Problem 2 - The Calculus with Analytic Geometry

Question

In Exercises 1 through 12 , find an equation of the tangent plane and equations of the normal line to the given surface at the indicated point. \(4 x^{2}+y^{2}+2 z^{2}=26 ;(1,-2,3)\)

Step-by-Step Solution

Answer

Tangent plane: 8x - 4y + 12z - 52 = 0Normal line: x = 1 + 8t, y = -2 - 4t, z = 3 + 12t

Step 1 - Find the gradient of the surface function

Given the surface function: 4x^{2} + y^{2} + 2z^{2} = 26 We need to find the partial derivatives with respect to x, y, and z to get the gradient vector, ∇F = ( ∂F/∂x, ∂F/∂y, ∂F/∂z )

Step 2 - Calculate partial derivatives

Compute ∂F/∂x: ∂/∂x (4x^{2} + y^{2} + 2z^{2}) = 8x Compute ∂F/∂y: ∂/∂y (4x^{2} + y^{2} + 2z^{2}) = 2y Compute ∂F/∂z: ∂/∂z (4x^{2} + y^{2} + 2z^{2}) = 4z

Step 3 - Evaluate the gradient at the given point

Given point: (1, -2, 3), Evaluate the gradient vector at this point: ∇F(1, -2, 3) = (8*1, 2*(-2), 4*3) = (8, -4, 12)

Step 4 - Write the equation of the tangent plane

The equation of the tangent plane to the surface at (1, -2, 3) can be written as: (∂F/∂x)|_{(1, -2, 3)} (x - 1) + (∂F/∂y)|_{(1, -2, 3)} (y + 2) + (∂F/∂z)|_{(1, -2, 3)} (z - 3) = 0 Substituting the values: 8(x - 1) - 4(y + 2) + 12(z - 3) = 0 Simplify the equation: 8x - 8 - 4y - 8 + 12z - 36 = 0 8x - 4y + 12z - 52 = 0

Step 5 - Write the equations of the normal line

The normal line can be parameterized using a direction vector (∇F(1, -2, 3)) and the given point: (1, -2, 3). So, the parametric equations are: x = 1 + 8t y = -2 - 4t z = 3 + 12t

Key Concepts

partial derivatives

Understanding partial derivatives is essential for finding the tangent plane and normal line of a surface. A partial derivative measures how a function changes as we vary one variable while keeping the others constant. For our surface equation, let's find the partial derivatives with respect to x, y, and z.
For the given function \[4x^{2} + y^{2} + 2z^{2} = 26\],
the partial derivatives are calculated as follows:
\[\frac{\text{∂F}}{\text{∂x}} = 8x, \frac{\text{∂F}}{\text{∂y}} = 2y, \frac{\text{∂F}}{\text{∂z}} = 4z\]
Evaluating these at the point \((1, -2, 3)\), we get:

• \(\frac{\text{∂F}}{\text{∂x}}|_{(1, -2, 3)} = 8 \times 1 = 8\)
• \(\frac{\text{∂F}}{\text{∂y}}|_{(1, -2, 3)} = 2 \times -2 = -4\)
• \(\frac{\text{∂F}}{\text{∂z}}|_{(1, -2, 3)} = 4 \times 3 = 12\)

These partial derivatives help us form the gradient vector, which is crucial for determining the tangent plane and normal line.

gradient vector

The gradient vector of a function gives the direction of the steepest ascent at any point on a surface. This vector contains the partial derivatives of the function with respect to each variable.
For our function \[4x^{2} + y^{2} + 2z^{2} = 26\],
we denote the gradient vector as\(abla F = \bigg(\frac{\text{∂F}}{\text{∂x}}, \frac{\text{∂F}}{\text{∂y}}, \frac{\text{∂F}}{\text{∂z}}\bigg)\).
By evaluating the partial derivatives at the point \((1, -2, 3)\), we have:
\(abla F(1, -2, 3) = (8, -4, 12)\).
This gradient vector is perpendicular to the tangent plane at the given point and serves as the direction vector for the normal line.

normal line parameterization

To find the equations of the normal line, we use the gradient vector as the direction vector. The normal line passes through the given point and extends in the direction of the gradient vector.
The given point is \((1, -2, 3)\) and the gradient (direction) vector is \((8, -4, 12)\).

• The parametric equation for x is: \(x = 1 + 8t\)
• The parametric equation for y is: \(y = -2 - 4t\)
• The parametric equation for z is: \(z = 3 + 12t\)

Here, \(t\) is the parameter that allows us to find any point on the normal line. By substituting different values of \(t\), we get different points on this line.
These parametric equations are helpful because they clearly define the path of the normal line using simple linear functions.