/*! 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 56 Show that the hyperbolic arc \(y... [FREE SOLUTION] | 91影视

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Chapter 11: Problem 56

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

Laplace transforms. Let \(f\) b: continuous on \([0,\infty)\). The Laplace transform of \(f\) is the function \(F\) defined by setting $$F(s)=\int_{0}^{\infty} e^{-s x} f(x) d x$$ The domain of \(F\) is the set of numbers \(s\) for which the improper integral converges. Find the Laplace transform \(F\) of each of the following functions specifying the domain of \(F\). $$f(x)=1$$

Set $$a_{n}=\frac{1}{n^{2}}+\frac{2}{n^{2}}+\frac{3}{n^{2}}+\dots+\frac{n}{n^{2}}$$,Show that \(a_{n}\) is a Riemann sum for \(\int_{0}^{1} x d x .\) Docs the sequence \(a_{1}, a_{2}, \cdots\) converge? If so, to what?

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

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Let \(\Omega\) be the region bounded below by \(y\left(x^{2}+1\right)=x,\) above by \(x y=1,\) and to the left by \(x=1\). (a) Find the area of \(\Omega\) (b) Show that the configuration obtained by revolving \(\Omega\) about the \(x\) -axis has finite volume. (c) Calculate the volume generated by revolving \(\Omega\) about the \(y\) -axis.
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