91Ó°ÊÓ

Gradient vectors are essential in calculus for finding directions in multivariable functions. When dealing with functions like the one in this exercise, where the function is implicitly defined as \( F(x, y, z) = xy^2 - xz^2 = 0 \), the gradient vector \( abla F \) consists of all the partial derivatives of \( F \).

• The partial derivative with respect to \(x\) is \( \frac{\partial F}{\partial x} = y^2 - z^2 \).
• The partial derivative with respect to \(y\) is \( \frac{\partial F}{\partial y} = 2xy \).
• The partial derivative with respect to \(z\) is \(\frac{\partial F}{\partial z} = -2xz\).

The gradient vector at any given point provides a vector pointing in the steepest ascent direction. By plugging in the values of \(x, y,\) and \(z\) from point \(P = (2, 1, -1)\), you can evaluate the gradient to see how the function changes near that point. For this exercise, the gradient evaluated at \(P\) turns out to be \( \langle 0, 4, 4 \rangle \), indicating the directions in which the function increases or decreases are primarily in the \(y\) and \(z\) directions.

The tangent plane is a linear approximation of a surface at a given point, and its equation can be derived using the gradient vector. In our exercise, the equation of the tangent plane at point \(P\) is formed by using the dot product of the gradient vector evaluated at \(P\), \(abla F(2, 1, -1) = \langle 0, 4, 4 \rangle \), with a vector \((x-2, y-1, z+1)\) pointing from \(P\) to any point \((x, y, z)\) on the plane.To derive the equation, we set up the dot product expression: \[ abla F(2, 1, -1) \cdot (x - 2, y - 1, z + 1) = 0 \]When you expand and solve the dot product, you find: - \(0(x - 2)\) does not influence the outcome - \(4(y - 1) + 4(z + 1) = 0\), simplifying to \(4y + 4z = 0\), or simply \(y + z = 0\). This equation represents the tangent plane to the surface at point \(P = (2, 1, -1)\). It captures the first-order changes around \(P\), providing a good approximation to the actual surface there.

The normal line is perpendicular to the tangent plane at a given point on a surface. In this task, it passes through the point \(P\) and is aligned with the gradient vector \(abla F\) you've already calculated. The direction of the normal line is given by the vector \(\langle 0, 4, 4 \rangle\), calculated at \(P = (2, 1, -1)\). To express the normal line mathematically, we use a parametric form:

• \( x = 2 + 0t = 2 \)
• \( y = 1 + 4t \)
• \( z = -1 + 4t \)

Notice that the value of \(x\) is constant because its gradient component was zero; this means the direction is parallel to the \(y\) and \(z\) axes. By manipulating parameter \(t\), you can find any point along this normal line that extends infinitely in both directions from \(P\). Understanding the normal line is crucial for analyzing the orientation of a plane or surface.

Partial derivatives measure how a function changes as each of its input variables is varied, holding the other variables constant. They are the building blocks for understanding gradients, which are vectors of partial derivatives. For the implicit function given, \( F(x, y, z) = xy^2 - xz^2 \), we calculated the following partial derivatives:

• With respect to \(x: \frac{\partial F}{\partial x} = y^2 - z^2 \)
• With respect to \(y: \frac{\partial F}{\partial y} = 2xy \)
• With respect to \(z: \frac{\partial F}{\partial z} = -2xz \)

Partial derivatives are vital because they indicate how a function like \( F \) behaves when one variable changes and others stay fixed. For instance, \(\frac{\partial F}{\partial x}\) tells us how \( F \) changes as \( x \) changes, with \(y\) and \(z\) held constant. Evaluating these derivatives at a specific point gives components of the gradient vector that can guide derivations of the tangent plane and normal line, as shown in this exercise. Mastering partial derivatives gives insight into surface behavior and interactions between variables.