not directly computing the directional derivative but understanding its concept which is needed to grasp why the maximum value of this derivative aligns with a certain vector direction.

Maximum Value of Directional Derivative

The gradient vector not only points in the direction of the greatest rate of increase of the function but its magnitude also tells us the maximum value of the directional derivative. This value represents the steepest ascent of the function at the given point.

In the context of the exercise, \(abla f\) at the point \(\(2,0,-4\)\) is \(\<1, -8, 0\>\). The magnitude of this gradient vector, calculated as \(\sqrt{(1)^2 + (-8)^2 + (0)^2}\), is \(\sqrt{65}\). Therefore, the maximum value of the directional derivative of \(f\) at the point \(\(2,0,-4\)\) is \(\sqrt{65}\). This value represents the greatest rate of change of the function if one were to move from \(\(2,0,-4\)\) in the direction of the gradient vector.

Grasping these concepts helps in optimizing functions and understanding their behavior in three-dimensional space, which is deeply connected to fields such as physics, engineering, and economics, where maximum efficiency or performance is often sought.