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 \(\nabla g(1,2,2)\).

Short answer

The parametric equations of the tangent line are \( x=1+44t \), \( y=2-35t \), \( z=2-17t \).

Step-by-step solution

Find the gradient of both surfaces

The gradients of the surfaces give us the normal vectors at any point on each surface. For the first surface, compute \( abla f(x, y, z) = \Big( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \Big) = (18x, 8y, 8z) \). For \( (x, y, z) = (1, 2, 2) \), \( abla f(1, 2, 2) = (18, 16, 16) \). For the second surface, compute \( abla g(x, y, z) = (4x, -2y, 6z) \). For \( (x, y, z) = (1, 2, 2) \), \( abla g(1, 2, 2) = (4, -4, 12) \).

Find the direction vector using the cross product

The direction of the tangent line is perpendicular to both gradients, \( abla f(1, 2, 2) \) and \( abla g(1, 2, 2) \). Calculate the cross product \( abla f \times abla g \). Use the formula for the cross product: \[ abla f \times abla g = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 18 & 16 & 16 \ 4 & -4 & 12 \end{vmatrix} \] leading to \( = (288 + 64, -216 - 64, -72 - 64) = (352, -280, -136) \).

Simplify the direction vector

The vector \( (352, -280, -136) \) can be simplified by dividing by their greatest common divisor (8), giving \( (44, -35, -17) \).

Write the parametric equations of the tangent line

Using \( (1, 2, 2) \) as a point on the line and \( (44, -35, -17) \) as the direction vector, the parametric equations of the tangent line are: \[ x = 1 + 44t \]\[ y = 2 - 35t \]\[ z = 2 - 17t \].

Key concepts

Gradient Vectors

In multivariable calculus, the gradient vector points in the direction of the greatest rate of increase of a function. It acts like a compass showing where the function increases most rapidly. For a function of the form \( f(x, y, z) \), the gradient, denoted \( abla f \), is a vector composed of its partial derivatives: \( \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) \).

To find the gradient vector at a given point, you simply evaluate these partial derivatives at that point. In our exercise, \( abla f(1, 2, 2) = (18, 16, 16) \) and \( abla g(1, 2, 2) = (4, -4, 12) \). These vectors are perpendicular to their respective surfaces at that point, serving as normal vectors. When two surfaces intersect, the vectors normal to each surface can be used to find the direction of a line tangent to their curve of intersection by finding a vector that is orthogonal to both gradients.

Tangent Line

A tangent line to a curve at a given point is the line that "just touches" the curve at that point. It represents the direction the curve is heading at that point and is crucial in approximations and understanding the curve's behavior locally.

For parametric equations, a line can be described as \( x = x_0 + at \), \( y = y_0 + bt \), \( z = z_0 + ct \), where \((x_0, y_0, z_0)\) is a point on the line and \((a, b, c)\) is a direction vector. The direction vector is found by determining the vector orthogonal to gradient vectors of both surfaces from which the tangent line emanates. In our case, after calculations, the direction vector becomes \((44, -35, -17)\).

This gives us the parametric equations for the tangent line at point \((1, 2, 2)\) along the direction of this vector.

Cross Product

The cross product is a vector operation used in three-dimensional space to find a vector perpendicular to two given vectors. This is very useful in calculus, especially when dealing with normal vectors and tangent lines.

The cross product of vectors \(\mathbf{A} = (a_1, a_2, a_3)\) and \(\mathbf{B} = (b_1, b_2, b_3)\) is given by:

• Find the determinant of the matrix formed by \( \mathbf{i}, \mathbf{j}, \mathbf{k} \), and the components of \( \mathbf{A} \) and \( \mathbf{B} \).
• This results in a new vector \((c_1, c_2, c_3)\), where \(c_1\), \(c_2\), and \(c_3\) are the assigned determinants for each coordinate axis component.

In the exercise, the cross product of \( abla f(1, 2, 2) \) and \( abla g(1, 2, 2) \) yields a vector which tells us the direction of the tangent line at the intersection: \((352, -280, -136)\), which we simplified to \((44, -35, -17)\).

This cross product ensures the direction is perpendicular to both normal vectors, a critical component in tracing the curve's tangent line.

Multivariable Calculus

Multivariable calculus extends single-variable calculus concepts to higher dimensions. It focuses on functions that have more than one independent variable and provides tools to analyze surfaces and curves in three-dimensional space.

Key operations in multivariable calculus include:

• Partial differentiation which helps in determining gradients and local linear approximations of functions.
• Line integrals and surface tension, which are important in physics and engineering applications, like calculating work done by a force field along a curve.
• Understanding and calculating tangent lines, which are essential when analyzing intersections and boundaries of geometric shapes.

In this exercise, multivariable calculus is at play as we find the tangent line on the curve formed by two intersecting surfaces. Concepts like the gradient vector and cross product are fundamental in performing the calculations necessary to determine the tangent line in three-dimensional space, highlighting the diverse applications of multivariable calculus.