Problem 34 - Calculus

Question

(for professors) Compare the lengths of the parabola \(y=x^{2}\) and the line \(y=b x\) from (0,0) to \(\left(b, b^{2}\right) .\) Does the difference approach a limit as \(b \rightarrow \infty\) ?

Step-by-Step Solution

Answer

The difference in lengths diverges to infinity as \( b \rightarrow \infty \).

Step 1: Understand the problem

We need to compare the lengths of two curves: the parabola \( y = x^2 \) and the line \( y = bx \) from the origin \((0,0)\) to the point \((b, b^2)\). Additionally, we have to determine if the difference in their lengths approaches a limit as \( b \rightarrow \infty \).

Step 2: Find the Length of the Parabola

The length of a curve from \(a\) to \(b\) is given by the integral \(\int_{a}^{b} \sqrt{1 + (dy/dx)^2} \, dx.\) For the parabola \(y = x^2\), \(dy/dx = 2x\). So, the length is \( \int_{0}^{b} \sqrt{1 + (2x)^2} \, dx = \int_{0}^{b} \sqrt{1 + 4x^2} \, dx.\)

Step 3: Find the Length of the Line

For the line \( y = bx \), \( dy/dx = b \). Thus, the length of the line from \((0,0)\) to \((b, b^2)\) is \( \int_{0}^{b} \sqrt{1 + b^2} \, dx = \sqrt{1 + b^2} * b.\)

Step 4: Compute the Lengths

For practical calculation, the integral for the parabola requires substitution or numerical methods: Let \( x = \frac{1}{2} \sinh(u) \) i.e., \( 4x^2 = \sinh^2(u) \), then the arc length becomes \( \int_{0}^{\sinh^{-1}(2b)} \cosh(u) \, du = \frac{1}{2}\sinh(2u) + u\Big|_{0}^{\sinh^{-1}(2b)}.\) Calculate this at the point at \( b.\)

Step 5: Compare the Lengths as b Approaches Infinity

Let's consider the asymptotic behavior. The length of the parabola (calculated using elliptic integrals) grows like \( \approx \frac{1}{2} 4b = 2b \) for large \( b \), whereas the length of the line is \( \sqrt{1 + b^2} \, b \approx b^2 \). Since \( 2b \) and \( b^2 \) diverge, their difference \( b^2 - 2b \rightarrow \infty \) as \( b \rightarrow \infty \).

Step 6: Conclusion

The difference in lengths does not approach a fixed limit as \( b \rightarrow \infty \). The length of the line grows much quicker (\( b^2 \)) than that of the parabola (\( 2b \)). Thus, as \( b \rightarrow \infty \), the difference diverges to infinity.

Key Concepts

Parabola Length Calculation

When calculating the length of a parabola, we use the fundamental arc length formula from calculus:

• The formula is: \[ L = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \]
• This integral helps capture how the slope changes along the curve.

For the parabola \( y = x^2 \), the derivative \( \frac{dy}{dx} = 2x \) is central to finding the length. Thus, the problem transforms into calculating:\[ L = \int_{0}^{b} \sqrt{1 + (2x)^2} \, dx = \int_{0}^{b} \sqrt{1 + 4x^2} \, dx \]This integral is complex and often requires specific techniques like substitution or numerical methods to solve. Don't be discouraged by the complexity—it is a common challenge in calculus.

Curve Comparison

Comparing the lengths of different curves involves evaluating each independently and then analyzing their differences.

• The parabola \( y = x^2 \) and the line \( y = bx \) have unique characteristics.
• While a parabola curves continuously, a line remains straight.

To compare them from the origin \((0,0)\) to \((b, b^2)\):

• Calculate the length of the parabola using the respective integral for \( y = x^2 \).
• For the line, derive its length from the simpler integral with \( \text{constant slope} \), resulting in \( \sqrt{1 + b^2} \cdot b \).

By analyzing these, we realize they behave very differently as \( b \) increases.

Integral Calculus

Integral calculus plays a crucial role in these length calculations. It deals with accumulation—like summing small lengths to find the total. For curves:

• An integral allows us to add up small intervals along the curve, each computed using the arc length formula.
• In our problem, we often substitute with functions like hyperbolic sine and cosine for the parabola calculation.

Making a variable substitution (like \( x = \frac{1}{2} \sinh(u) \)) assists in simplifying the integral:\[ L = \int_{0}^{\sinh^{-1}(2b)} \cosh(u) \, du = \frac{1}{2}\sinh(2u) + u\Big|_{0}^{\sinh^{-1}(2b)} \]While seemingly intricate, understanding this substitution approach is vital as it transforms complex integrals into more manageable forms.

Asymptotic Analysis

Asymptotic analysis helps to understand the behavior of functions as variables grow large. In this problem, it is about how lengths change as \( b \rightarrow \infty \).

• For the parabola, the length grows approximately like \( 2b \).
• The length of the line approximates to \( b^2 \) for larger \( b \).

This analysis is especially helpful in comparing functions when they extend towards infinity:

• The parabola's \( 2b \) growth is linear, slower than the quadratic growth of the line's \( b^2 \).
• Thus, the difference between them, \( b^2 - 2b \), keeps increasing and diverges to infinity.

This shows the practical use of asymptotic analysis in calculus—offering insights into the infinite behavior of different mathematical entities.