/*! 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 53 Show that the hyperbolic arc \(y... [FREE SOLUTION] | 91影视

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Chapter 11: Problem 53

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

Let \(\Omega\) be the region bounded by the curve \(y=e^{-x}\) and the \(x\) -axis, \(x \geq 0\). (a) Sketch \(\Omega\). (b) Find the area of \(\Omega\). (c) Find the volume obtained by revolving \(\Omega\) about the \(x\) -axis. (d) Find the volume obtained by revolving \(\Omega\) about the \(y\) -axis. (e) Find the surface area of the configuration in part (c).

The sequence defined recursively by setting $$a_{\mathrm{n}+2}=a_{n+1}+a_{n} \quad \text { starting with } \quad a_{1}=a_{2}=1$$,is called the Fibonacci sequence. (a) Calculate \(a_{3}, a_{4}, \cdots, a_{10}\) (b) Define $$r_{n}=\frac{a_{n+1}}{a_{n}}$$,$$\text { Calculate } r_{1}, r_{2}, \cdots, r_{6}$$, (c) Assume that \(r_{n} \rightarrow L\), and find \(L\). HINT: Rclatc \(r\), to \(r_{n}, 1\).

Suppose that \(S\) is a nonempty bounded set of real numbers and \(T\) is a nonempty subset of \(S\). (a) Show that \(T\) is bounded. (b) Show that glb \(S \leq g\) lb \(T \leq\) lub \(T:\) lub \(S\).

Use a graphing utility to draw the graph of the integrand. Then use a CAS to determine whether the integral converges or diverges. (a) \(\int_{0}^{\infty} \frac{x}{\left(16+x^{2}\right)^{2}} d x\) (b) \(\int_{0}^{\infty} \frac{x^{2}}{\left(16+x^{2}\right)^{2}} d x\) (c) \(\int_{0}^{\infty} \frac{x}{16+x^{4}} d x\) (d) \(\int_{0}^{\infty} \frac{x}{16+x^{2}} d x\)

The set \(S\) of rational numbers \(x\) with \(x^{2}<2\) has rational upper bounds but no least rational upper bound. The argument goes like this. Suppose that \(S\) has a least rational upper bound and call it \(x_{0} .\) Then either $$x_{0}^{2}=2, \quad \text { or } \quad x_{0}^{2}>2, \quad \text { or } \quad x_{0}^{2} < 2.$$ (a) Show that \(x_{0}^{2}=2\) is impossible by showing that if \(x_{0}^{2}=\) 2, then \(x_{0}\) is not rational. (b) Show that \(x_{0}^{2}>2\) is impossible by showing that if \(x_{0}^{2}>2,\) then there is a positive integer \(n\) for which \(\left(x_{0}-\frac{1}{n}\right)^{2}>2,\) which makes \(x_{0}-\frac{1}{n}\) a rational upper bound for \(S\) that is less than the least rational upper bound \(x_{0}\). (c) Show that \(x_{0}^{2}<2\) is impossible by showing that if \(x_{0}^{2}<2,\) then there is a positive integer \(n\) for which \(\left(x_{0}+\frac{1}{n}\right)^{2}<2 .\) This places \(x_{0}+\frac{1}{n}\) in \(S\) and show's that \(x_{0}\) cann0ot be an upper bound for \(S\).
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