Problem 22 - Calculus Early Transcendentals

Question

Find an equation of the plane tangent to the following surfaces at the given points. $$z=\ln (1+x y) ;(1,2, \ln 3) \text { and }(-2,-1, \ln 3)$$

Step-by-Step Solution

Answer

Answer: The equations of the tangent planes at the given points are: $$z - \ln 3 = \frac{2}{3}(x - 1) + \frac{1}{3}(y - 2)$$ and $$z - \ln 3 = \frac{-1}{3}(x + 2) + \frac{-2}{3}(y + 1)$$

Step 1: Calculate the partial derivatives with respect to x and y

First, we need to find the partial derivatives of the function $$z = \ln(1 + xy)$$ with respect to x and y. We'll denote these partial derivatives as $$\frac{\partial z}{\partial x}$$ and $$\frac{\partial z}{\partial y}$$. For $$\frac{\partial z}{\partial x}$$, we have: $$\frac{\partial z}{\partial x} = \frac{\partial \ln(1+xy)}{\partial x} = \frac{y}{1+xy}$$ Similarly, for $$\frac{\partial z}{\partial y}$$, we have: $$\frac{\partial z}{\partial y} = \frac{\partial \ln(1+xy)}{\partial y} = \frac{x}{1+xy}$$

Step 2: Evaluate the partial derivatives at the given points

Now, we need to find the gradient, or the partial derivatives, at each of the given points (1,2, ln 3) and (-2,-1, ln 3). We'll start with the point (1,2, ln 3). At (1,2, ln 3), $$\frac{\partial z}{\partial x} = \frac{y}{1+xy} = \frac{2}{1+1\cdot 2} = \frac{2}{3}$$ $$\frac{\partial z}{\partial y} = \frac{x}{1+xy} = \frac{1}{1+1\cdot 2} = \frac{1}{3}$$ At (-2,-1, ln 3), $$\frac{\partial z}{\partial x} = \frac{y}{1+xy} = \frac{-1}{1+(-2)(-1)} = \frac{-1}{3}$$ $$\frac{\partial z}{\partial y} = \frac{x}{1+xy} = \frac{-2}{1+(-2)(-1)} = \frac{-2}{3}$$

Step 3: Construct the tangent plane equations

Using the gradient at each point, we can now construct the equation of the tangent planes at the given points. The equation of the tangent planes is given by: $$z - z_0 = \frac{\partial z}{\partial x}(x - x_0) + \frac{\partial z}{\partial y}(y - y_0)$$ For (1,2, ln 3), the tangent plane equation is: $$z - \ln 3 = \frac{2}{3}(x - 1) + \frac{1}{3}(y - 2)$$ For (-2,-1, ln 3), the tangent plane equation is: $$z - \ln 3 = \frac{-1}{3}(x + 2) + \frac{-2}{3}(y + 1)$$ Thus, the equations of the tangent planes at the given points are: $$z - \ln 3 = \frac{2}{3}(x - 1) + \frac{1}{3}(y - 2)$$ and $$z - \ln 3 = \frac{-1}{3}(x + 2) + \frac{-2}{3}(y + 1)$$

Key Concepts

Partial Derivatives

Partial derivatives are a crucial part of multivariable calculus, helping us understand the behavior of functions with more than one variable. When you have a function like \(z = \ln(1 + xy)\), you're looking at a surface instead of a simple curve. To find how the function changes with respect to one variable while keeping the other constant, we use partial derivatives.
For this example, the function depends on both \(x\) and \(y\). To find the partial derivative with respect to \(x\), we treat \(y\) as a constant. Similarly, to find \(\frac{\partial z}{\partial y}\), we consider \(x\) a constant.

• The partial derivative with respect to \(x\) is found to be \(\frac{y}{1+xy}\).
• The partial derivative with respect to \(y\) is \(\frac{x}{1+xy}\).

These derivatives tell us how steep the surface is in the direction of the respective variables, providing critical information needed to find equations of tangent planes.

Surface Gradient

The gradient gives the direction of the steepest ascent on the surface at a particular point. Think of it as an arrow pointing "uphill." For functions of two variables, the gradient is a vector consisting of the partial derivatives with respect to each variable.
For the function \(z = \ln(1 + xy)\), the gradient \(abla z\) at a point \((x, y)\) is given by:

• \(\left(\frac{\partial z}{\partial x}, \frac{\partial z}{\partial y}\right) = \left(\frac{y}{1+xy}, \frac{x}{1+xy}\right)\)

Evaluating this at, for instance, \((1, 2, \ln 3)\) yields the gradient \(\left(\frac{2}{3}, \frac{1}{3}\right)\). This vector indicates the direction and rate of the steepest increase of \(z\) starting from the given point. At \((-2, -1, \ln 3)\), the gradient is \(\left(\frac{-1}{3}, \frac{-2}{3}\right)\).
The gradient not only helps in forming the tangent plane but also provides insights into how the surface behaves at specific points.

Multivariable Calculus

Multivariable calculus extends the ideas of single-variable calculus to functions of multiple variables, like \(f(x, y)\). It is especially useful in fields that require modeling of complex systems, such as physics, engineering, and economics.
With more than one input variable, we encounter concepts like gradients, tangent planes, and partial derivatives. The tangent plane is especially important in multivariable calculus. It provides a linear approximation of the surface near a given point, similar to how a tangent line approximates a curve in single-variable calculus.
To construct a tangent plane equation, we take the value of the function at a particular point and combine it with the partial derivatives, forming a surface that "touches" the original at just that point. This is captured in the formula:

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

This equation represents a flat surface that best matches the behavior of the original multi-variable function at and around the chosen point.