Problem 68 - 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. $$f(x, y)=e^{1-x y} ; P(1,0, e)$$

Step-by-Step Solution

Answer

Answer: The function has no change in any direction parallel to the x-axis in the xy-plane.

Step 1: Find the gradient of the function

To find the gradient of the function, we need to find the partial derivatives with respect to both x and y. $$ \frac{\partial f}{\partial x} = -ye^{1-xy} $$ $$ \frac{\partial f}{\partial y} = -xe^{1-xy} $$ Now we'll evaluate these partial derivatives at point P(1, 0, e): $$ \frac{\partial f}{\partial x}(1,0) = -0e^1 = 0 $$ $$ \frac{\partial f}{\partial y}(1,0) = -1e^1 = -e $$ Thus, the gradient of the function at point P is: $$ \text{grad } f(1,0) = \begin{bmatrix} 0 \\ -e \end{bmatrix} $$

Step 2: Set up the equation for zero change

We are interested in the directions in which the function has zero change. This occurs when the directional derivative is equal to zero, or when the dot product of the gradient vector and the direction vector is zero. Let the direction vector be $$\begin{bmatrix}u\\v\end{bmatrix}$$ where u and v are the x and y components of the direction vector. The dot product of grad f and the direction vector is: $$ \begin{bmatrix} 0 \\ -e \end{bmatrix} \cdot \begin{bmatrix} u \\ v \end{bmatrix} = 0u - ev = 0 $$

Step 3: Solve for the direction vector

To find the direction vector, we need to solve the equation $$-ev = 0$$ from step 2: $$ v = \frac{0}{-e} = 0 $$ So, the direction vector can be any vector with y-component equal to 0, such as $$\begin{bmatrix} 1 \\ 0 \end{bmatrix}$$, $$\begin{bmatrix} -1 \\ 0 \end{bmatrix}$$, or $$\begin{bmatrix} k \\ 0 \end{bmatrix}$$ where k is any scalar value.

Step 4: Express the direction vector in terms of unit vectors

Finally, we need to express the direction vector in terms of unit vectors. Because the direction vector can be any vector with a y-component of 0, we can write it as a multiple of the unit vector in the x-direction, $$\mathbf{i}$$: $$ \begin{bmatrix} k \\ 0 \end{bmatrix} = k\mathbf{i} $$ This means, the function has a zero change in any direction parallel to the x-axis in the xy-plane.

Key Concepts

Partial Derivatives

Partial derivatives are a fundamental concept when it comes to understanding how a function changes in different directions. Consider a function of two variables, such as \( f(x, y) \). To explore how this function changes with respect to one variable, while keeping the other constant, we use partial derivatives.

• The partial derivative with respect to \( x \), denoted as \( \frac{\partial f}{\partial x} \), tells us how the function \( f \) changes as \( x \) changes, with \( y \) held constant.
• Similarly, the partial derivative with respect to \( y \), \( \frac{\partial f}{\partial y} \), describes the change in \( f \) as \( y \) changes, with \( x \) held constant.

In our exercise's case, these derivatives help us comprehend the instantaneous rate of change for the function \( f(x, y) = e^{1-xy} \).Knowing these rates of change is crucial for finding the gradient vector, which is like a compass that points in the direction of the steepest ascent on the function's surface. The partial derivatives at a specific point, like Point \( P(1, 0, e) \), help form this gradient.

Directional Derivative

The directional derivative allows us to measure the rate of change of a function in any direction specified by a vector rather than just straight along the axes. This concept extends the idea of partial derivatives by incorporating any direction into the rate of change calculation.

• In mathematical terms, if \( \mathbf{u} \) is our direction vector, the directional derivative of a function \( f \) at a point \( P \) in the direction of \( \mathbf{u} \) is the dot product of the gradient and \( \mathbf{u} \).
• For a function \( f(x, y) \), this is represented by: \( abla f \cdot \mathbf{u} \)
• When this product equals zero, it indicates that the function has no change in that direction at the given point.

In our problem, we explore directions in the \( xy \)-plane where the change is zero. Setting up the directional derivative with zero shows us which directions lead to no alteration in the function's value at point \( P(1, 0, e) \). It turns out that this direction aligns with the x-axis,

Unit Vectors

Unit vectors play a vital role in representing directions in space. They have a magnitude of 1 and are used to indicate direction without implying any specific length.

• In the Cartesian plane, common unit vectors are \( \mathbf{i} \) and \( \mathbf{j} \), which point in the positive \( x \)-direction and \( y \)-direction, respectively.
• When we express a direction as a unit vector, we're essentially saying, "this is the direction, but of standard length 1," to simplify calculations and representations.

In our solution, we are interested in directions in which the function doesn't change, expressed in terms of unit vectors. As our calculations show, any direction parallel to the \( x \)-axis can be expressed in terms of the unit vector \( \mathbf{i} \). Thus, the direction vector that maintains zero change in the function \( f \) at point \( P \) must be a multiple of \( \mathbf{i} \). This clever use of unit vectors simplifies how we interpret and manipulate directional problems in multiple dimensions.