Problem 59 - Calculus with Concepts in Calculus

Question

Use implicit differentiation to find \(\partial z / \partial x\) and \(\partial z / \partial y\) at the given point. Then find an equation of the plane tangent to the level surface at that point. $$ x^{2}-y^{2}-z^{2}=1 ;(\sqrt{2}, 0,1) $$

Step-by-Step Solution

Answer

Partial derivatives are: \( \frac{\partial z}{\partial x} = \sqrt{2} \), \( \frac{\partial z}{\partial y} = 0 \). Tangent plane equation: \( z = \sqrt{2}x - 1 \).

Step 1: Understand the Equation

The given equation is a level surface: \( x^2 - y^2 - z^2 = 1 \). We want to find the partial derivatives \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \) using implicit differentiation.

Step 2: Differentiate with Respect to x

Differentiate \( x^2 - y^2 - z^2 = 1 \) with respect to \( x \). Using implicit differentiation, we get:\[ 2x - 0 - 2z \frac{\partial z}{\partial x} = 0 \]Solving for \( \frac{\partial z}{\partial x} \), we have:\[ \frac{\partial z}{\partial x} = \frac{x}{z} \]

Step 3: Differentiate with Respect to y

Differentiate \( x^2 - y^2 - z^2 = 1 \) with respect to \( y \). Using implicit differentiation, we get:\[ 0 - 2y - 2z \frac{\partial z}{\partial y} = 0 \]Solving for \( \frac{\partial z}{\partial y} \), we have:\[ \frac{\partial z}{\partial y} = \frac{y}{z} \]

Step 4: Evaluate at the Given Point

The given point is \((\sqrt{2}, 0, 1)\). Evaluate the partial derivatives at this point:\[ \frac{\partial z}{\partial x} \Big|_{(\sqrt{2}, 0, 1)} = \frac{\sqrt{2}}{1} = \sqrt{2} \]\[ \frac{\partial z}{\partial y} \Big|_{(\sqrt{2}, 0, 1)} = \frac{0}{1} = 0 \]

Step 5: Find the Tangent Plane Equation

The equation of the tangent plane at \((x_0, y_0, z_0)\) is:\[ z - z_0 = \frac{\partial z}{\partial x} (x - x_0) + \frac{\partial z}{\partial y} (y - y_0) \]Substituting the given point and partial derivatives, the equation becomes:\[ z - 1 = \sqrt{2} (x - \sqrt{2}) + 0 (y - 0) \]Simplifying, the tangent plane is:\[ z = \sqrt{2}x + 1 - 2 \]

Step 6: Simplify the Tangent Plane Equation

Simplify the equation from Step 5:\[ z = \sqrt{2}x - 1 \]This represents the equation of the tangent plane to the level surface at the given point.

Key Concepts

Partial Derivatives

To fully grasp partial derivatives, think of slicing a function into simpler parts. If you have a multi-variable function, say \( f(x, y, z) \), partial derivatives help you understand how the function behaves as you change one variable, keeping the others constant. Here, we're examining a level surface given by the equation \( x^2 - y^2 - z^2 = 1 \). To find the partial derivatives \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \), we use implicit differentiation.

This process involves differentiating each term of the equation with respect to one variable while treating others as constants. When differentiating implicitly with respect to \( x \), we derived \( \frac{\partial z}{\partial x} = \frac{x}{z} \), indicating that the change of \( z \) concerning \( x \) depends on both \( x \) and \( z \). Similarly, with \( y \), we found \( \frac{\partial z}{\partial y} = \frac{y}{z} \). It shows how \( z \) changes with alterations in \( y \), again reliant on both variables. This process becomes particularly useful in multi-variable calculus and when evaluating functions of several variables.

Level Surface

A level surface is a fascinating concept in multivariable calculus. It's essentially a three-dimensional space where a function remains constant. Imagine a mountain in 3D space; the surface pointing to the same altitude forms a level surface. In our exercise, the level surface is determined by the equation \( x^2 - y^2 - z^2 = 1 \). This describes a hyperboloid, a surface characterized by its double cone shape.

A level surface can help visualize how a function behaves in three dimensions. It's a snapshot of the function's constant value across a particular slice of its domain. This concept ties directly to implicit differentiation, as varying values along this surface require care when computing derivatives. Through implicit differentiation, we ascertain how \( x, y, \) and \( z \) interrelate on this surface without explicitly solving for \( z \). Looking at level surfaces can provide insight into the geometry of complex equations.

Tangent Plane Equation

Imagine gently resting a flat sheet of paper on a curved surface; the sheet represents a tangent plane. In multivariable calculus, the tangent plane is crucial for approximating surfaces around a specific point. In our example, we derive the tangent plane to the level surface defined by \( x^2 - y^2 - z^2 = 1 \) at the point \((\sqrt{2}, 0, 1)\).

To find the tangent plane equation, we need the point's coordinates and the partial derivatives at that point. We use these to write:

• \( z - z_0 = \frac{\partial z}{\partial x}(x - x_0) + \frac{\partial z}{\partial y}(y - y_0) \)

The derivatives \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \) we found are substituted into this equation. After simplification, our tangent plane equation with constants becomes \( z = \sqrt{2}x - 1 \). This plane provides a linear approximation of the surface at the given point, offering insights into the immediate neighborhood of the level surface.