Problem 19 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the parametric equations of the line that is tangent to the curve of intersection of the surfaces $$ f(x, y, z)=9 x^{2}+4 y^{2}+4 z^{2}-41=0 $$ and $$ g(x, y, z)=2 x^{2}-y^{2}+3 z^{2}-10=0 $$ at the point \((1,2,2)\). Hint: This line is perpendicular to \(\nabla f(1,2,2)\) and \(\nabla g(1,2,2)\).

Step-by-Step Solution

Answer

The parametric equations are: \(x = 1 + 256t\), \(y = 2 - 176t\), \(z = 2 - 128t\).

Step 1: Compute the Gradient of f

The gradient of a function \( f(x, y, z) \) is given by \( abla f(x, y, z) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) \). For \( f(x, y, z) = 9x^2 + 4y^2 + 4z^2 - 41 \), compute the gradient: \[ abla f(x, y, z) = (18x, 8y, 8z) \].Substitute the point \( (1, 2, 2) \):\[ abla f(1, 2, 2) = (18, 16, 16) \].

Step 2: Compute the Gradient of g

The gradient of the function \( g(x, y, z) \) is given by \( abla g(x, y, z) = \left( \frac{\partial g}{\partial x}, \frac{\partial g}{\partial y}, \frac{\partial g}{\partial z} \right) \). For \( g(x, y, z) = 2x^2 - y^2 + 3z^2 - 10 \), compute the gradient: \[ abla g(x, y, z) = (4x, -2y, 6z) \].Substitute the point \( (1, 2, 2) \):\[ abla g(1, 2, 2) = (4, -4, 12) \].

Step 3: Find the Direction Vector of the Tangent Line

The direction vector of the tangent line at the point \((1,2,2)\) is perpendicular to both \(abla f(1, 2, 2)\) and \(abla g(1, 2, 2)\). To find this vector, compute the cross product of \(abla f(1, 2, 2)\) and \(abla g(1, 2, 2)\):\[ \mathbf{v} = abla f \times abla g = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 18 & 16 & 16 \ 4 & -4 & 12 \end{vmatrix} \].Calculating the determinant, we get:\[ \mathbf{v} = \langle 256, -176, -128 \rangle \].

Step 4: Express the Parametric Equations

The parametric equations of a line with a point \((x_0, y_0, z_0)\) and direction vector \(\mathbf{v} = \langle a, b, c \rangle\) are:\[ x = x_0 + at, \quad y = y_0 + bt, \quad z = z_0 + ct \].Substituting \((x_0, y_0, z_0) = (1, 2, 2)\) and the direction vector \(\langle 256, -176, -128 \rangle\):\[ x = 1 + 256t \]\[ y = 2 - 176t \]\[ z = 2 - 128t \].

Key Concepts

Gradient

When working with functions of several variables, the gradient plays a crucial role in understanding how the function behaves at any point in its domain. For a three-variable function like \( f(x, y, z) \), the gradient, denoted as \( abla f(x, y, z) \), is a vector composed of the first partial derivatives of the function with respect to each variable:

• \( \frac{\partial f}{\partial x} \) gives us the rate of change of the function in the \(x\)-direction.
• \( \frac{\partial f}{\partial y} \) indicates how the function changes with respect to \(y\).
• \( \frac{\partial f}{\partial z} \) reveals the change along the \(z\) direction.

By calculating these components, the gradient vector provides a direction in which the function increases most rapidly.
In our problem, we've calculated the gradients for two functions at the point (1, 2, 2). These vectors are \( abla f = (18, 16, 16) \) and \( abla g = (4, -4, 12) \), which highlight the directions of steepest ascent for \( f \) and \( g \) respectively.
The tangent line will be perpendicular to these gradients, as they represent the normal direction to the surfaces at the point of interest.

Cross Product

The cross product is a mathematical operation used in vector algebra, ideal for finding a vector that is perpendicular to two given vectors in three-dimensional space. This property is particularly useful when dealing with problems involving tangent lines to curves formed by the intersection of two surfaces.
To find the direction vector of the tangent line, we compute the cross product of the gradients \( abla f \) and \( abla g \). The cross product \( \mathbf{v} = abla f \times abla g \) is calculated as follows:

• Setting up a 3x3 determinant which includes unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) and the components of \( abla f \) and \( abla g \).
• Evaluating this determinant:
  \[ \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \18 & 16 & 16 \4 & -4 & 12 \end{vmatrix} \]
• Results in the direction vector \( \langle 256, -176, -128 \rangle \).

This vector is orthogonal to both \( abla f \) and \( abla g \), making it a suitable direction vector for the tangent line.

Tangent Line

In three-dimensional geometry, a tangent line to a curve at a given point is a line that just "touches" the curve at that point. This line is vital for understanding the curve's direction at closely examined locations. To find a tangent line to the curve of intersection of two surfaces, we need both a specific point and a direction.
In our context, the intersection of two surfaces \( f \) and \( g \) at the point (1, 2, 2), gives us the location point. The direction of the tangent line is derived from the cross product of the gradients \( abla f \) and \( abla g \).
The parametric equations describing this tangent line are:

• \( x(t) = 1 + 256t \)
• \( y(t) = 2 - 176t \)
• \( z(t) = 2 - 128t \)

These equations indicate that as \( t \) varies, the point (x, y, z) traces out the line, defining the exact path along the tangent direction.

3D Surfaces

In mathematics, 3D surfaces are intricate structures extending through three-dimensional space and can be represented by equations involving variables \(x, y, \) and \(z\). Each point on these surfaces satisfies the given equation of the surface.
For the intersection of surfaces, consider functions like \( f(x, y, z) = 9x^2 + 4y^2 + 4z^2 - 41 =0 \) and \( g(x, y, z) = 2x^2 - y^2 + 3z^2 -10 =0 \).
The variety of shapes and forms they create, ranging from spheres to complex hyperboloids, highlights the diverse mathematical models they can represent.
When these two surfaces intersect, the result is a 3D curve. Finding tangent lines to such curves at particular points, like (1, 2, 2) in our problem, helps analyze the behavior of these curves and understand the spatial relationships between intersection surfaces.