Problem 8 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-8, find the directional derivative of \(f\) at the point \(\mathbf{p}\) in the direction of \(\mathbf{a}\). \(f(x, y, z)=x^{2}+y^{2}+z^{2} ; \mathbf{p}=(1,-1,2)\);. \(\mathbf{a}=\sqrt{2} \mathbf{i}-\mathbf{j}-\mathbf{k}\)

Step-by-Step Solution

Answer

The directional derivative is \( \sqrt{2} - 1 \).

Step 1: Find the Gradient of f

The gradient of a function \( f(x, y, z) \) is given by \( abla f = \left< \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right> \). For \( f(x, y, z) = x^2 + y^2 + z^2 \), calculate each partial derivative:\[ \frac{\partial f}{\partial x} = 2x, \quad \frac{\partial f}{\partial y} = 2y, \quad \frac{\partial f}{\partial z} = 2z \]Thus, the gradient is \( abla f = \left< 2x, 2y, 2z \right> \).

Step 2: Evaluate the Gradient at Point p

Substitute \( \mathbf{p} = (1,-1,2) \) into the gradient to find \( abla f(\mathbf{p}) \):\[ abla f(1, -1, 2) = \left< 2(1), 2(-1), 2(2) \right> = \left< 2, -2, 4 \right> \]

Step 3: Normalize the Direction Vector a

The direction vector \( \mathbf{a} = \sqrt{2} \mathbf{i} - \mathbf{j} - \mathbf{k} \) must be normalized. Calculate the magnitude of \( \mathbf{a} \):\[ ||\mathbf{a}|| = \sqrt{(\sqrt{2})^2 + (-1)^2 + (-1)^2} = \sqrt{2 + 1 + 1} = \sqrt{4} = 2 \]The normalized vector \( \mathbf{\hat{a}} \) is:\[ \mathbf{\hat{a}} = \left( \frac{\sqrt{2}}{2}, \frac{-1}{2}, \frac{-1}{2} \right) \]

Step 4: Calculate the Directional Derivative

The directional derivative of \( f \) in the direction of a unit vector \( \mathbf{\hat{a}} \) is given by the dot product:\[ D_{\mathbf{a}}f(\mathbf{p}) = abla f(\mathbf{p}) \cdot \mathbf{\hat{a}} \]Substitute \( abla f(\mathbf{p}) = \left< 2, -2, 4 \right> \) and \( \mathbf{\hat{a}} = \left( \frac{\sqrt{2}}{2}, \frac{-1}{2}, \frac{-1}{2} \right) \):\[ D_{\mathbf{a}}f(\mathbf{p}) = \left< 2, -2, 4 \right> \cdot \left< \frac{\sqrt{2}}{2}, \frac{-1}{2}, \frac{-1}{2} \right> = 2\left(\frac{\sqrt{2}}{2}\right) + (-2)\left(\frac{-1}{2}\right) + 4\left(\frac{-1}{2}\right) \]Simplify this expression:\[ = \sqrt{2} + 1 - 2 = \sqrt{2} - 1 \]

Step 5: Conclusion

The directional derivative of \( f \) at the point \( \mathbf{p} = (1, -1, 2) \) in the direction of \( \mathbf{a} \) is \( \sqrt{2} - 1 \).

Key Concepts

gradient

The gradient is an essential concept when studying multivariable functions. Think of it like a compass showing the direction of the steepest ascent from any point on a surface. Given a function of several variables, the gradient at each point becomes a vector.
The components of this vector represent the rates of change of the function in each direction: x, y, and z. This means the gradient vector summarizes the partial derivatives.

• For a function \( f(x, y, z) \), the gradient \( abla f \) is \( \left< \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right> \).
• It will point in the direction where the function increases the fastest.
• Its magnitude gives the maximum rate of increase.

Calculating it is a crucial step to understand how the function behaves around a point, making it key for finding directional derivatives.

partial derivatives

Partial derivatives are a way to understand how a function changes when you tweak one variable at a time while keeping others constant.
The idea is to take derivatives, similar to single-variable calculus, but apply them to only one of the function's variables.

• For example, in a function \( f(x, y, z) = x^2 + y^2 + z^2 \), the partial derivatives with respect to x, y, and z are: \( \frac{\partial f}{\partial x} = 2x \), \( \frac{\partial f}{\partial y} = 2y \), and \( \frac{\partial f}{\partial z} = 2z \).
• They indicate the rate of change of the function along each coordinate direction.

Understanding them is pivotal for finding the gradient, which in turn helps measure how the function moves in multiple dimensions.

vector normalization

Vector normalization is the process of converting a vector to a unit vector, meaning it maintains its direction but now has a length of one.
This is achieved by dividing each component of the vector by its magnitude. It is vital when calculating directional derivatives because only unit vectors can properly indicate direction alone. Here's how it works:

• Start with a vector, such as \( \mathbf{a} = \sqrt{2} \mathbf{i} - \mathbf{j} - \mathbf{k} \).
• Calculate its magnitude: \(||\mathbf{a}|| = \sqrt{2^2 + (-1)^2 + (-1)^2} = 2\).
• Divide each component by this magnitude to get the unit vector: \( \mathbf{\hat{a}} = \left( \frac{\sqrt{2}}{2}, \frac{-1}{2}, \frac{-1}{2} \right) \).

Vector normalization ensures that the direction is standardized, making it easier to use in further calculations, like the dot product for directional derivatives.

dot product

The dot product is a mathematical operation that takes two vectors and returns a scalar. It is particularly significant in finding how much one vector extends in the direction of another. In the context of directional derivatives, it's used to combine the gradient vector with the normalized direction vector.

• Consider two vectors \( \mathbf{u} = \left< u_1, u_2, u_3 \right> \) and \( \mathbf{v} = \left< v_1, v_2, v_3 \right> \).
• Their dot product is \( \mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 + u_3v_3 \).
• For directional derivatives, we use \( abla f(\mathbf{p}) \cdot \mathbf{\hat{a}} \) to find the rate of change in the direction of \( \mathbf{a} \).

This operation is simple but powerful, distilling the relationship between directions to a single numerical value.