Problem 30 - University Calculus: Early Transcendentals

Question

Find the slope of the curve at the given points. $$\left(x^{2}+y^{2}\right)^{2}=(x-y)^{2} \quad \text { at } \quad(1,0) \text { and }(1,-1)$$

Step-by-Step Solution

Answer

The slope is -1 at (1,0) and 1 at (1,-1).

Step 1: Differentiate Implicitly

Given the equation \((x^2 + y^2)^2 = (x-y)^2\). We need to find the derivative with respect to \(x\). Differentiating both sides: \[ 2(x^2 + y^2) \cdot (2x + 2y \cdot \frac{dy}{dx}) = 2(x-y)(1 - \frac{dy}{dx}). \] This simplifies to: \[ (x^2 + y^2)(2x + 2y \cdot \frac{dy}{dx}) = (x-y)(1 - \frac{dy}{dx}). \]

Step 2: Solve for \(\frac{dy}{dx}\)

Rearrange the terms to solve for \(\frac{dy}{dx}\): \[ 2x(x^2 + y^2) + 2y(x^2 + y^2) \cdot \frac{dy}{dx} = (x-y) - (x-y) \cdot \frac{dy}{dx}. \] Reorganize and solve the equation: \[ 2y(x^2 + y^2) \cdot \frac{dy}{dx} + (x-y) \cdot \frac{dy}{dx} = (x-y) - 2x(x^2 + y^2). \] \[ \frac{dy}{dx} = \frac{(x-y) - 2x(x^2 + y^2)}{2y(x^2 + y^2) + (x-y)}. \]

Step 3: Calculate Slope at Point (1, 0)

Substitute point \((1, 0)\) into the derivative formula: \[ \frac{dy}{dx} = \frac{(1-0) - 2 \cdot 1(1^2 + 0^2)}{2 \cdot 0(1^2 + 0^2) + (1-0)} = \frac{1 - 2}{0 + 1} = \frac{-1}{1} = -1. \] The slope of the curve at \((1, 0)\) is \(-1\).

Step 4: Calculate Slope at Point (1, -1)

Substitute point \((1, -1)\) into the derivative formula: \[ \frac{dy}{dx} = \frac{(1 - (-1)) - 2 \cdot 1(1^2 + (-1)^2)}{2(-1)(1^2 + (-1)^2) + (1 - (-1))}. \] Simplifying: \[ \frac{dy}{dx} = \frac{2 - 4}{-4 + 2} = \frac{-2}{-2} = 1. \] The slope of the curve at \((1, -1)\) is \(1\).

Key Concepts

Derivative of Implicit Functions

When dealing with equations where the dependent variable, typically \(y\), is not isolated (like in explicit functions), we must use a technique called implicit differentiation. In implicit differentiation, both the dependent and independent variables are mixed on either side of the equation. This technique is essential in calculus because it enables us to find the derivative of implicit functions.

To apply implicit differentiation, follow these key steps:

• Differentiate both sides with respect to an independent variable like \(x\), treating \(y\) as a function of \(x\).
• Apply the chain rule for terms involving \(y\). This means every time you differentiate a function involving \(y\), multiply by \(\frac{dy}{dx}\) because \(y\) is treated as \(y(x)\).
• Once differentiated, collect all terms involving \(\frac{dy}{dx}\) on one side of the equation.
• Solve for \(\frac{dy}{dx}\) to find the derivative.

Implicit differentiation is particularly helpful when the equation of a curve cannot be easily rearranged to solve for \(y\) as a function of \(x\). This makes it an indispensable tool for curves where explicit formulas are cumbersome or impossible to obtain.

Slope of a Curve

Understanding the slope of a curve is fundamental in calculus and relates to how steep a line is at any given point on the curve. The slope tells us the rate at which \(y\) is changing with respect to \(x\).

For a curve described by an implicit equation, like \((x^2 + y^2)^2 = (x-y)^2\), the slope at a specific point is obtained using the derivative we've found through implicit differentiation. At a point \((x_0, y_0)\), substitute \(x_0\) and \(y_0\) into \(\frac{dy}{dx}\). This gives the slope of the tangent line to the curve at that point.

• A positive slope indicates the curve is rising as \(x\) increases, while a negative slope means it is falling.
• If the slope is zero, the curve is flat at that point.
• Infinite or undefined slopes indicate a vertical tangent, occurring when the denominator in \(\frac{dy}{dx}\) becomes zero.

Using implicit differentiation, we've been able to find specific slopes at given points. For instance, at \((1,0)\), the slope is \(-1\), resulting in a decreasing curve. At \((1,-1)\), it is \(1\), indicating an increasing curve.

Calculus Problem Solving

Solving calculus problems involving implicit differentiation can appear complex at first, but they become manageable with practice and a step-by-step approach. The goal is always to find the derivative, which in turn helps determine the behavior of the function, such as calculating slopes at specific points.

Here’s how you can tackle these problems effectively:

• Understand the problem: Carefully read the problem statement, identifying what is being asked. Here, it was to find the slope at specific points on the curve.
• Implicit differentiation: Use it to derive an expression for \(\frac{dy}{dx}\). Follow the rules of differentiation and don't forget chain rules for \(y\).
• Simplify and substitute: Simplify the derived expression as much as possible. Then substitute the given points into \(\frac{dy}{dx}\) to find slopes at those points.
• Interpret your results: Think about what the slopes imply about the curve's shape and direction at the given points.

These methods help make calculus problems, including finding derivatives and slopes from implicit functions, straightforward and solvable. Practicing these steps not only enhances problem-solving skills but also builds confidence in handling more complex calculus scenarios.