Problem 177 - 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 \(y . )\) $$x^{3}+y^{3}=3 x y z, P\left(1,2, \frac{3}{2}\right)$$

Step-by-Step Solution

Answer

The equation of the tangent plane is \(-6x + 7.5y - 6z = 0\).

Step 1: Isolate the variable z

First, identify how we can solve the equation for \( z \) in terms of \( x \) and \( y \). The given equation is \( x^3 + y^3 = 3xy z \). Isolate \( z \) to get \( z = \frac{x^3 + y^3}{3xy} \).

Step 2: Compute the partial derivatives

To find the equation of the tangent plane, compute the partial derivatives \( f_x \), \( f_y \), and \( f_z \) at the point \( P(1, 2, \frac{3}{2}) \). Start with the function \( f(x, y, z) = x^3 + y^3 - 3xyz = 0 \).

Step 3: Calculate partial derivative with respect to x

The partial derivative with respect to \( x \) is \( f_x = 3x^2 - 3yz \). Evaluate \( f_x \) at the point \( (1, 2, \frac{3}{2}) \) to get \( f_x(1, 2, \frac{3}{2}) = 3(1)^2 - 3(2)(\frac{3}{2}) = 3 - 9 = -6 \).

Step 4: Calculate partial derivative with respect to y

The partial derivative with respect to \( y \) is \( f_y = 3y^2 - 3xz \). Evaluate \( f_y \) at the point \( (1, 2, \frac{3}{2}) \) to get \( f_y(1, 2, \frac{3}{2}) = 3(2)^2 - 3(1)(\frac{3}{2}) = 12 - 4.5 = 7.5 \).

Step 5: Calculate partial derivative with respect to z

The partial derivative with respect to \( z \) is \( f_z = -3xy \). Evaluate \( f_z \) at the point \( (1, 2, \frac{3}{2}) \) to get \( f_z(1, 2, \frac{3}{2}) = -3(1)(2) = -6 \).

Step 6: Form the equation of the tangent plane

The equation of the tangent plane at point \( (x_0, y_0, z_0) \) is given by: \( f_x(x_0, y_0, z_0)(x - x_0) + f_y(x_0, y_0, z_0)(y - y_0) + f_z(x_0, y_0, z_0)(z - z_0) = 0 \). Substitute \( x_0 = 1 \), \( y_0 = 2 \), \( z_0 = \frac{3}{2} \), and the partial derivatives: \( -6(x - 1) + 7.5(y - 2) - 6(z - \frac{3}{2}) = 0 \).

Step 7: Simplify the equation

Distribute and simplify the equation: \( -6x + 6 + 7.5y - 15 - 6z + 9 = 0 \). Combine like terms to get the final equation of the tangent plane: \( -6x + 7.5y - 6z = 0 \).

Key Concepts

Partial Derivatives

Partial derivatives are a fundamental concept in multivariable calculus, especially in understanding functions with more than one variable. Simply put, a partial derivative measures how a function changes with respect to one variable while keeping the others constant.
For a function like \( f(x, y, z) = x^3 + y^3 - 3xyz \), finding the partial derivatives with respect to each variable involves calculating:

• The rate of change in \( x \), \( f_x = 3x^2 - 3yz \)
• The rate of change in \( y \), \( f_y = 3y^2 - 3xz \)
• The rate of change in \( z \), \( f_z = -3xy \)

To obtain these, differentiate the function while leaving the other variables constant to focus on the relationship between our variables.
The evaluated partial derivatives give key insights into how the surface behaves locally around a specific point, like \( P(1, 2, \frac{3}{2}) \), which can then be used to construct a tangent plane, a linear approximation of the surface at that point.

Equation of a Surface

The equation of a surface in three-dimensional space is often expressed by relating variables \( x \), \( y \), and \( z \) through a function, such as \( x^3 + y^3 = 3xyz \). In calculus exercises, solving this equation for one variable, like \( z \), in terms of the others is a crucial step to simplify subsequent computations.
Through solving, we isolate \( z \) and rewrite the equation as \( z = \frac{x^3 + y^3}{3xy} \). This puts the surface in an explicit form for \( z \), which is useful in both visualizing the surface and computing derivatives.
The equation of a surface helps describe a wide variety of geometric shapes ranging from simple planes to complex non-linear forms. Understanding how to manipulate and rewrite these equations enables us to tackle more advanced problems in calculus, such as finding tangent planes and normal lines.

Calculus 3

Calculus 3 extends the concepts from previous calculus courses into multiple dimensions, dealing with functions of two or more variables. This includes concepts such as partial derivatives, multiple integrals, and vector calculus.
In the context of this exercise, Calculus 3 is necessary for finding tangent planes to surfaces. A tangent plane linearly approximates a surface at a given point, crucial for visualizing and understanding the behavior of surfaces in 3D space.
These problems require synthesizing various Calculus 3 techniques:

• Determining partial derivatives to understand local behavior
• Solving surface equations in terms of one variable to simplify problems
• Constructing tangent planes using derivatives to approximate surfaces

Mastering these aspects of Calculus 3 opens up the world of multivariable calculus, allowing us to solve complex real-world problems involving multiple variables, much like this example.