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The gradient of a function is like a compass for finding the steepest path up a hill. For functions with several variables, the gradient is a vector that points towards the direction of greatest increase. It combines all partial derivatives of the function into a single vector. Each component of this gradient vector corresponds to how much the function changes if you nudge a variable while keeping others constant.

In mathematical terms, if you have a function \( f(x, y, z) \), its gradient \( abla f \) is given by:

• The partial derivative with respect to \( x \), noted as \( f_x \)
• The partial derivative with respect to \( y \), noted as \( f_y \)
• The partial derivative with respect to \( z \), noted as \( f_z \)

In this problem, we calculated the gradient \( abla f \) as the vector \( \left( \frac{1}{8}, \frac{1}{4}, \frac{3}{8} \right) \) at point \( P \). This vector directly gives us our direction guide to where \( f(x, y, z) \) grows fastest.

Partial derivatives are a crucial concept when dealing with multivariable functions. They represent how a function changes when we alter one variable, keeping the others constant.

For the function \( f(x, y, z) = \sqrt{2x + 4y + 6z} \), we calculated:

• \( f_x = \frac{1}{\sqrt{2x + 4y + 6z}} \)
• \( f_y = \frac{2}{\sqrt{2x + 4y + 6z}} \)
• \( f_z = \frac{3}{\sqrt{2x + 4y + 6z}} \)

These partial derivatives provide insights into the function’s sensitivity to changes in \( x \), \( y \), and \( z \), respectively. By considering these, we form the components of the gradient.

The magnitude of a vector gives you the length or size of the vector. In our context, finding the magnitude of the gradient vector gives the maximum rate of increase of the function, known as the directional derivative.

To find the magnitude of a vector \( \mathbf{v} = (a, b, c) \), we calculate:\[ \| \mathbf{v} \| = \sqrt{a^2 + b^2 + c^2} \]For \( abla f(P) = \left( \frac{1}{8}, \frac{1}{4}, \frac{3}{8} \right) \), its magnitude is:\[ \sqrt{\left(\frac{1}{8}\right)^2 + \left(\frac{1}{4}\right)^2 + \left(\frac{3}{8}\right)^2} = \frac{1}{2} \]This result means that \( \frac{1}{2} \) is the upper bound for how steeply the function \( f \) increases at point \( P \).

Normalization transforms a vector to have a length of one while maintaining its direction. To normalize a vector \( \mathbf{v} = (a, b, c) \) with magnitude \( \| \mathbf{v} \| \), divide each component by its magnitude.In this exercise, we wanted the direction of maximum increase in the function. This is the direction of the gradient provided it’s a unit vector. For \( abla f(P) = \left( \frac{1}{8}, \frac{1}{4}, \frac{3}{8} \right) \):

• Calculate \( \| abla f(P) \| = \frac{1}{2} \)
• Normalize by dividing each component by \( \frac{1}{2} \)

This gives us the normalized vector \( \left( \frac{1}{4}, \frac{1}{2}, \frac{3}{4} \right) \). This unit vector shows precisely the way to proceed when aiming for the steepest increase of the function from point \( P \).