Problem 20 - The Calculus with Analytic Geometry

Question

Given \(b^{2} x^{2}-a^{2} y^{2}=a^{2} b^{2}\left(a\right.\) and \(b\) are constants), find \(D_{x}{\underline{\phantom{xx}}}^{2} y\) in simplest form.

Step-by-Step Solution

Answer

-\(\frac{b}{a\sqrt{x^2-a^2}}\)

Step 1: Identify the Given Equation

The given equation is: \[ b^{2} x^{2} - a^{2} y^{2} = a^{2} b^{2} \] where \(a\) and \(b\) are constants.

Step 2: Express y as a function of x

Rearrange the equation to solve for \(y^2\):\[ b^{2} x^{2} - a^{2} y^{2} = a^{2} b^{2} \]\[ -a^{2} y^{2} = a^{2} b^{2} - b^{2} x^{2} \]\[ y^{2} = \frac{b^{2} x^{2} - a^{2} b^{2}}{a^{2}} \].

Step 3: Simplify the expression

Simplify the right side of the equation:\[ y^{2} = \frac{b^{2}(x^{2} - a^{2})}{a^{2}} \]\[ y = \pm \sqrt{\frac{b^{2}(x^{2} - a^{2})}{a^{2}}} \]Now, let’s choose the positive root for simplicity: \[ y = \frac{b \sqrt{x^{2} - a^{2}}}{a} \].

Step 4: Find the First Derivative of y

Compute the first derivative \( \frac{dy}{dx} \):\[ y = \frac{b \sqrt{x^{2} - a^{2}}}{a} \]Using the chain rule,\[ \frac{dy}{dx} = \frac{b}{a} \cdot \frac{1}{2 \sqrt{x^{2} - a^{2}}} \cdot 2x = \frac{bx}{a\sqrt{x^{2} - a^{2}}} \].

Step 5: Find the Second Derivative of y

Compute the second derivative \( \frac{d^{2}y}{dx^2} \):\[ \frac{dy}{dx} = \frac{bx}{a\sqrt{x^{2} - a^{2}}} \]Using the quotient rule,\[ \frac{d}{dx} \left( \frac{bx}{a\sqrt{x^{2} - a^{2}}} \right) = \frac{b \left(a \sqrt{x^{2} - a^{2}} \cdot 1 - x \cdot \frac{d}{dx}(a \sqrt{x^{2} - a^{2}}) \right) }{ a (x^{2}-a^{2})} \]Convert and simplify the derivatives further to obtain the result.

Step 6: Final Calculation

Simplify using derivatives:\[ \frac{d}{dx}(a \sqrt{x^{2}-a^{2}}) = \frac{a \cdot x}{\sqrt{x^{2}-a^{2}}} \]So,\[ \frac{d_{x}^2y}{dx^2} = \frac{b \left(a \sqrt{x^{2} - a^{2}} - x \cdot \frac{ax}{\sqrt{x^{2} - a^{2}}} \right)}{a (x^{2} - a^{2})} \]\[ = \frac{b \left(a \sqrt{x^{2}-a^{2}} - \frac{a x^2}{\sqrt{x^{2}-a^{2}}} \right)}{a (x^{2}-a^{2})} \]\[ = \frac{ba (x^2 - a^2) - a x^2}{a(a \sqrt{x^{2}-a^{2}} (x^{2}-a^{2}))}\]Which simplifies to:\[ \frac{-ba}{a^2 (\√x^2 − a^2)}\]Thus,\[ \boxed{-\frac{b}{a\sqrt{x^2-a^2}}} \].

Key Concepts

Implicit Differentiation

Implicit differentiation is useful when you cannot solve an equation explicitly for one variable. For instance, given the equation \(b^2 x^2 - a^2 y^2 = a^2 b^2\), isolating \(y\) in terms of \(x\) directly can be complex. Instead, take derivatives directly on both sides of the equation with respect to \(x\), treating \(y\) as a function of \(x\). This way, we effectively differentiate \(y\) as we would any function, but using the chain rule to incorporate \(\frac{dy}{dx}\). This technique allows us to find the slopes of curves that are difficult to handle algebraically by explicitly solving for \(y\).
Example:
For \( x^2 + y^2 = 1 \), differentiate both sides with respect to \( x\):
\(2x + 2y\frac{dy}{dx} = 0\)
Then, solve for \( \frac{dy}{dx}\):
\( \frac{dy}{dx} = -\frac{x}{y}\)

Quotient Rule

The quotient rule is used when differentiating a function that is the quotient of two other functions. Formally, for functions \( u(x) \) and \( v(x) \), if we have a function \( f(x) = \frac{u(x)}{v(x)} \), the derivative \( \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) \) is:
\[ \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} \] This rule is crucial when dealing with rational functions and allows us to compute derivatives in a structured, rule-based manner instead of using longer algebraic methods. In our example:
Given \( y = \frac{b \sqrt{x^2 - a^2}}{a} \) we use the quotient rule to find \( \frac{dy}{dx} \).
Here, the numerator \( u(x) = b \sqrt{x^2 - a^2} \) and the denominator \( v(x) = a\). Thus, using the quotient rule we derive:
\( \frac{dy}{dx} = \frac{ (b) * (\frac{x}{\sqrt{x^2 - a^2}}) - (b \sqrt{x^2 - a^2})(0)}{a^2}\)

Chain Rule

The chain rule is essential when differentiating composite functions. If a function \( y \) depends on \( u \), and \( u \) depends on \( x \), then \( y \) indirectly depends on \( x \). Mathematically, if \( y = f(u) \) and \( u = g(x) \), then:
\( \frac{dy}{dx} = \frac{dy}{du} * \frac{du}{dx} \)
This rule breaks down the complex process of differentiating compositions into simpler steps.
Take our initial function:
\( y = \frac{b \sqrt{x^2 - a^2}}{a}\) If we differentiate this using the chain rule, we need to recognize that \( y = \frac{b}{a} * (x^2 - a^2)^{1/2} \). First, let\( u = x^2 - a^2 \). Hence, \( y = \frac{b}{a} * u^{1/2} \). The chain rule then gives us \( dy/dx \) by first computing \( dy/du \) and multiplying it by \( du/dx \).

Calculus Problem

Solving a calculus problem often requires a combination of multiple differentiation rules and techniques. The given exercise asks us to find the second derivative of \( y \) with respect to \( x \) for the equation \( b^2 x^2 - a^2 y^2 = a^2 b^2 \). This entails several steps:

• Rewriting the equation to isolate \( y \).
• Using implicit differentiation to find the first derivative.
• Simplifying and then applying the quotient rule for the second derivative.

Calculus problems often test the application of various methods learned in calculus courses. More complex problems, like finding higher-order derivatives, require sequential application of these rules. Each step builds on your understanding of fundamental calculus principles like the chain rule and the quotient rule as demonstrated in the example problem.