/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 55 Show that the hyperbolic arc \(y... [FREE SOLUTION] | 91影视

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Chapter 11: Problem 55

Show that the hyperbolic arc \(y=(b / a) \sqrt{x^{2}-a^{2}}\) is asymptotic to the line \(y=(b / a) x\) as \(x \rightarrow \infty\)

Short Answer

Expert verified
The hyperbolic arc \(y = \frac{b}{a} \sqrt{x^2 - a^2}\) is asymptotic to the line \(y = \frac{b}{a} x\) as \(x \rightarrow \infty\) because the limit of their ratio as \(x\) approaches infinity equals 1: \[\lim_{x \rightarrow \infty} \frac{\frac{b}{a} \sqrt{x^2 - a^2}}{\frac{b}{a} x} = \lim_{x \rightarrow \infty} \frac{\sqrt{x^2 - a^2}}{x} = 1\]

Step by step solution

01

Write the ratio of the functions

We want to check if the following limit exists: \[\lim_{x \rightarrow \infty} \frac{\frac{b}{a} \sqrt{x^2 - a^2}}{\frac{b}{a} x}\]
02

Simplify the expression

We can simplify the expression inside the limit as follows: \[\frac{\frac{b}{a} \sqrt{x^2 - a^2}}{\frac{b}{a} x} = \frac{\sqrt{x^2 - a^2}}{x}\]
03

Apply the limit

Now we need to find the limit of the simplified expression as \(x \rightarrow \infty\): \[\lim_{x \rightarrow \infty} \frac{\sqrt{x^2 - a^2}}{x}\]
04

Rationalize the numerator

To simplify the limit further, we can rationalize the numerator by multiplying the numerator and the denominator by the conjugate of the numerator: \[\lim_{x \rightarrow \infty} \frac{\sqrt{x^2 - a^2}}{x} \times \frac{\sqrt{x^2 + a^2}}{\sqrt{x^2 + a^2}}\] This gives us: \[\lim_{x \rightarrow \infty} \frac{x^2 - a^2}{x \sqrt{x^2 + a^2}}\]
05

Simplify the limit

We can simplify the limit further by factoring out an \(x^2\) term from the numerator and the denominator: \[\lim_{x \rightarrow \infty} \frac{x^2(1 - \frac{a^2}{x^2})}{x^2 \sqrt{1 + \frac{a^2}{x^2}}}\] Now, we can cancel out the \(x^2\) terms: \[\lim_{x \rightarrow \infty} \frac{1 - \frac{a^2}{x^2}}{\sqrt{1 + \frac{a^2}{x^2}}}\]
06

Evaluate the limit

As \(x\) approaches infinity, the terms \(\frac{a^2}{x^2}\) approach 0. Thus, the limit becomes: \[\lim_{x \rightarrow \infty} \frac{1}{\sqrt{1}}\] Which simplifies to: \[1\] Since the limit of the ratio of the functions is 1 as \(x \rightarrow \infty\), it means that the hyperbolic arc \(y = \frac{b}{a} \sqrt{x^2 - a^2}\) is asymptotic to the line \(y = \frac{b}{a} x\) as \(x \rightarrow \infty\).

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Most popular questions from this chapter

Set \(a_{n}=\frac{1}{n(n+1)}, n=1.2 .3, \cdots,\) and form the sequence \(S_{1}=a_{1}\) \(S_{2}=a_{1}+a_{2}\) \(S_{1}=a_{1}+a_{2}+a_{3}\) . . . \(S_{n}=a_{1}+a_{2}+a_{3}+\dots+a_{n}\) . . . Find a formula for \(S\), that does not involve adding up the terms \(a_{1}, a_{2}, a_{3}, \cdots\). HINT: Use partial fractions to write \(t$$/[k(k+1)]\) as the sum of two fractions.

Below are some sequences defined recursively. Determine in each case whether the sequence converges and, if so, find the limit. Start each sequence with \(a_{1}=1\). $$a_{n+1}=\frac{1}{2} a_{n}+1$$

Use comparison test (11.7.2) to determine whether the integral converges. $$\int_{a}^{\infty} \frac{\sin ^{2} 2 x}{x^{2}} d x$$

Give the first six terms of the sequence and then give the \(n\) th term. $$a_{1}=3 ; \quad a_{n+1}=4-a_{n}$$.

Show that $$ \frac{2^{n}}{n !} \rightarrow 0 $$ by showing that $$ \frac{2^{n}}{n !} \leq \frac{4}{n} $$
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