Problem 56 - Calculus Single Variable

Question

Suppose that \(a>b>0 .\) Let \(\varepsilon=\sqrt{a^{2}-b^{2}} / a\). Show that, for any \(0<\xi

Step-by-Step Solution

Answer

The length \(L(\xi)\) can be expressed as \(a \cdot E(\varepsilon, \text{arcsin}(\xi/a))\), an elliptic integral dependent on \(\xi\).

Step 1: Understanding Arc Length

To find the length of an arc of the ellipse over the interval \([0, \xi]\), we use the formula for arc length: \[ L = \int \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \] For the given ellipse \((x^2/a^2 + y^2/b^2 = 1), y = b\sqrt{1 - x^2/a^2}\). We need to find \( \frac{dy}{dx} \).

Step 2: Differentiating to Find dy/dx

Differentiate \(y = b\sqrt{1 - x^2/a^2}\) with respect to \(x\):\[ \frac{dy}{dx} = \frac{d}{dx}(b\sqrt{1 - x^2/a^2}) \]Use the chain rule:\[ \frac{dy}{dx} = b \cdot \frac{1}{2\sqrt{1 - x^2/a^2}} \cdot (-2x/a^2) \]\[ = -\frac{bx}{a^2\sqrt{1 - x^2/a^2}} \]

Step 3: Substituting dy/dx into Arc Length Integrand

Substitute \(\frac{dy}{dx}\) into the arc length formula:\[ L(\xi) = \int_0^\xi \sqrt{1 + \left(-\frac{bx}{a^2\sqrt{1 - x^2/a^2}}\right)^2} \, dx \]Simplify the expression inside the square root:\[ L(\xi) = \int_0^\xi \sqrt{1 + \frac{b^2x^2}{a^4(1-x^2/a^2)}} \, dx \]\[ = \int_0^\xi \sqrt{\frac{a^2 - \varepsilon^2 x^2}{a^2 - x^2}} \, dx \]

Step 4: Change of Variable for Elliptic Integral

To express the integral as an elliptic integral, make the substitution \((x = a \sin(t))\), which implies \((dx = a \cos(t) \, dt)\).Notice that \(x/a = \sin(t)\) implies the range for \(t\) is \(0\) to \( ext{arcsin}(\xi/a)\).Substitute and simplify:\[ L(\xi) = \int_0^{\text{arcsin}(\xi/a)} \sqrt{\frac{a^2 - \varepsilon^2 a^2 \sin^2(t)}{a^2 - a^2 \sin^2(t)}} \, a \cos(t) \, dt \]\[ = a \int_0^{\text{arcsin}(\xi/a)} \sqrt{1 - \varepsilon^2 \sin^2(t)} \, dt \]

Step 5: Expressing in terms of Elliptic Integral

Now, the integral matches the standard form of the elliptic integral of the second kind \(E(\varepsilon, \eta)\):\[ L(\xi) = a \cdot E(\varepsilon, \text{arcsin}(\xi/a)) \] Here, \(E(\varepsilon, \eta) = \int_0^{\eta} \sqrt{1 - \varepsilon^2 \sin^2(t)} \, dt\).

Key Concepts

Arc Length of Ellipse

Understanding the arc length of an ellipse is essential in calculus, especially when dealing with non-linear shapes such as ellipses. An ellipse is defined mathematically by the equation \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), where \(a\) and \(b\) are the semi-major and semi-minor axes, respectively. Calculating the arc length over a portion of the ellipse involves using integral calculus.

• Firstly, the arc length \(L\) can be obtained with the standard arc length formula \(L = \int \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx\).
• Substituting the ellipse equation into the arc length formula requires deriving \(\frac{dy}{dx}\).
• As a result, the integral simplifies to an expression that seems complex, showcasing the unique intricacies of working with ellipses.

Determining the arc length of an ellipse numerically often involves elliptic integrals, especially when derived expressions become non-elementary.

Calculus Problem Solving

Solving calculus problems systematically involves breaking down complex expressions into manageable parts. For the arc length problem of an ellipse, it is crucial to:

• Identify the overall goal, which is to express the length over a specific interval.
• Use differentiation to find rates of changes such as \(\frac{dy}{dx}\) for the ellipse's equation.
• Apply integrals to accumulate continuous sum, which, in this case, gives the arc length over the desired interval.

By concentrating on each step, the conversion from geometry to calculus, and then to a solvable mathematical expression, becomes more approachable. Advanced problems often require both analytical skills and computational tools, especially when dealing with special functions like elliptic integrals.

Change of Variable Technique

The change of variable technique is a powerful method in calculus to simplify integrals, particularly those with complex expressions. In the context of finding the arc length of an ellipse that translates to elliptic integrals, this technique helps by simplifying the integral's limits and integrand:

• We made the substitution \(x = a \sin(t)\), which effectively changes the variable and transforms limits from \(0\) to \(\arcsin(\xi/a)\).
• This substitution also transforms the differential \(dx\) into \(a \cos(t) \, dt\), making the integration process more straightforward.
• Such a technique often reduces complicated integrals into known forms, letting us express them in terms of standard mathematical functions like elliptic integrals.

Using change of variables in calculus is akin to a shortcut or simplification process that exposes the inherent structure within complex problems, making them manageable.

Derivatives in Calculus

Derivatives are central to understanding changes in calculus, especially when calculating the arc length of curves like ellipses. In this exercise:

• We differentiate \(y = b\sqrt{1 - \frac{x^2}{a^2}}\) to find \(\frac{dy}{dx}\).
• The differentiation involves the chain rule, reflecting how complex dependencies in functions influence the derivative outcome.
• The derived expression for \(\frac{dy}{dx} = -\frac{bx}{a^2\sqrt{1-x^2/a^2}}\) is substituted back into the arc length equation, emphasizing differentiation's role in transforming geometric problems into analytical calculations.

Understanding derivatives enables the assessment of path slopes, enabling the accurate calculation of arc lengths in varied mathematical contexts. Differentiation is a tool that translates motion and changes into quantifiable metrics in the realm of calculus.