Problem 41 - Calculus with Concepts in Calculus

Question

Find an equation of the plane tangent to the given surface at the given point. $$ x y z=1 ;\left(\frac{1}{2},-2,-1\right) $$

Step-by-Step Solution

Answer

The tangent plane is \(4x - y - 2z = 6\).

Step 1: Understand the Problem

We need to find the equation of a plane tangent to the surface defined by the equation \(xyz = 1\) at the point \(\left(\frac{1}{2}, -2, -1\right)\). This requires us to first determine the normal vector to the surface at this point.

Step 2: Find the Gradient Vector

The surface \(xyz = 1\) can be treated as a level surface of the function \(f(x, y, z) = xyz\). The gradient vector \(abla f\) of \(f\) will give us the normal vector to the surface. Compute \(abla f = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\right) = (yz, xz, xy)\).

Step 3: Evaluate the Gradient at the Point

Substitute \(x = \frac{1}{2}\), \(y = -2\), and \(z = -1\) into the gradient to find the normal vector at the point: \(abla f\left(\frac{1}{2}, -2, -1\right) = (-2\cdot(-1), \frac{1}{2}\cdot(-1), \frac{1}{2}\cdot(-2)) = (2, -\frac{1}{2}, -1)\).

Step 4: Use the Point-Normal Form of a Plane

A plane with a normal vector \((a, b, c)\) that passes through a point \((x_0, y_0, z_0)\) can be defined by the equation: \(a(x - x_0) + b(y - y_0) + c(z - z_0) = 0\). Use this to set up the equation with \(a = 2\), \(b = -\frac{1}{2}\), \(c = -1\) and the point \(\left(\frac{1}{2}, -2, -1\right)\).

Step 5: Substitute and Simplify the Equation

Substitute \((x_0, y_0, z_0) = \left(\frac{1}{2}, -2, -1\right)\) and \((a, b, c) = (2, -\frac{1}{2}, -1)\) into the point-normal form equation: \[ 2(x - \frac{1}{2}) - \frac{1}{2}(y + 2) - 1(z + 1) = 0 \]Simplify it: \[ 2x - 1 - \frac{1}{2}y - 1 - z - 1 = 0 \]\[ 2x - \frac{1}{2}y - z - 3 = 0 \].

Step 6: Finalize the Equation

Multiply the entire equation by 2 to eliminate the fraction: \[ 4x - y - 2z - 6 = 0 \]. Thus, the tangent plane at \(\left(\frac{1}{2}, -2, -1\right)\) is \(4x - y - 2z = 6\).

Key Concepts

Gradient Vector

The gradient vector is a fundamental concept in differential calculus, often represented by \( abla f \) when derived from a function \( f(x, y, z) \). Simply put, it is a vector pointing in the direction where the function increases most steeply. For a function \( f(x, y, z) = xyz \), its gradient consists of the partial derivatives with respect to each variable:

• The partial derivative with respect to \( x \) is \( yz \).
• With respect to \( y \), it is \( xz \).
• And for \( z \), it is \( xy \).

This gives us the gradient vector \( abla f = (yz, xz, xy) \).

When evaluated at a specific point, for instance \( \left(\frac{1}{2}, -2, -1\right) \) on the surface defined by \( xyz = 1 \), the gradient vector yields the values \( (2, -\frac{1}{2}, -1) \). This vector not only indicates the fastest change in the function but also serves as a normal to the surface at that point.

Normal Vector

A normal vector to a surface is perpendicular to the tangent plane of the surface at a given point. In the context of our problem, it is derived using the gradient vector.

For the surface defined by \( xyz = 1 \), the normal vector at any point \((x, y, z)\) can be found by computing the gradient vector \( abla f = (yz, xz, xy) \). Evaluating this at \(\left(\frac{1}{2}, -2, -1\right)\), we obtain the normal vector \( (2, -\frac{1}{2}, -1) \).

The normal vector has significant implications:

• It defines the orientation of the tangent plane.
• Helps in establishing equations involving the plane.
• Acts as a critical component in various applications such as physics and geometry, where understanding the behavior of planes relative to surfaces is essential.

Thus, this vector is indispensable as it both defines and describes the plane tangent at a point on a surface.

Point-Normal Form of a Plane

The point-normal form of a plane is a powerful formula used to express the equation of a plane given a normal vector and a point on the plane. The general form is:
\[ a(x - x_0) + b(y - y_0) + c(z - z_0) = 0 \]
Here, \( (a, b, c) \) is the normal vector and \((x_0, y_0, z_0)\) is a specific point on the plane.

Applying this to our problem where we have a normal vector \((2, -\frac{1}{2}, -1)\) and a point \(\left(\frac{1}{2}, -2, -1\right)\), we substitute these into the point-normal form. This results in:
\[ 2(x - \frac{1}{2}) - \frac{1}{2}(y + 2) - 1(z + 1) = 0 \]
By simplifying, we get the equation of the plane:
\[ 4x - y - 2z = 6 \]
The point-normal form is particularly helpful because:

• It relates geometrical configurations directly to algebraic expressions.
• Allows simplification and manipulation to find pivotal properties like inclination angles and distances.
• Makes it easier to visualize the spatial orientation of the plane.

This form is a cornerstone in the study of geometry, yielding crucial insights into the relationship between different geometric entities.