91Ó°ÊÓ

When calculating a gradient, we are dealing with a vector that points in the direction of the greatest rate of increase of a function. For a function of two variables, like the one given in the exercise, the gradient is

• an ordered pair of partial derivatives.

A partial derivative is taken with respect to one variable while keeping the others constant.
In the formula for the gradient, \(abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)\), each component represents a derivative showing how the function changes along each axis.
In our example of \( f = e^x \cos(y) \), the gradient becomes \( abla f = ( e^x \cos(y), -e^x \sin(y) ) \).
This gives the rise of the function in both the x and y directions.

Partial derivatives focus on the change of multivariable functions with respect to one variable at a time. Consider how \( f(x, y) = e^x \cos(y) \) is another such function. Here, to find \( \frac{\partial f}{\partial x} \), we differentiate only with respect to \( x \), treating \( y \) as a constant. Similarly, the partial with respect to \( y \) requires differentiating in terms of \( y \) treating \( x \) as a constant:

• \( \frac{\partial f}{\partial x} = e^x \cos(y) \)
• \( \frac{\partial f}{\partial y} = -e^x \sin(y) \)

These derivatives are the building blocks for constructing the gradient, guiding us when understanding directional shifts.

The dot product is a method used to find the directional derivative. It is a scalar result measuring how two vectors align.
The operation is \(\mathbf{A} \cdot \mathbf{B} = A_1B_1 + A_2B_2\) for vectors of 2 dimensions. It is simple: multiply and sum corresponding components from each vector. For the directional derivative, choose \( abla f \) and the unit direction vector \( \mathbf{u} \).
At this point in our solution, calculate \(abla f(1, \pi/4) = \left( e \frac{\sqrt{2}}{2}, -e \frac{\sqrt{2}}{2} \right)\), and the direction \( \mathbf{u} = \left( \frac{\sqrt{3}}{2}, \frac{1}{2} \right) \). A dot product gives the desired directional derivative.

In mathematics, trigonometric functions relate angles of triangles to sides' ratios, pivotal in providing direction vectors. These functions: cosine, sine, and tangent are essential in converting angles into useful numerical values.
For our angle \(30^\circ \), knowing its trigonometric function values

• \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \)
• \( \sin(30^\circ) = \frac{1}{2} \)

helps define the direction vector \( \mathbf{u} = \left( \frac{\sqrt{3}}{2}, \frac{1}{2} \right) \). Trigonometric functions thus transform angular data into forms ideal for vector calculations.