Problem 59 - Calculus Early Transcendentals

Question

Find parametric equations for the tangent line to the curve of intersection of the paraboloid \(z=x^{2}+y^{2}\) and the ellipsoid \(4 x^{2}+y^{2}+z^{2}=9\) at the point \((-1,1,2)\) .

Step-by-Step Solution

Answer

Parametric equations: \(x = -1 + 8t\), \(y = 1 + 8t\), \(z = 2 + 12t\).

Step 1: Finding the Gradient of the Paraboloid

The paraboloid is given by the equation \( z = x^2 + y^2 \). To find its gradient, we use partial derivatives: \( \frac{\partial z}{\partial x} = 2x \) and \( \frac{\partial z}{\partial y} = 2y \). At the point \((-1, 1, 2)\), the gradient is \( abla z = (2(-1), 2(1)) = (-2, 2) \).

Step 2: Finding the Gradient of the Ellipsoid

The ellipsoid is given by \( 4x^2 + y^2 + z^2 = 9 \). Differentiating with respect to each variable gives the gradient as: \( \left( \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \right) = (8x, 2y, 2z) \). At \((-1, 1, 2)\), this becomes \( (8(-1), 2(1), 2(2)) = (-8, 2, 4) \).

Step 3: Finding the Normal Vector

The line of intersection will be tangent to both surfaces but perpendicular to their gradients. We take the cross product \( \mathbf{n} = abla z \times abla (4x^2 + y^2 + z^2) = (-2, 2, 0) \times (-8, 2, 4) \). The result is \( (8, 8, 12) \).

Step 4: Write the Parametric Equations

With the normal vector \( (8, 8, 12) \) and point \((-1, 1, 2)\), the parametric equations are: \( x = -1 + 8t \), \( y = 1 + 8t \), and \( z = 2 + 12t \). These describe the tangent line.

Key Concepts

Parametric Equations

Parametric equations are a powerful tool used to express a line or curve where each point on the line or curve is described using one or more parameters. These equations allow us to methodically determine points along a surface or through a space using algebraic expressions for each coordinate.

For example, if you have a point \((x_0, y_0, z_0)\) and a direction vector \((a, b, c)\), the parametric equations for a line can be given by:

• \( x = x_0 + at \)
• \( y = y_0 + bt \)
• \( z = z_0 + ct \)

Here, \(t\) is the parameter, and as it changes, it traces the path of the line. This is exactly what we applied in the exercise to find the tangent line at the given point of intersection.

Gradient Vector

The gradient vector is an essential concept in multivariable calculus, often helping us understand slopes and normal directions of surfaces. It consists of the partial derivatives of a function with respect to its variables.

For a function \( f(x, y, z) \), its gradient \( abla f \) is given by the vector:

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

In physical terms, it points in the direction of the steepest ascent from any given point. For the paraboloid and ellipsoid in our problem, their gradients tell us the normal directions at the intersection point, which aids in finding the tangent line by applying vector operations.

Curve of Intersection

A curve of intersection represents the set of points that two surfaces have in common in a three-dimensional space. Identifying this curve involves determining points that simultaneously satisfy the equations describing both surfaces.

In our example, the curve of intersection between the paraboloid \( z = x^2 + y^2 \) and the ellipsoid \( 4x^2 + y^2 + z^2 = 9 \) is characterized by finding where the equations meet at a given point. Understanding the intersection is crucial because it allows us to describe lines tangent to this curve. By employing gradients or tangent surfaces, we discern the behavior and direction of these curves.

Vector Cross Product

The vector cross product is a fundamental operation in vector algebra used to find a vector that is perpendicular to two given vectors. This process is also crucial in determining tangent lines to intersecting surfaces.

For vectors \( \mathbf{a} = (a_1, a_2, a_3) \) and \( \mathbf{b} = (b_1, b_2, b_3) \), their cross product \( \mathbf{a} \times \mathbf{b} \) is calculated as:

• \( \mathbf{a} \times \mathbf{b} = (a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1) \)

This result gives a vector orthogonal to both \( \mathbf{a} \) and \( \mathbf{b} \). In the context of the exercise, we used the cross product of the gradients of the paraboloid and ellipsoid to find a vector normal to both, which then helped in finding the parametric tangent line.