Problem 66 - Calculus Early Transcendentals

Question

Find the directions in the xy-plane in which the following functions have zero change at the given point. Express the directions in terms of unit vectors. $$f(x, y)=x^{2}-4 y^{2}-8 ; P(4,1,4)$$

Step-by-Step Solution

Answer

Express your answer in terms of unit vectors. Answer: The direction in which there is zero change in the function $$f(x, y)$$ at point $$P(4, 1, 4)$$ is given by $$\mathbf{v} \propto (\mathbf{i} + \mathbf{j})$$.

Step 1: Calculate the gradient vector of the function

To calculate the gradient vector of the function, we need to take the partial derivatives with respect to x and y: $$\nabla f(x, y) = \frac{\partial f}{\partial x} \mathbf{i} + \frac{\partial f}{\partial y} \mathbf{j}$$ Where, $$\frac{\partial f}{\partial x} = 2x \quad \text{and} \quad \frac{\partial f}{\partial y} = -8y$$ So, $$\nabla f(x, y) = 2x \mathbf{i} - 8y \mathbf{j}$$

Step 2: Find the value of the gradient vector at point $$P(4, 1, 4)$$

Now, we need to find the value of the gradient vector at the point $$P(4, 1, 4)$$. So, let's plug in the values of x and y into the gradient vector: $$\nabla f(4, 1) = 2(4) \mathbf{i} - 8(1) \mathbf{j} = 8\mathbf{i} - 8\mathbf{j}$$

Step 3: Calculate the directional derivative

Now, considering any direction in the xy-plane in the vector form $$\mathbf{v} = a\mathbf{i} + b\mathbf{j}$$, where a and b are the components of the direction vector. The directional derivative is given by the dot product of the gradient vector and direction vector: $$D_{\mathbf{v}}f(x,y) = \nabla f(x, y) \cdot \mathbf{v} = (8\mathbf{i} - 8\mathbf{j}) \cdot (a\mathbf{i} + b\mathbf{j})$$

Step 4: Set the directional derivative to zero

To find the direction in which there is no change, the directional derivative must be zero: $$D_{\mathbf{v}(a,b)}f(x,y) = (8\mathbf{i} - 8\mathbf{j}) \cdot (a\mathbf{i} + b\mathbf{j}) = 0$$ $$8a - 8b = 0$$

Step 5: Solve for the direction components

Now, let's solve for a and b: $$a - b = 0$$ $$a = b$$ So, the direction vector is: $$\mathbf{v} = a\mathbf{i} + b\mathbf{j} = a\mathbf{i} + a\mathbf{j} = a(\mathbf{i} + \mathbf{j})$$ Here, $$a$$ can be any constant.

Step 6: Express the direction in terms of unit vectors

The direction vector can now be expressed in terms of unit vectors $$\mathbf{i}$$ and $$\mathbf{j}$$: $$\mathbf{v} = a(\mathbf{i} + \mathbf{j})$$ Where $$a$$ is a constant, and the direction in which there is zero change in the function $$f(x, y)$$ at point $$P(4, 1, 4)$$ is given by $$\mathbf{v} \propto (\mathbf{i} + \mathbf{j})$$.

Key Concepts

Gradient Vector

The gradient vector is a crucial concept in multivariable calculus. It's essentially a vector composed of all the partial derivatives of a function, and it points in the direction of the steepest increase of the function. For example, if you have a function such as \( f(x, y) = x^2 - 4y^2 - 8 \), the gradient vector would be calculated as the partial derivatives with respect to \( x \) and \( y \).
To find the gradient vector, we differentiate: \( \frac{\partial f}{\partial x} = 2x \) and \( \frac{\partial f}{\partial y} = -8y \).
As a result, the gradient vector is \( abla f(x, y) = 2x \mathbf{i} - 8y \mathbf{j} \). This vector indicates the direction in which the function increases most rapidly. Calculating the gradient vector at a particular point gives insight into how the function behaves around that point.

Partial Derivatives

Partial derivatives are a concept that extends regular derivatives to multivariable functions. They measure the rate of change of a function as one of its variables is changed, while the other variables are held constant. In our problem with \( f(x, y) = x^2 - 4y^2 - 8 \), we calculated two partial derivatives.

• The partial derivative with respect to \( x \): \( \frac{\partial f}{\partial x} = 2x \)
• The partial derivative with respect to \( y \): \( \frac{\partial f}{\partial y} = -8y \)

These derivatives help in constructing the gradient vector, which then dictates how the function's value changes as you move through the plane. Knowing how to calculate partial derivatives is essential for analyzing multivariable functions.

Unit Vectors

A unit vector is a vector with a length of one that points in a specific direction. They are often used to indicate direction without considering the magnitude. In the context of finding directions with zero change in our function \( f(x, y) = x^2 - 4y^2 - 8 \), unit vectors \( \mathbf{i} \) and \( \mathbf{j} \) are used to express direction in the \( xy \)-plane.
When we derive that the zero-change direction is along \( \mathbf{v} \propto (\mathbf{i} + \mathbf{j}) \), it means that the unit vector provides a standardized way to express this direction. Therefore, the unit vector in the direction of \( \mathbf{i} + \mathbf{j} \) would be \( \frac{1}{\sqrt{2}} (\mathbf{i} + \mathbf{j}) \), ensuring that its length is precisely one.

Zero Change Direction

The zero change direction is where the function has no increase or decrease in value, meaning the directional derivative equals zero. For the problem \( f(x, y) = x^2 - 4y^2 - 8 \) at point \( P(4, 1, 4) \), we found the gradient \( 8\mathbf{i} - 8\mathbf{j} \).
To find directions of zero change, solve the equation given by the dot product of the gradient vector and a direction vector \( (a\mathbf{i} + b\mathbf{j}) \) set to zero: \( 8a - 8b = 0 \).
It simplifies to \( a = b \), meaning any direction \( \mathbf{v} = a(\mathbf{i} + \mathbf{j}) \) will result in zero change. This insight shows that moving in a path where \( x \) and \( y \) change equally will keep the function constant at that point, emphasizing the directional derivative concept.