Problem 170 - Calculus 3

Question

For the following exercises, find the equation for the tangent plane to the surface at the indicated point. (Hint: Solve for \(z\) in terms of \(x\) and \(y . )\) $$-8 x-3 y-7 z=-19, P(1,-1,2)$$

Step-by-Step Solution

Answer

The tangent plane equation is: \( z = \frac{-8}{7}x - \frac{3}{7}y + \frac{33}{7} \).

Step 1: Solve the equation for z

We start by rearranging the equation to solve for \( z \) in terms of \( x \) and \( y \). Given the surface equation is \(-8x - 3y - 7z = -19\). We can express \( z \) as: \[ z = \frac{-8x - 3y + 19}{7} \]

Step 2: Find partial derivatives

To find the tangent plane, we need the partial derivatives of \( z \) with respect to \( x \) and \( y \). The partial derivative with respect to \( x \) is: \( \frac{\partial z}{\partial x} = \frac{-8}{7} \).The partial derivative with respect to \( y \) is: \( \frac{\partial z}{\partial y} = \frac{-3}{7} \).

Step 3: Write the tangent plane equation formula

The equation of the tangent plane at a point \( (x_0, y_0, z_0) \) is given by:\( z - z_0 = \frac{\partial z}{\partial x}(x - x_0) + \frac{\partial z}{\partial y}(y - y_0) \).Substitute the point \( P(1, -1, 2) \) and partial derivatives into the equation.

Step 4: Substitute partial derivatives and point into the tangent plane equation

Using the values at point \( P(1, -1, 2) \), we substitute:\[ z - 2 = \frac{-8}{7}(x - 1) + \frac{-3}{7}(y + 1) \]Simplify the expression to get the tangent plane equation:\[ z = \frac{-8}{7}x - \frac{3}{7}y + \frac{33}{7} \]

Key Concepts

Partial Derivatives

When we discuss partial derivatives, we're looking at the rate of change of a function concerning one variable while keeping the other variables constant. This helps us understand how a surface behaves locally. In the context of the exercise, we solve the surface equation for \( z \) in terms of \( x \) and \( y \). Then, we find the partial derivative of \( z \) with respect to \( x \), denoted as \( \frac{\partial z}{\partial x} \), which in this case equals \( \frac{-8}{7} \). The other partial derivative, \( \frac{\partial z}{\partial y} \), equals \( \frac{-3}{7} \).

• A partial derivative tells us how \( z \) changes when one variable changes, while the other stays constant.
• For the surface \(-8x - 3y - 7z = -19\), finding these derivatives helps us to establish the tangent's slope in the directions of \( x \) and \( y \).
• The tangent plane uses these derivatives to approximate the surface near the given point \( P(1, -1, 2) \).

Surface Equation

The surface equation is a mathematical expression that represents a surface in three-dimensional space. Often, this is given in terms of \( x \), \( y \), and \( z \). In this exercise, our surface is initially expressed as \(-8x - 3y - 7z = -19\). For convenience and further calculations, we "solve for \( z \)" to get it in the form \( z = f(x,y) \). Applying this to our equation:\[ z = \frac{-8x - 3y + 19}{7} \]

• Rewriting the equation in terms of \( z \) helps us use differentiation techniques easily.
• This form clarifies the function we are working with, making it easier to compute the partial derivatives.
• By solving for \( z \), we can better visualize and work with the surface as a function of \( x \) and \( y \).

Plane Equation

Once we have the partial derivatives, we use them to find the tangent plane equation at the specified point. The tangent plane is very useful as it provides a linear approximation of the surface near a point. The formula for the tangent plane is:\[ z - z_0 = \frac{\partial z}{\partial x}(x - x_0) + \frac{\partial z}{\partial y}(y - y_0) \]Substituting \( (x_0, y_0, z_0) = (1, -1, 2) \) as well as our partial derivatives \( \frac{-8}{7} \) and \( \frac{-3}{7} \) into this formula, we get:\[ z - 2 = \frac{-8}{7}(x - 1) + \frac{-3}{7}(y + 1) \]With simplification, this becomes:\[ z = \frac{-8}{7}x - \frac{3}{7}y + \frac{33}{7} \]

• This equation describes the plane that just "touches" the surface at the point \( P(1, -1, 2) \).
• The equation is derived using the slopes given by the partial derivatives.
• The result is a linear plane equation that approximates our surface near the point accurately.