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The gradient vector plays a crucial role in finding the directional derivative of a function. It is formed by the partial derivatives of the function with respect to each variable. Essentially, the gradient vector points in the direction of the steepest ascent of the function from a given point.

For the function in the exercise, which is given by \( f(x, y) = x^2 + 3y^2 \), the gradient vector \( abla f \) is derived by calculating the partial derivatives:

• The partial derivative with respect to \( x \) is \( \frac{\partial f}{\partial x} = 2x \).


• The partial derivative with respect to \( y \) is \( \frac{\partial f}{\partial y} = 6y \).

Combining these, the gradient vector becomes \( abla f = (2x, 6y) \). By substituting the point \( P(1,1) \), we get \( abla f(1,1) = (2, 6) \). This vector provides information about how the function changes around the point \( P \). It is important as it points in the direction of greatest increase of the function value.

Partial derivatives are used to find the slope of a function in relation to each of its variables separately. Consider them like taking a slice through a multi-dimensional graph of the function, holding all other variables constant.

In our exercise, we deal with two variables, \( x \) and \( y \). To find out how the function \( f(x, y) = x^2 + 3y^2 \) changes with respect to these variables, we compute:

• The partial derivative \( \frac{\partial f}{\partial x} = 2x \) shows how the function changes as \( x \) changes while \( y \) is kept constant.


• The partial derivative \( \frac{\partial f}{\partial y} = 6y \) shows how the function changes as \( y \) changes while \( x \) is kept constant.

These derivatives help us construct the gradient vector, which is key to understanding the behavior of multivariable functions.

A unit vector is a vector with a length of one. It’s used to indicate direction without affecting magnitude. To convert any non-zero vector into a unit vector, you normalize it by dividing the vector by its magnitude.

For the direction vector \( \overrightarrow{PQ} = (3, 4) \) obtained in the exercise, its magnitude is calculated as \( \sqrt{3^2 + 4^2} = 5 \). The unit vector \( \mathbf{u} \) in the same direction is then \( \left(\frac{3}{5}, \frac{4}{5}\right) \).

This normalized vector is crucial in the context of directional derivatives because it ensures that only the direction affects the result of the derivative, not the magnitude of \( \overrightarrow{PQ} \) itself. This ensures that we measure the rate of change of the function in just the specified direction.

The dot product is a mathematical operation that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number. This operation is key when calculating directional derivatives.

In our exercise, after finding the gradient vector \( abla f = (2, 6) \) and the unit direction vector \( \mathbf{u} = \left(\frac{3}{5}, \frac{4}{5}\right) \), the dot product is computed as follows:
The dot product is given by \( abla f \cdot \mathbf{u} = 2 \cdot \frac{3}{5} + 6 \cdot \frac{4}{5} \). Calculating this, we find \( \frac{6}{5} + \frac{24}{5} = \frac{30}{5} = 6 \).

The result shows the rate at which the function \( f \) is increasing in the direction specified by the unit vector. The directional derivative thus offers critical insight into how the function behaves as we move along a particular direction.