91Ó°ÊÓ

The gradient vector is a crucial concept in multivariable calculus and vector calculus. It is a vector of all the partial derivatives of a function. For a function of three variables, such as our function \(f(x, y, z) = xy^2 + yz\), the gradient vector is written as \(abla f = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\right)\). The gradient vector points in the direction of the greatest rate of increase of the function. In the exercise, we found the partial derivatives and constructed the gradient vector: \(abla f = (1, 4, 1)\). This gradient vector represents how the function changes at the point (1, 1, 2).

Partial derivatives are used to find how a function changes as each individual variable changes, while keeping the other variables constant. In the given problem, we have the function \(f(x, y, z) = xy^2 + yz\). To find the partial derivatives, we compute:

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

These derivatives give us rates of change in the directions of \(x\), \(y\), and \(z\) respectively. Evaluating these at the point (1,1,2), we get:

• \(\frac{\partial f}{\partial x} = 1\)
• \(\frac{\partial f}{\partial y} = 4\)
• \(\frac{\partial f}{\partial z} = 1\)

Vector normalization is the process of converting a vector into a unit vector, which has a magnitude of 1. This is important for directional derivatives because we usually want to find the rate of change of a function in a specific direction without changing its magnitude. The given direction vector is \(2i - j + 2k\). To normalize it, we first find its magnitude:
\[ |v| = \sqrt{2^2 + (-1)^2 + 2^2} = 3 \]
To get the unit vector, we divide each component by this magnitude:
\(\frac{2}{3}i - \frac{1}{3}j + \frac{2}{3}k\).
This gives us the normalized vector, which is then used in the dot product with the gradient vector to find the directional derivative.

The dot product is a way to multiply two vectors that gives a scalar (number) as a result. It is calculated by multiplying corresponding components and then summing these products. For the vectors \(\boldsymbol{a} = (a_1, a_2, a_3)\) and \(\boldsymbol{b} = (b_1, b_2, b_3)\), the dot product is:
\[ \boldsymbol{a} \cdot \boldsymbol{b} = a_1 b_1 + a_2 b_2 + a_3 b_3 \]
In our exercise, we computed the dot product of the gradient vector \((1, 4, 1)\) and the normalized direction vector \((\frac{2}{3}, -\frac{1}{3}, \frac{2}{3})\):
\[ 1 \times \frac{2}{3} + 4 \times -\frac{1}{3} + 1 \times \frac{2}{3} = 2/3 - 4/3 + 2/3 = 0 \]
This result is the directional derivative, which in this case, turns out to be zero, indicating no change in the function in that specific direction.