Problem 21 - APEX Calculus

Question

An implicitly defined function of \(x, y\) and \(z\) is given along with a point \(P\) that lies on the surface. Use the gradient \(\nabla F\) to: (a) find the equation of the normal line to the surface at \(P,\) and (b) find the equation of the plane tangent to the surface at \(P\). \(\frac{x^{2}}{8}+\frac{y^{2}}{4}+\frac{z^{2}}{16}=1,\) at \(P=(1, \sqrt{2}, \sqrt{6})\)

Step-by-Step Solution

Answer

Normal line: \(x = 1 + \frac{1}{4}t, y = \sqrt{2} + \frac{\sqrt{2}}{2}t, z = \sqrt{6} + \frac{\sqrt{6}}{8}t\). Tangent plane: \(\frac{x}{4} + \frac{y\sqrt{2}}{2} + \frac{z\sqrt{6}}{8} = 1\).

Step 1: Find Partial Derivatives

Given the function \(F(x, y, z) = \frac{x^2}{8} + \frac{y^2}{4} + \frac{z^2}{16} - 1\), we need to find the partial derivatives to compute the gradient. Compute: \[ \frac{\partial F}{\partial x} = \frac{x}{4}, \quad \frac{\partial F}{\partial y} = \frac{y}{2}, \quad \text{and} \quad \frac{\partial F}{\partial z} = \frac{z}{8}. \]

Step 2: Evaluate Gradient at Point P

Substitute the coordinates of point \(P = (1, \sqrt{2}, \sqrt{6})\) into the partial derivatives to find the gradient at \(P\).\[ abla F(1, \sqrt{2}, \sqrt{6}) = \left( \frac{1}{4}, \frac{\sqrt{2}}{2}, \frac{\sqrt{6}}{8} \right) \]

Step 3: Equation of the Normal Line

The equation of a normal line at the point \(P\) can be derived using the gradient. It is given by:\[ x = 1 + \frac{1}{4}t, \quad y = \sqrt{2} + \frac{\sqrt{2}}{2}t, \quad z = \sqrt{6} + \frac{\sqrt{6}}{8}t. \]

Step 4: Equation of the Tangent Plane

The equation of the tangent plane can be determined using the gradient evaluated at \(P\). It is:\[ \frac{1}{4}(x - 1) + \frac{\sqrt{2}}{2}(y - \sqrt{2}) + \frac{\sqrt{6}}{8}(z - \sqrt{6}) = 0. \]Simplifying, we get:\[ x(\frac{1}{4}) + y(\frac{\sqrt{2}}{2}) + z(\frac{\sqrt{6}}{8}) = 1. \]

Key Concepts

Gradient

In multivariable calculus, the gradient vector is a vital tool that helps understand how a function changes at different points in space. The gradient, denoted by \( abla F \), is a vector comprised of the partial derivatives of a function. When we encounter a function like \( F(x, y, z) \), the gradient vector is defined as:

• \( abla F = \left( \frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z} \right) \).

Here, the gradient points in the direction of the greatest rate of increase of the function.
In our exercise, calculating the gradient at point \( P \) helps us find both the direction of the normal line and the tangent plane. These gradients give insight into how the surface behaves near the point \( P \).

The calculated gradient \( abla F(1, \sqrt{2}, \sqrt{6}) = \left( \frac{1}{4}, \frac{\sqrt{2}}{2}, \frac{\sqrt{6}}{8} \right) \) shows the direction at which changes occur fastest at that point on the surface.

Partial Derivatives

Partial derivatives are the derivatives of a function with respect to one variable, treating other variables as constants. They provide crucial information about how a multivariable function changes as each variable is altered independently. For the function \( F(x, y, z) = \frac{x^2}{8} + \frac{y^2}{4} + \frac{z^2}{16} - 1 \):

• \( \frac{\partial F}{\partial x} = \frac{x}{4} \)
• \( \frac{\partial F}{\partial y} = \frac{y}{2} \)
• \( \frac{\partial F}{\partial z} = \frac{z}{8} \)

These calculations help us construct the gradient.
At point \( P = (1, \sqrt{2}, \sqrt{6}) \), the partial derivatives give us specific rates of change for each variable at that point, helping in the formation of the tangent plane and normal line.
Understanding individual effects is key to grasping how multivariable systems operate.

Tangent Plane

The tangent plane to a surface at a point provides an approximate "flat" representation of the surface near that point. It's like laying a piece of paper on the surface at point \( P \). The gradient vector, \( abla F \), is perpendicular to this plane.
For the given problem, the tangent plane equation is derived from:

• \( \frac{1}{4}(x - 1) + \frac{\sqrt{2}}{2}(y - \sqrt{2}) + \frac{\sqrt{6}}{8}(z - \sqrt{6}) = 0 \)

Simplifying results in the equation:

• \( x\left(\frac{1}{4}\right) + y\left(\frac{\sqrt{2}}{2}\right) + z\left(\frac{\sqrt{6}}{8}\right) = 1 \)

This tangent plane equation allows us to understand how closely the surface resembles a plane at that specific point, and gives a local linear approximation of the surface.

Normal Line

The normal line represents a line that is perpendicular to the surface at a given point. In multivariable calculus, it provides an understanding of how the surface extends or sharpens away from that point.
For a surface described by a function, the gradient \( abla F \) at a point denotes the direction of the normal line.
For our exercise, the normal line at point \( P \) is given by:

• \( x = 1 + \frac{1}{4}t \)
• \( y = \sqrt{2} + \frac{\sqrt{2}}{2}t \)
• \( z = \sqrt{6} + \frac{\sqrt{6}}{8}t \)

The parameter \( t \) progresses along the line in both the positive and negative directions, indicating how the surface projects outward or inward.
The normal line is crucial in understanding the orientation and behavior of the surface relative to a specific location.