Problem 15 - Advanced Engineering Mathematics

Question

Find an equation of the tangent plane to the graph of the given equation at the indicated point. $$ x^{2}+y^{2}+z^{2}=9 ;(-2,2,1) $$

Step-by-Step Solution

Answer

The equation of the tangent plane is \(-4x + 4y + 2z = 18\).

Step 1: Understand the Problem

We are required to find the equation of the tangent plane to the surface given by the equation \(x^2 + y^2 + z^2 = 9\) at the point \((-2, 2, 1)\). This type of problem involves understanding the application of partial derivatives in calculus.

Step 2: Differentiate the Given Equation

Firstly, consider the function \(F(x, y, z) = x^2 + y^2 + z^2 - 9\). The tangent plane at a point \((x_0, y_0, z_0)\) can be found using the gradient vector \(abla F(x_0, y_0, z_0)\). We need to find the partial derivatives \(\frac{\partial F}{\partial x}\), \(\frac{\partial F}{\partial y}\), and \(\frac{\partial F}{\partial z}\).

Step 3: Calculate Partial Derivatives

For \(F(x, y, z) \), the partial derivatives are calculated as follows:- \(\frac{\partial F}{\partial x} = 2x\),- \(\frac{\partial F}{\partial y} = 2y\),- \(\frac{\partial F}{\partial z} = 2z\).

Step 4: Evaluate Gradient at Given Point

Evaluate the gradient vector at the point \((-2, 2, 1)\):- \(\frac{\partial F}{\partial x}(-2, 2, 1) = 2(-2) = -4\),- \(\frac{\partial F}{\partial y}(-2, 2, 1) = 2(2) = 4\),- \(\frac{\partial F}{\partial z}(-2, 2, 1) = 2(1) = 2\).The gradient vector \(abla F(-2, 2, 1) = (-4, 4, 2)\).

Step 5: Formulate the Tangent Plane Equation

The general equation of the tangent plane is given by:\[\frac{\partial F}{\partial x}(x_0, y_0, z_0)(x - x_0) + \frac{\partial F}{\partial y}(x_0, y_0, z_0)(y - y_0) + \frac{\partial F}{\partial z}(x_0, y_0, z_0)(z - z_0) = 0\]Substituting our values, the equation becomes:\[-4(x + 2) + 4(y - 2) + 2(z - 1) = 0\]

Step 6: Simplify the Tangent Plane Equation

Distribute and combine like terms in the equation from Step 5:\[ -4x - 8 + 4y - 8 + 2z - 2 = 0 \]This simplifies to:\[ -4x + 4y + 2z = 18 \]

Key Concepts

Partial Derivatives

Partial derivatives help us to understand how a function changes when we vary one variable, keeping others constant. In multivariable calculus, a partial derivative of a function is the derivative with respect to one of those variables, treating all other variables as constants.
For the function in the problem, we consider it as:

• Function: \( F(x, y, z) = x^2 + y^2 + z^2 - 9 \)

To find a tangent plane, we begin by calculating the partial derivatives:

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

By using these derivatives, we can explore how the surface behaves as we move in different directions along the coordinate axes.

Gradient Vector

The gradient vector, \( abla F \), is a crucial concept when working with surfaces in multivariable calculus. Simply put, it is a vector composed of all partial derivatives of a function and points in the direction of the steepest increase of the function.
For our function \( F(x, y, z) = x^2 + y^2 + z^2 - 9 \), the gradient vector is:

• \( abla F = \left( \frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z} \right) = (2x, 2y, 2z) \)

At the point \( (-2, 2, 1) \), the gradient becomes:

• \( abla F(-2, 2, 1) = (-4, 4, 2) \)

This vector is perpendicular to the tangent plane at the point on the surface, essentially guiding us to the tangent plane equation.

Surface Equation

A surface equation in 3D geometry represents the set of all points \( (x, y, z) \) that satisfy a specific relationship. For our problem, the equation \( x^2 + y^2 + z^2 = 9 \) describes the surface of a sphere with a radius of 3 centered at the origin.
This means any point \( (x, y, z) \) lying on this surface maintains a constant distance from the origin, i.e., 3 units.

• The sphere's symmetry can reveal insights when evaluating derivatives or solving problems related to points on the surface.
• Knowing the equation helps us form the function \( F(x, y, z) \) used in the partial derivative calculations.

Realizing how these equations link to geometrical shapes can be a potent tool in understanding and manipulating complex surfaces.

Calculus Applications

In calculus, especially multivariable calculus, understanding how to apply derivative concepts to solve real-world problems opens vast potential in scientific and engineering fields.
The tangent plane, in our context, provides a linear approximation of the surface at a point. This is extremely useful because it simplifies a complex problem involving curved surfaces in a localized area.

• Applications of this include optimizing functions where determining local maxima or minima is necessary.
• It can also apply in navigation, where linear approximations are needed for non-linear pathways.

By mastering how to calculate and use the tangent plane, you not only solve mathematical problems but also apply these principles to practical challenges, such as computer graphics or physics simulations.