91Ó°ÊÓ

A gradient is an essential concept in multivariable calculus, especially in the context of directional derivatives. The gradient of a function \( f(x, y) \) is a vector that contains all of its partial derivatives. This vector points in the direction of the greatest rate of increase of the function. For the function \( f(x, y) = x^2 + \sin(xy) \), the gradient \( abla f(x, y) \) consists of two components:

• \( \frac{\partial f}{\partial x} = 2x + y\cos(xy) \)
• \( \frac{\partial f}{\partial y} = x\cos(xy) \)

At the specific point of interest, which is \((1,0)\), the gradient becomes \((2,1)\). This tells us that at this point, the steepest ascent of the function is in the direction of the vector \( (2, 1) \). The size of the gradient also indicates how steep the hill is.

In mathematics, a unit vector is a vector with a length of exactly one unit. These vectors are crucial when discussing directional derivatives because they provide a directional sense without magnitude affecting the result. When you want to measure how a function, such as \( f(x, y) \), changes in a particular direction, you specify this direction with a unit vector \( \mathbf{u} = (u_1, u_2) \).The condition that \( \mathbf{u} \) is a unit vector can be written as \( u_1^2 + u_2^2 = 1 \). This ensures that the vector is normalized and is only providing directional data. In the context of our problem, the unit vectors \( (0,1) \) and \( \left( \frac{4}{5}, -\frac{3}{5} \right) \) were used. Both satisfy the unit vector condition since their lengths are both equal to 1.

Partial derivatives are the building blocks of gradients and are a way to see how a function changes as one of its variables changes while keeping the other variables constant. For the function \( f(x, y) = x^2 + \sin(xy) \), we compute partial derivatives with respect to each variable. They are:

• \( \frac{\partial f}{\partial x} = 2x + y\cos(xy) \)
• \( \frac{\partial f}{\partial y} = x\cos(xy) \)

This means that \( \frac{\partial f}{\partial x} \) tells us how \( f \) changes as \( x \) changes, with \( y \) constant, while \( \frac{\partial f}{\partial y} \) does the same as \( y \) changes, with \( x \) constant. These provide the components for the gradient \( abla f(x, y) \), enabling us to determine the rate of change in any direction.

Solving a system of equations is required when calculating directional derivatives in specific directions. In the given exercise, two equations need to be satisfied:

• The constraint of the directional derivative: \( 2u_1 + u_2 = 1 \)
• The unit vector condition: \( u_1^2 + u_2^2 = 1 \)

Together, these equations form a system that needs to be solved to find the directions contributing to the desired directional derivative of one. By substitution, \( u_2 \) is expressed as \( 1-2u_1 \), which simplifies solving the problem. This leads us to solve the quadratic equation \( 5u_1^2 - 4u_1 = 0 \), resulting in two valid solutions: \( u_1 = 0 \) and \( u_1 = \frac{4}{5} \). From these, we find \( u_2 \) values, identifying the specific unit vectors that result in the directional derivative being equal to one.