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The gradient vector of a function is a crucial tool in multivariable calculus. It provides a way to measure the rate of change of the function in space. For any function of several variables, such as \(f(x, y, z) = xy + z^2\), the gradient is a vector that consists of its partial derivatives.

• For the given function, the gradient \(abla f(x, y, z) = (y, x, 2z)\).
• This means for each variable \(x\), \(y\), and \(z\), you take a partial derivative.

The gradient vector \(abla f\) points in the direction of the greatest rate of increase of the function, and its magnitude tells us how steep the slope is in that direction.In the exercise, after computing the gradient vector, it's evaluated at the specific point \((1, 1, 1)\) to obtain \((1, 1, 2)\). This gives the direction and rate of change of the function at that point.

Partial derivatives help us understand how a multivariable function changes with respect to one variable, while the other variables are held constant. This is like slicing through one direction of the function and examining its incline or decline at a certain spot.For the function \( f(x, y, z) = xy + z^2 \):

• \( \frac{\partial f}{\partial x} = y \) tells us how \(f\) changes as \(x\) slightly changes, with \(y\) and \(z\) fixed.
• \( \frac{\partial f}{\partial y} = x \) shows \(f\)'s sensitivity to changes in \(y\), keeping other variables steady.
• \( \frac{\partial f}{\partial z} = 2z \) indicates how \(f\) changes with \(z\), observing no change in \(x\) and \(y\).

Calculating partial derivatives is a fundamental step in obtaining the gradient vector of a function. This makes them essential when exploring the directional derivatives.

A unit vector is crucial when dealing with directional derivatives. It provides direction while ensuring the length is exactly one. This normalization is critical because it helps isolate direction from magnitude.To find the unit vector, follow these steps:- Calculate the preliminary direction vector. In our problem, it is obtained from the difference \((5, -3, 3) - (1, 1, 1) = (4, -4, 2)\).- Compute the magnitude or length of this vector: \( \sqrt{4^2 + (-4)^2 + 2^2} = 6 \).- Divide each component by the magnitude to get a unit vector: \( \left( \frac{4}{6}, -\frac{4}{6}, \frac{2}{6} \right) = \left( \frac{2}{3}, -\frac{2}{3}, \frac{1}{3} \right) \).The result is a vector pointing in the same direction as the initial direction vector but standardized to a unit length.

The dot product operation is used to compute the directional derivative by relating two vectors. Specifically, it measures how much one vector goes in the direction of another. This is particularly useful when working with gradients and unit vectors.In this exercise, the dot product gives us the directional derivative:- Compute \( abla f(1, 1, 1) = (1, 1, 2) \) dot \( \left( \frac{2}{3}, -\frac{2}{3}, \frac{1}{3} \right) \).- Multiply corresponding components: \(1 \cdot \frac{2}{3} + 1 \cdot \left(-\frac{2}{3}\right) + 2 \cdot \frac{1}{3}\).- Add the results: \(\frac{2}{3} - \frac{2}{3} + \frac{2}{3} = \frac{2}{3}\).The dot product here shows how the gradient vector \((1, 1, 2)\) aligns with the unit vector, determining the rate of increase of \(f\) in the given direction.