Problem 47 - Advanced Engineering Mathematics

Question

The graphs of the members of the one-parameter family \(x^{3}+\) \(y^{3}=3 c x y\) are called folia of Descartes. Verify that this family is an implicit solution of the first-order differential equation $$ \frac{d y}{d x}=\frac{y\left(y^{3}-2 x^{3}\right)}{x\left(2 y^{3}-x^{3}\right)} $$

Step-by-Step Solution

Answer

Yes, the solution family satisfies the given differential equation.

Step 1: Differentiate the Folium of Descartes Equation

Start with the implicit function given by the folium of Descartes: \(x^3 + y^3 = 3cxy\). Differentiating both sides of this equation implicitly with respect to \(x\) using derivative rules, we get:\\(3x^2 + 3y^2 \frac{dy}{dx} = 3c \left(y + x \frac{dy}{dx}\right)\)

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

Rearrange the differentiated equation from Step 1 to solve for \(\frac{dy}{dx}\):\\(3y^2 \frac{dy}{dx} - 3cx \frac{dy}{dx} = 3cy - 3x^2\)\Factor out \(\frac{dy}{dx}\):\\(\frac{dy}{dx}(3y^2 - 3cx) = 3cy - 3x^2\)\So, \(\frac{dy}{dx} = \frac{3cy - 3x^2}{3y^2 - 3cx}\)

Step 3: Simplify the Result

Simplify \(\frac{dy}{dx} = \frac{3cy - 3x^2}{3y^2 - 3cx}\) by dividing both the numerator and the denominator by 3:\\(\frac{dy}{dx} = \frac{cy - x^2}{y^2 - cx}\)

Step 4: Verify with Given Differential Equation

The given differential equation for verification is \(\frac{dy}{dx} = \frac{y(y^3 - 2x^3)}{x(2y^3 - x^3)}\). Replace \(c\) using the implicit folium equation \(x^3 + y^3 = 3cxy\) to find \(c = \frac{x^3 + y^3}{3xy}\). Substitute this \(c\) into the simplified form found in Step 3: \(\frac{dy}{dx} = \frac{\frac{(x^3+y^3)y}{3xy} - x^2}{y^2 - \frac{x^3+y^3}{3xy} x}\), which simplifies to match the given equation after algebraic rearrangements, verifying that the implicit solution holds true for the given differential equation.

Key Concepts

Implicit Differentiation

Implicit differentiation is a handy technique used in calculus when you're dealing with equations that can't be expressed easily as a function of one variable. Instead of solving the equation for one variable and differentiating, implicit differentiation lets us differentiate both sides of the equation with respect to a variable directly. This is essential when dealing with equations like the Folium of Descartes.

When you have an equation like the Folium of Descartes: \[ x^3 + y^3 = 3cxy \]Attempting to solve for either \(x\) or \(y\) directly is complex or even impossible in simple terms. By differentiating each term with respect to \(x\), and applying the chain rule where needed, we introduce \( \frac{dy}{dx} \) to capture how \(y\) changes relative to \(x\). This allows us to determine the derivative without explicit separation.

• Use the power rule on terms like \(x^3\) to get \(3x^2\).
• Use the chain rule for \(y^3\) to end up with \(3y^2 \frac{dy}{dx}\).
• Chain and product rules help with terms like \(3cxy\), leading to differentiation of the parts.

By rearranging the resulting equation, you can solve for \(\frac{dy}{dx}\) and find a relationship between the rates of change of \(x\) and \(y\). This method is critical in finding solutions for more complex relationships in calculus.

Folium of Descartes

The Folium of Descartes is an elegant curve known for its leaf-like shape, named after the philosopher and mathematician René Descartes. This particular curve is represented by the equation:\[ x^3 + y^3 = 3cxy \]This cubic curve is special because it is a member of a family of curves defined over one parameter, \(c\). The shape of the folium changes depending on the value of \(c\). When \(c=1\), the folium has a distinctive loop.

• It consists of an asymptotic line at 45 degrees, extending from each axis.
• The equation is symmetric with respect to the line \(y=x\).
• The curve passes through the origin, emphasizing its distinctive loop pattern.

The Folium of Descartes is often studied using implicit differentiation since \(y\) cannot be easily expressed in terms of \(x\). We use this implicitly defined curve to explore more robust solutions in calculus and investigate relationships between parameters in geometric terms.

First-order Differential Equation

First-order differential equations are equations that involve the first derivative of a function. These equations play a crucial role in modeling and solving problems where rates of change are involved. A first-order differential equation can often be expressed in the form:\[ \frac{dy}{dx} = f(x, y) \]This notation indicates that the rate of change of \(y\) with respect to \(x\) is dependent on both variables. In the context of the Folium of Descartes, our task is to show that the curve is indeed a solution to a given differential equation:\[ \frac{dy}{dx} = \frac{y(y^3 - 2x^3)}{x(2y^3 - x^3)} \]Understanding this involves:

• Using implicit differentiation to find \(\frac{dy}{dx}\) from \(x^3 + y^3 = 3cxy\).
• Substituting the expression for \(c\) from the folium equation to verify against the given differential formula.
• Simplifying the resulting expression to ensure it matches the differential equation precisely.

First-order differential equations like this are foundational in defining dynamic systems and modeling real-world phenomena. They provide insight into how changing conditions affect a system's trajectory, which is mirrored by the behavior of curves like the Folium of Descartes.