Problem 23 - Calculus Early Transcendentals

Question

(a) Find a slope field whose integral curve through \((1,1)\) satisfies \(x y^{3}-x^{2} y=0\) by differentiating this equation implicitly. (b) Prove that if \(y(x)\) is any integral curve of the slope field in part (a), then \(x[y(x)]^{3}-x^{2} y(x)\) will be a constant function. (c) Find an equation that implicitly defines the integral curve through \((-1,-1)\) of the slope field in part (a).

Step-by-Step Solution

Answer

The slope field equation is \(\frac{dy}{dx} = \frac{2xy - y^3}{x (3y^2 - x)}\). The function \(x y^3 - x^2 y\) is constant for any integral curve. The curve through \((-1, -1)\) satisfies \(x y^3 - x^2 y = 0\).

Step 1: Differentiate Implicitly

Given the equation \(x y^{3} - x^{2} y = 0\), we will differentiate both sides with respect to \(x\). The left side becomes \(d/dx\left(x y^{3}\right) - d/dx\left(x^{2} y\right)\). Use the product rule for differentiation: \((u v)' = u'v + uv'\).ewline\[\frac{d}{dx}(xy^3) = x \cdot 3y^2 \cdot \frac{dy}{dx} + y^3 \] ewline\[\frac{d}{dx}(x^2y) = 2xy + x^2 \cdot \frac{dy}{dx} \] ewlineSet the derivatives equal: ewline\[ x \cdot 3y^2 \cdot \frac{dy}{dx} + y^3 - (2xy + x^2 \cdot \frac{dy}{dx}) = 0 \]

Step 2: Simplify and Solve for dy/dx

To find \( \frac{dy}{dx} \), reorganize the equation: ewline\[ x \cdot 3y^2 \cdot \frac{dy}{dx} - x^2 \cdot \frac{dy}{dx} = 2xy - y^3 \] ewlineFactor \( \frac{dy}{dx} \) and solve: ewline\[ \frac{dy}{dx} (3xy^2 - x^2) = 2xy - y^3 \] ewline\[ \frac{dy}{dx} = \frac{2xy - y^3}{x (3y^2 - x)} \]

Step 3: Prove Constant Function

To show \( x[y(x)]^{3} - x^{2} y(x) \) is constant, use the fact that the derivative with respect to \(x\) of a constant is zero: ewlineCalculate the derivative: ewline\( f(x) = x y^3 - x^2 y \). ewlineUsing the product rule as before, differentiate: ewline\[ \frac{d}{dx}(xy^3) - \frac{d}{dx}(x^2y) = 0 \] ewlineFrom Step 1, this expression is already zero because it equals the expression derived for \( \frac{dy}{dx} \). Thus, any integral curve will satisfy this condition.

Step 4: Integral Curve Through a Point

To find the equation of the integral curve going through \((-1, -1)\), use the provided equation \(x y^3 - x^2 y = 0\). Because \((-1, -1)\) is given as a point on the curve, substitute it into the equation: ewline\[ (-1)(-1)^3 - (-1)^2(-1) = 0 \] ewline\[ -1 + 1 = 0 \].ewlineThis fits, confirming \(x y^3 - x^2 y = 0\) implicitly defines the curve through \((-1, -1)\).

Key Concepts

Implicit Differentiation

When we talk about implicit differentiation, it's like finding the hidden relationships between variables of an equation. In our exercise, we have a tricky situation where everything can't be easily expressed as a simple 'y equals' formula. Instead, we have a mix of x and y: \(x y^{3} - x^{2} y = 0\). So, how do we tackle this?

• We differentiate both sides with respect to \(x\).
• We apply the product rule: this helps break down multiplication into manageable parts.
• For \(xy^3\), think of \(x\) as one part and \(y^3\) as another. We get \((3xy^2\frac{dy}{dx} + y^3)\).
• Similarly, for \(x^2y\), it becomes \((2xy + x^2\frac{dy}{dx})\).
• After simplifying, you find \(\frac{dy}{dx} = \frac{2xy - y^3}{x (3y^2 - x)}\).

This approach lets us evaluate changes in the slope field, unveiling the behavior of the integral curve through differentiation.

Integral Curve

An integral curve essentially traces out a path following the slope field's direction. Imagine small arrows indicating slope at any given point. The curve gliding along these arrows is your integral curve.

• In part (a) of the exercise, the goal was to find the curve passing through (1,1) for the given equation.
• Implicit differentiation gave us a way to find \(\frac{dy}{dx}\), the slope of this curve at any point.
• Integrating this slope information constructs the integral curve: a visual representation of solutions passing through specific points (like (1,1) or (-1,-1)).

This curve is unique to the initial condition, demonstrating how powerful curves like "integral curves" can project the full picture from local slope details.

Constant Function

A constant function is a special type of function where the output value doesn't change. It's like a flat line on a graph. In part (b) of the exercise, we proved a special property about our integral curve.

• The task was to show that \(x[y(x)]^3 - x^2y(x)\) is constant along the integral curve.
• This involves taking the derivative of the entire expression and showing it remains zero, meaning no change occurs while moving along the curve.
• By using the result from implicit differentiation, we ensured that this expression actually represents a constant.

Finding a constant function is essential because it tells us that the integral curve stabilizes under these conditions, staying at a fixed value without variability.