Problem 3 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems \(1-8\), find the equation of the tangent plane to the given surface at the indicated point. $$ x^{2}-y^{2}+z^{2}+1=0 ;(1,3, \sqrt{7}) $$

Step-by-Step Solution

Answer

The tangent plane is \(2x - 6y + 2\sqrt{7}z + 2 = 0\).

Step 1: Understand the Problem

We need to find the equation of the tangent plane to the surface given by the equation \(x^2 - y^2 + z^2 + 1 = 0\) at the point \((1, 3, \sqrt{7})\).

Step 2: Identify the Implicit Function

Our surface is defined by the implicit function \(F(x, y, z) = x^2 - y^2 + z^2 + 1\). The equation of the tangent plane at a point \((x_0, y_0, z_0)\) can be derived using the gradient of this function.

Step 3: Compute the Gradient at the Point

Calculate the gradient of the function \(F(x, y, z)\):\[abla F = \left( \frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z} \right)\]That gives us:\[abla F = (2x, -2y, 2z)\]Evaluate \(abla F\) at \((1, 3, \sqrt{7})\):\[abla F = (2(1), -2(3), 2(\sqrt{7})) = (2, -6, 2\sqrt{7})\]

Step 4: Use the Gradient to Find the Tangent Plane Equation

The equation of the tangent plane can be written as:\[F_x(x_0, y_0, z_0)(x - x_0) + F_y(x_0, y_0, z_0)(y - y_0) + F_z(x_0, y_0, z_0)(z - z_0) = 0\]Substitute the values:\[2(x - 1) - 6(y - 3) + 2\sqrt{7}(z - \sqrt{7}) = 0\]

Step 5: Simplify the Tangent Plane Equation

Simplify the equation:Distribute each component:\[2x - 2 - 6y + 18 + 2\sqrt{7}z - 14 = 0\]Combine like terms:\[2x - 6y + 2\sqrt{7}z + 2 = 0\]

Key Concepts

Gradient Calculus

In gradient calculus, we deal with functions that have multiple variables. The gradient, represented by the symbol \( abla \), gives us a vector of partial derivatives. This vector points in the direction of the greatest rate of increase of the function. It's a key concept because it helps us understand how a function changes as we move through its domain.

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

• \( \frac{\partial F}{\partial x} \) - the partial derivative with respect to \( x \)
• \( \frac{\partial F}{\partial y} \) - the partial derivative with respect to \( y \)
• \( \frac{\partial F}{\partial z} \) - the partial derivative with respect to \( z \)

In the exercise, we found \( abla F(x, y, z) = (2x, -2y, 2z) \). This gradient helps us find the equation of the tangent plane at a specific point.

Implicit Differentiation

Implicit differentiation is a technique used when dealing with equations that define one variable implicitly by the others, rather than explicitly. Especially in multivariable calculus, it allows us to differentiate such functions without having to solve explicitly for one variable.
When we have a function like \( F(x, y, z) = 0 \), every variable depends on the others, hidden in the surface equation. Here, the gradient plays a critical role by providing the necessary derivatives that describe how each variable affects the other.

In this problem, the surface is represented implicitly by \( x^2 - y^2 + z^2 + 1 = 0 \). We use the gradient of this implicit function to derive the necessary derivatives, which help in determining tangent planes and other related quantities.

Partial Derivatives

Partial derivatives help us understand how a function changes when we change one variable at a time, keeping the others constant. This concept is crucial in multivariable calculus as it allows us to examine the function's behavior more deeply.

For the function \( F(x, y, z) = x^2 - y^2 + z^2 + 1 \), partial derivatives are computed:

• \( \frac{\partial F}{\partial x} = 2x \)
• \( \frac{\partial F}{\partial y} = -2y \)
• \( \frac{\partial F}{\partial z} = 2z \)

These derivatives give us the components of the gradient. Evaluating these at a point provides us the necessary coefficients to formulate equations such as the tangent plane.

Multivariable Calculus

Multivariable calculus extends the principles of calculus to functions with more than one variable. It opens up new possibilities for analyzing surfaces and curves in higher-dimensional spaces.

In this exercise, we deal with a surface equation \( x^2 - y^2 + z^2 + 1 = 0 \), which is inherently multivariable. Finding the tangent plane involves leveraging concepts like gradients, implicit differentiation, and partial derivatives, all of which come under the umbrella of multivariable calculus.

• We analyze how each variable contributes to changes in the function.
• The solution involves computing derivatives with respect to each variable.
• We use this information to solve for relationships in higher dimensions, such as the tangent plane equation.

Mastering multivariable calculus allows us to model and solve complex problems involving multiple factors and variables, like the one in this exercise.