Question

Directions of zero change 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. $$f(x, y)=\sqrt{3+2 x^{2}+y^{2}} ; P(1,-2,3)$$

Short answer

Answer: The direction of zero change in the xy-plane for the function at point P is given by the unit vector $$\hat{v} = \frac{1}{\sqrt{2}}(1, 1)$$

Step-by-step solution

Find the gradient of the function

First, we need to find the gradient of the function. For a function of two variables, the gradient is: $$\nabla f(x,y) = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right)$$ We will calculate the partial derivatives of the function.

Calculate partial derivatives

The partial derivatives with respect to x and y are: $$\frac{\partial f}{\partial x} = \frac{4x}{\sqrt{3+2x^2+y^2}} \quad \text{and} \quad \frac{\partial f}{\partial y} = \frac{2y}{\sqrt{3+2x^2+y^2}}$$

Evaluate the gradient at point P

Next, we need to evaluate the gradient at the given point P(1,-2,3). $$\nabla f(1, -2) = \left(\frac{4(1)}{\sqrt{3+2(1)^2+(-2)^2}}, \frac{2(-2)}{\sqrt{3+2(1)^2+(-2)^2}}\right) = \left(\frac{4}{\sqrt{11}}, \frac{-4}{\sqrt{11}}\right)$$

Determine the direction vector

Let's use the vector \(\vec{v}=(a, b)\) as the direction vector. The condition for zero change is the gradient and direction vector being orthogonal (having a null dot product). Therefore: $$\nabla f(1, -2) \cdot \vec{v} = 0$$ Substituting the values we found earlier: $$\left(\frac{4}{\sqrt{11}}, \frac{-4}{\sqrt{11}}\right) \cdot (a, b) = \frac{4}{\sqrt{11}}a + \frac{-4}{\sqrt{11}}b = 0$$

Express the direction vector in terms of unit vectors

To express the direction vector in terms of unit vectors, let's set a = 1 and find the value of b: $$\frac{4}{\sqrt{11}}(1) + \frac{-4}{\sqrt{11}}b = 0$$ Solving for b, we get b = 1. So, the direction vector is \(\vec{v}=(1, 1)\). Now we need to convert it to a unit vector. $$\hat{v} = \frac{\vec{v}}{\|\vec{v}\|} = \frac{(1, 1)}{\sqrt{1^2 + 1^2}} = \frac{1}{\sqrt{2}}(1, 1)$$ The direction of zero change in the xy-plane for the function at point P is given by the unit vector $$\hat{v} = \frac{1}{\sqrt{2}}(1, 1)$$

Key concepts

Gradient of a Function

Understanding the gradient of a function is crucial when analyzing how a function behaves in space. In essence, the gradient points in the direction of the steepest ascent within a function's graph. It is represented as a vector made up of partial derivatives of the function with respect to each of its variables.

For instance, in a two-dimensional space for a function f(x, y), the gradient would be written as abla f(x,y) = \bigg(\frac{\(\partial f\)}{\partial x}, \frac{\(\partial f\)}{\partial y}\bigg), where each component of this vector signifies how much f changes with a small change in either x or y while keeping the other constant. If you move in the direction of the gradient, you are moving in the direction of greatest increase of the function's value.

Partial Derivatives

Partial derivatives are the building blocks of the gradient. A partial derivative of a function of several variables is its derivative with respect to one of those variables, with the others held constant. In simpler terms, they measure the rate at which the function changes as one of its input variables changes, while the other input variables stay fixed.

As seen in the given exercise, the computation involved \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\) for the function f(x, y). These represent how f changes in the x-direction and y-direction respectively. The partial derivative with respect to x is a measure of how much the function's output changes when x is varied, and similarly for y with \(\frac{\partial f}{\partial y}\).

Dot Product

The dot product is a way to multiply two vectors that results in a scalar (a single number). The formula for the dot product of two vectors \(\vec{a}\) and \(\vec{b}\) is \(\vec{a} \(\cdot\) \vec{b} = a_1b_1 + a_2b_2 + \(\dots\) + a_nb_n\), where \(a_1\) to \(a_n\) and \(b_1\) to \(b_n\) are the components of the vectors.

In terms of geometry, the dot product can tell us about the angle between two vectors. If the dot product of two nonzero vectors is zero, the vectors are orthogonal, meaning they are at a right angle to each other. This concept is exactly what was used in the step-by-step solution to find directions of zero change, where the gradient vector's dot product with the direction vector must be zero.

Unit Vector

A unit vector is a vector that has a magnitude (length) of 1. It effectively provides the direction of a vector without regard to its magnitude. When working in a coordinate system, common unit vectors used as references are \(\hat{i}\), \(\hat{j}\), and \(\hat{k}\), which point along the x, y, and z-axis respectively.

To turn any non-zero vector into a unit vector, you divide the vector by its own magnitude. The exercise you encountered showed converting a direction vector into a unit vector by dividing by its magnitude \(\|\vec{v}\|\). This is exactly what gives us \(\hat{v}\), the direction of zero change as a unit vector. Using unit vectors helps standardize directions so that they are easier to compare and apply in different situations.