Problem 175 - 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 \(y . )\) $$x y+y z+z x=11, P(1,2,3)$$

Step-by-Step Solution

Answer

The tangent plane equation is \(z = x + y\).

Step 1: Solve for z

To find the tangent plane, we first express the surface equation in terms of a single variable. We start with the equation \(x y + y z + z x = 11\). Solve for \(z\): \(z = \frac{11 - xy}{x+y}\).

Step 2: Calculate Partial Derivatives

We need to find the partial derivatives \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\). Using implicit differentiation, we get \(\frac{\partial z}{\partial x} = \frac{y+z-y}{y+x}\) and \(\frac{\partial z}{\partial y} = \frac{x+z-x}{x+y}\).

Step 3: Evaluate at Point P

Evaluate the partial derivatives at the given point \(P(1,2,3)\). Substitute in the values: \(\frac{\partial z}{\partial x}|_{(1,2,3)} = 1\) and \(\frac{\partial z}{\partial y}|_{(1,2,3)} = 1\).

Step 4: Equation of the Tangent Plane

The equation of the tangent plane is given by \(z - z_0 = \frac{\partial z}{\partial x}(x-x_0) + \frac{\partial z}{\partial y}(y-y_0)\). Substitute \(z_0 = 3\), \(x_0 = 1\), \(y_0 = 2\), \(\frac{\partial z}{\partial x} = 1\), \(\frac{\partial z}{\partial y} = 1\) to obtain: \(z - 3 = 1(x - 1) + 1(y - 2)\).

Step 5: Simplify the Equation

Simplify the tangent plane equation: \(z - 3 = x - 1 + y - 2\). Rearrange to the form \(z = x + y\).

Key Concepts

Partial Derivatives

Partial derivatives are a fundamental concept in calculus. They allow us to determine how a function changes when varying one variable while keeping others constant. Basically, they help us analyze changes in multi-variable functions.

When dealing with surfaces, like the one in this exercise, we use partial derivatives to study how the surface behaves as we change one of the variables, either \(x\) or \(y\), holding the other fixed.

• For the derivative with respect to \(x\), consider \(y\) constant.
• For the derivative with respect to \(y\), consider \(x\) constant.

Here, given the surface equation \(xy + yz + zx = 11\), we differentiate with respect to \(x\) and \(y\). This process helps us find the slope of the tangent line in respective directions, ultimately contributing to the equation of the tangent plane.

Implicit Differentiation

Implicit differentiation is a technique used when a function is not solved explicitly for one variable in terms of another. It is essential when we have equations involving multiple variables, like the ones seen in surfaces.

Instead of solving for one variable straight away, we take derivatives with respect to one of the variables while treating the others as implicit functions. In this exercise, implicit differentiation helps us find \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \):

• For \( \frac{\partial z}{\partial x} \), differentiate the entire equation as if \(x\) is the variable and \(y\) is treated as a constant.
• For \( \frac{\partial z}{\partial y} \), differentiate similarly, considering \(y\) as variable, and \(x\) as constant.

Through implicit differentiation, we obtain the results needed to inform our tangent plane equation.

Surface Equation

The surface equation in this problem is a combination of terms involving \(x\), \(y\), and \(z\). It represents a three-dimensional surface in space.

Understanding how to manipulate and solve surface equations is key in multi-variable calculus. Generally, before finding a tangent plane, we rearrange the equation to solve for one variable, if possible.

In our case, we solved for \(z\) in terms of \(x\) and \(y\): \[ z = \frac{11 - xy}{x + y} \]This form makes it easier to analyze changes in \(z\) as \(x\) and \(y\) vary. Once restated, we can differentiate the function partially to derive required expressions.

Evaluate at a Point

When working on problems involving derivatives or tangent planes, evaluating at a specific point is crucial. This step gives us concrete values that describe how the surface behaves around a point of interest.

Here, the point \(P(1,2,3)\) is where we will find the tangent plane. By substituting \(x = 1\), \(y = 2\), and \(z = 3\) into our partial derivative expressions, we can calculate precise slopes at that point.

• \( \frac{\partial z}{\partial x} \) evaluates to \(1\)
• \( \frac{\partial z}{\partial y} \) evaluates to \(1\)

These values contribute to the equation of the tangent plane by defining its orientation in space at the point \(P\). The result? A clear approach to approximating the surface near this specific point through the equation \(z = x + y\).