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A gradient vector is an important concept in multivariable calculus. It represents both the direction and rate of the steepest ascent of a function. For a function of two variables, like our function \( f(x, y) = x - 3y \), the gradient vector is composed of its partial derivatives with respect to each variable.
To find the gradient, denote it as \( abla f \), we calculate the partial derivatives \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \). The gradient vector is then \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \).

• Points in the direction of greatest increase of the function.
• Magnitude shows rate of increase.
• Orthogonal to level curves (or equipotential lines).

In our case, the gradient vector \( abla f = (1, -3) \) is constant, suggesting that the steepest ascent of the function \( f \) is in the direction \( (1, -3) \) across the entire plane.

Equipotential lines, also known as level curves, are lines where the function has a constant value \( c \). For example, if we have \( f(x, y) = x - 3y \), the equipotential lines are where \( x - 3y = c \). Solving this for \( y \), we get \( y = \frac{x - c}{3} \).

• Represent sets of points where the function value is the same.
• In our example, these are lines with a slope of \( \frac{1}{3} \), showing that the function remains constant along these lines.
• The gradient vector is always orthogonal to these lines, indicating the direction of greatest change perpendicular to the equipotential lines.

Understanding equipotential lines helps visualize potential changes in functions and how they interact with their gradients, providing insight into the function’s behavior over its domain.

Partial derivatives help us understand how a multivariable function changes when we tweak one variable at a time. For the function \( f(x, y) = x - 3y \), finding \( \frac{\partial f}{\partial x} \) involves treating \( y \) as a constant.
The process:

• Differentiate \( f(x, y) \) with respect to \( x \).
• The derivative of \( x \) is 1.
• The term \( -3y \) remains unaffected, being constant w.r.t. \( x \).

This yields \( \frac{\partial f}{\partial x} = 1 \), indicating that for each unit increase in \( x \), the function \( f(x, y) \) increases by one unit. This result showcases how changes in \( x \) alone influence the overall function value.

Just like partial differentiation with respect to \( x \), taking the partial derivative with respect to \( y \) allows us to see the effect of changing \( y \) while keeping \( x \) constant. Using our function \( f(x, y) = x - 3y \), we find \( \frac{\partial f}{\partial y} \).
Here's how it works:

• Treat \( x \) as a constant and differentiate with respect to \( y \).
• The derivative of \( -3y \) is \(-3\).
• Since \( x \) is constant in the differentiation process, its contribution is 0.

Thus, \( \frac{\partial f}{\partial y} = -3 \), indicating that for each unit increase in \( y \), the function \( f(x, y) \) decreases by three units. This negative sign reflects the negative contribution of \( y \) to the function's overall value when it increases.