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The gradient vector is a crucial concept in calculus and vector analysis. It represents the direction and rate of steepest ascent of a scalar function. For a function of two variables, such as \(f(x, y) = 13e^{xy}\), the gradient is a vector composed of the partial derivatives of the function with respect to each variable.

• Partial Derivative with respect to \(x\): Denoted as \(\frac{\partial f}{\partial x}\), this gives the rate of change of the function as \(x\) changes, while \(y\) remains constant. For our function, it's \(13ye^{xy}\).


• Partial Derivative with respect to \(y\): Denoted as \(\frac{\partial f}{\partial y}\), this measures the rate of change of the function as \(y\) changes, keeping \(x\) constant. For our function, it's \(13xe^{xy}\).

To find the gradient vector, \(abla f\), evaluate these partial derivatives at a given point. For the point \(P(1,0)\), substituting \(x = 1\) and \(y = 0\) gives the gradient \(abla f(1,0) = \langle 0, 13 \rangle\). This gradient indicates how \(f\) increases most sharply around \(P\).

Partial derivatives are fundamental in multivariable functions, representing the rate at which functions change in one direction while holding other variables constant. They're essential for calculating the gradient vector.For example, in \(f(x, y) = 13e^{xy}\):

• \(\frac{\partial f}{\partial x} = 13ye^{xy}\): This derivative tells us how \(f\) changes with \(x\). By taking the derivative, we see that if \(y\) is zero, this becomes \(0\).


• \(\frac{\partial f}{\partial y} = 13xe^{xy}\): This derivative reveals how the function changes with \(y\). At \(P(1, 0)\), this simplifies to \(13\) as \(e^{x \cdot 0} = 1\).

The derivatives let us assess how the function \(f\) responds to small changes in \(x\) and \(y\), individually. Understanding these changes helps in constructing the gradient vector and determining the function's behavior around a specific point.

A unit vector is a vector with a magnitude of 1. It is crucial for determining the direction of vectors and simplifying calculations without altering direction.To convert any vector, such as \(\langle 5, 12 \rangle\), into a unit vector, first find the vector's magnitude. The magnitude is calculated using:\[||\langle a, b \rangle|| = \sqrt{a^2 + b^2}\]For \(\langle 5, 12 \rangle\), it turns out to be \(13\). To form the unit vector:\[\text{Unit Vector} = \frac{1}{13} \langle 5, 12 \rangle = \langle \frac{5}{13}, \frac{12}{13} \rangle \]This unit vector maintains the direction of \(\langle 5, 12 \rangle\) but with a standard size, aiding in calculations like the directional derivative.

The dot product is a way to multiply two vectors, resulting in a scalar. It is especially useful in finding the projection of one vector onto another and calculating directional derivatives.For two vectors \(\mathbf{a} = \langle a_1, a_2 \rangle\) and \(\mathbf{b} = \langle b_1, b_2 \rangle\), their dot product is given by:\[\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2\]In the context of directional derivatives, we use the dot product to combine the gradient vector and the unit direction vector. For our example, given the gradient \(\langle 0, 13 \rangle\) and the unit vector \(\langle \frac{5}{13}, \frac{12}{13} \rangle\), the dot product becomes:\[0 \cdot \frac{5}{13} + 13 \cdot \frac{12}{13} = 12\]This scalar value, \(12\), represents the rate of change of \(f\) in the direction of \(\langle 5, 12 \rangle\) at the point \(P(1, 0)\). It tells us how much \(f\) increases in that specific direction, completing the directional derivative calculation.