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The concept of a gradient is essential in understanding changes to a multivariable function. Think of the gradient as a multi-dimensional generalization of the derivative. It provides a vector that points in the direction of the steepest ascent of the function.
When we talk about gradients, we're referring to the vector that is composed of all partial derivatives of a function. For a function like \( g(x, y) = \ln\left(4 + x^2 + y^2\right) \), finding the gradient involves calculating the partial derivatives with respect to each variable involved - in this case, \( x \) and \( y \). This is why the gradient is sometimes represented as \( abla g(x, y) \).

• The gradient vector \( abla g(x, y) \) for our function is found based on the individual partial derivatives.
• Mathematically, it is expressed as \( abla g(x, y) = \left\langle \frac{2x}{4+x^2+y^2}, \frac{2y}{4+x^2+y^2} \right\rangle \).

This vector not only tells you the direction of the steepest increase but also its magnitude, which is the rate of the function's change.

Partial derivatives are a crucial tool in multivariable calculus. They allow us to focus on how a function changes with respect to one variable, while keeping the others constant.
In our function \( g(x, y) = \ln(4 + x^2 + y^2) \), the objective was to find how \( g \) changes as \( x \) or \( y \) changes on its own. This is achieved with partial derivatives, denoted as \( \frac{\partial}{\partial x} \) and \( \frac{\partial}{\partial y} \).

• For the partial derivative with respect to \( x \), apply the chain rule leading to \( \frac{2x}{4+x^2+y^2} \).
• Similarly, for \( y \), the derivative is \( \frac{2y}{4+x^2+y^2} \).

These calculations show how changes in each individual variable affect the function as a whole. Such insights are critical for optimization and modeling processes involving multiple variables.

The dot product, also called the scalar product, is a way to multiply two vectors, resulting in a single number (scalar). This is particularly useful when we are dealing with directional derivatives, as it helps to measure how much of one vector goes in the direction of another.
To find the directional derivative, calculate the dot product of the gradient vector with the unit direction vector. For our example, the directional derivative is the dot product \( abla g(-1,2) \cdot \mathbf{u} \).

• The gradient at a given point \((-1,2)\) is \( \left\langle -\frac{1}{3}, \frac{2}{3} \right\rangle \).
• The unit vector corresponding to \( \langle 2, 1 \rangle \) is \( \left\langle \frac{2}{\sqrt{5}}, \frac{1}{\sqrt{5}} \right\rangle \).
• When the dot product is \( -\frac{2}{3\sqrt{5}} + \frac{2}{3\sqrt{5}} = 0 \), it indicates that there is no change in the function in that direction at \( P(-1, 2) \).

The dot product's result can inform you whether a change in the function is happening in the considered direction. A result of zero, as in our case, indicates a balance where no increase or decrease occurs in that particular direction.