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When we want to understand how a function changes at a specific point, we use partial derivatives. Partial derivatives help us see how a function varies as we tweak one variable while keeping the other fixed. It’s like seeing how a plant reacts to changes in sunlight, assuming water levels stay constant.

Take the function \( f(x, y) = 4x^3y^2 \). If we want to know how \( f \) changes when only \( x \) changes, we find the partial derivative with respect to \( x \), denoted \( \frac{\partial f}{\partial x} \). For our function, we differentiate \( 4x^3y^2 \) with respect to \( x \) and get \( 12x^2y^2 \).

Similarly, the partial derivative with respect to \( y \), \( \frac{\partial f}{\partial y} \), shows how \( f \) changes with shifts in \( y \). We differentiate \( 4x^3y^2 \) by \( y \) to get \( 8x^3y \). By calculating these, we're building up to finding the gradient, a vector showing the direction of greatest increase in \( f \).

A unit vector plays a fundamental role in identifying direction without considering magnitude. Think of it as a pointer showing direction alone. It has a magnitude of 1, like pointing to a place without specifying how far it is.

Finding a unit vector involves taking a vector and scaling it down to a length of 1. In our case, after calculating the gradient which is \((12, -8)\), the job is to convert this into a unit vector. We do this by dividing each component of the vector by its magnitude.

• The magnitude of the vector \((12, -8)\) is \(\sqrt{144 + 64} = 4\sqrt{13}\).
• To scale down, divide 12 by \(4\sqrt{13}\) and -8 by \(4\sqrt{13}\).

This gives us the unit vector \(\left( \frac{3}{\sqrt{13}}, \frac{-2}{\sqrt{13}} \right)\). This unit vector tells us the direction in which the function \( f \) increases most rapidly, without multiplying that direction by any distance.

The rate of change refers to how quickly a function’s value is increasing (or decreasing) as you move in a certain direction. Imagine flying a drone uphill where the incline determines how quickly you gain altitude.

For the function \( f(x, y) = 4x^3y^2 \), the rate of change at a point \( P \), in the direction of the steepest ascent, is given by the magnitude of the gradient vector at that point.

We calculated the gradient at \( P(-1,1) \) to be \((12, -8)\). The magnitude of this vector, \(4\sqrt{13}\), tells us how sharply \( f \) increases in that direction. This is because the gradient vector not only shows the direction but also scales by how intense the increase is. So, \(4\sqrt{13}\) represents the fastest speed of increase for \( f \) at point \( P \).

• A larger magnitude means a steeper increase.
• The sign doesn't matter for magnitude since we're measuring speed, not direction.