Problem 41 - Calculus Early Transcendentals: Pearson New International Edition

Question

Show that the normal line to \(x^{3}+y^{3}=3 x y\) at \(\left(\frac{3}{2}, \frac{3}{2}\right)\) passes through the origin.

Step-by-Step Solution

Answer

The normal line passes through the origin.

Step 1: Find the normal line

Implicitly differentiate \( x^3+y^3=3xy \) to find \( dy/dx \) at \( (3/2, 3/2) \). The slope is \( -1 \) (by symmetry). Normal slope = 1.

Step 2: Check origin

Normal line: \( y - 3/2 = 1(x - 3/2) \), i.e., \( y = x \). At \( (0,0) \): \( 0 = 0 \). Yes, passes through origin. \( \blacksquare \)

Key Concepts

Implicit Differentiation

Implicit Differentiation is a technique used to find the derivative of a function that is not given explicitly. Instead of having y explicitly solved in terms of x, an implicitly defined function mixes x and y together. For example, given the equation \( x^3 + y^3 = 3xy \), y is not isolated.

When differentiating implicitly, follow these guidelines:

• Differentiate both sides of the equation with respect to x.
• Apply the chain rule for terms involving y because y is a function of x (e.g., \( \frac{d}{dx}[y^3] = 3y^2 \cdot \frac{dy}{dx} \)).
• Solve for \( \frac{dy}{dx} \) after differentiating all terms.

Using these steps, you handle rates of change in multivariable equations without needing to rearrange the equation to explicitly solve for y.

Normal Line

The Normal Line to a curve at a given point is perpendicular to the Tangent Line at that point. If the slope of the Tangent Line is m, then the slope of the Normal Line is \( -\frac{1}{m} \).

The Normal Line is critical in understanding the geometry of curves:

• It's helpful in optimizing problems and analyzing the behavior of graphs.
• It provides insight on how curves interact with nearby points and lines.

In the given problem, you're asked to find the Normal Line to the curve \( x^3 + y^3 = 3xy \) at \( (\frac{3}{2}, \frac{3}{2}) \). After finding the slope of the Tangent Line, you determine the slope of the Normal Line, which is needed to show that it passes through the origin.

Tangent Line

The Tangent Line at a given point on a curve represents the linear approximation or the "instantaneous" direction of the curve at that point. It touches the curve at exactly one point without crossing it at this point of contact.

To find the slope of the Tangent Line for an implicitly defined curve like \( x^3 + y^3 = 3xy \), perform implicit differentiation as discussed earlier.

Here's what a Tangent Line helps with:

• It provides the best linear approximation to the curve at a specific point.
• It is useful for solving related rates and optimization problems.
• It assists in calculating point slope form: \( y - y_1 = m(x - x_1) \), where m is the slope obtained from \( \frac{dy}{dx} \).

Derivative

A derivative represents the rate of change of a function concerning one of its variables. It is the slope of the Tangent Line to a function at a given point.

In calculus, derivatives are fundamental:

• They are used to determine how functions change, allowing for deeper analysis of functions.
• In implicit differentiation, like in our equation, derivatives are crucial for finding the Tangent and Normal Lines.
• The derivative in this context is \( \frac{dy}{dx} \), and applying it involves skills like using the chain rule effectively.

Differential Equations

Differential Equations are equations that involve an unknown function and its derivatives. They are essential for modeling real-world phenomena involving rates of change.

Solving a differential equation can involve various techniques:

• Separation of variables - separating terms involving different variables.
• Homogeneous and non-homogeneous equations - depending on whether the function is zero or not.

In the given problem, while it's framed around finding a normal line, handling equations like \( 3x^2 + 3y^2 \frac{dy}{dx} = 3(y + x \cdot \frac{dy}{dx}) \) shares similarities with the types of manipulations you would do with differential equations.