91Ó°ÊÓ

Partial derivatives are a fundamental concept in multivariable calculus, allowing us to understand how functions change with respect to one variable at a time, while keeping other variables constant.
Partial derivatives are denoted as \(\frac{\partial z}{\partial x}\) for derivatives concerning \(x\) and \(\frac{\partial z}{\partial y}\) for \(y\). Here, we are dealing with a function \(z = \frac{x - y}{x^2 + y^2}\), and we have found its partial derivatives.

• \(\frac{\partial z}{\partial x} = \frac{y^{2}-x^{2}+y}{(x^{2}+y^{2})^{2}}\) gives us information about how \(z\) changes when \(x\) is varied.
• \(\frac{\partial z}{\partial y} = \frac{-x^{2}+x+y^{2}}{(x^{2}+y^{2})^{2}}\) tells us how \(z\) changes with changes in \(y\).

By evaluating these derivatives at specific points, we gain valuable information about the behavior of \(z\) around those points, which is essential for constructing the tangent plane.

The gradient vector is a powerful tool that encapsulates the rate and direction of change of a function with respect to its variables.
For a function \(z = f(x, y)\), the gradient vector \(abla z\) is composed of its partial derivatives:
\(abla z = \left( \frac{\partial z}{\partial x}, \frac{\partial z}{\partial y} \right)\)
. This vector points in the direction of greatest increase of the function and its magnitude tells how steep that increase is.

The gradient vector is crucial when finding the equation of a tangent plane, as it is parallel to the normal vector of the tangent plane.

• In our example, the gradient vector at point \((1, 2)\) is \(\left( \frac{1}{3}, \frac{1}{3} \right)\).
• At point \((2, -1)\), it is \(\left( -\frac{1}{3}, \frac{1}{9} \right)\).

These vectors are calculated using the partial derivatives and are used to generate the normal vectors needed for the tangent planes.

Normal vectors are perpendicular to surfaces and are essential in defining a plane's orientation in space.
For a plane, the normal vector is of the form \((a, b, c)\). This is key when using the point-normal form to describe the tangent plane.

In the exercise, the normal vector is derived by extending the gradient vector to three dimensions:
\((\frac{\partial z}{\partial x}, \frac{\partial z}{\partial y}, -1)\).
This represents the direction perpendicular to the tangent plane.

• At \((1, 2, -\frac{1}{5})\), the normal vector is \(\vec{n_1} = \left(\frac{1}{3}, \frac{1}{3}, -1\right)\).
• At \((2, -1, \frac{3}{5})\), it is \(\vec{n_2} = \left(-\frac{1}{3}, \frac{1}{9}, -1\right)\).

These vectors directly define the orientation of the respective tangent planes.

The point-normal form is an elegant way of expressing the equation of a plane given a point on the plane and a normal vector.
This form is given by the equation:
\[a(x - x_0) + b(y - y_0) + c(z - z_0) = 0\]
Here, \((a, b, c)\) is the normal vector, and \((x_0, y_0, z_0)\) is a specific point on the plane.

In our tangent plane example, this allows us to use the normal vector calculated from the partial derivatives:

• For \((1, 2, -\frac{1}{5})\), the equation is \(\frac{1}{3}(x - 1) + \frac{1}{3}(y - 2) - (z + \frac{1}{5}) = 0\).
• And for \((2, -1, \frac{3}{5})\), it is \(-\frac{1}{3}(x - 2) + \frac{1}{9}(y + 1) - (z - \frac{3}{5}) = 0\).

Each tangent plane is uniquely defined by its normal vector and the point it touches on the surface, offering a geometric understanding of how the plane sits in space relative to the surface.