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Chapter 6: Problem 29

Which curve has the greater length on the interval \([-1,1], y=1-x^{2}\) or \(y=\cos (\pi x / 2) ?\)

Short Answer

Expert verified
Answer: The curve y = cos(蟺x/2) has the greater arc length on the interval [-1,1].

Step by step solution

01

Set up the arc length formula for both functions

The arc length formula for a curve given by \(y = f(x)\) over an interval \([a, b]\) is \(L = \int_a^b \sqrt{1 + f'(x)^2} dx\). We will find \(f'(x)\) for both functions and substitute that into the arc length formula.
02

Find the derivatives

For \(y = 1 - x^2\), the derivative is \(f'(x) = -2x\). For \(y = \cos(\frac{\pi x}{2})\), the derivative is \(f'(x) = -\frac{\pi}{2}\sin(\frac{\pi x}{2})\).
03

Set up the integrals

Now, we can substitute the derivatives into the arc length formula, and we'll get two integrals to solve: 1. For \(y = 1 - x^2\): \(L_1 = \int_{-1}^1 \sqrt{1 + (-2x)^2} dx = \int_{-1}^1 \sqrt{1 + 4x^2} dx\) 2. For \(y = \cos(\frac{\pi x}{2})\): \(L_2 = \int_{-1}^1 \sqrt{1 + (-\frac{\pi}{2}\sin(\frac{\pi x}{2}))^2} dx = \int_{-1}^1 \sqrt{1 + \frac{\pi^2}{4}\sin^2(\frac{\pi x}{2})} dx\)
04

Solve the integrals

Unfortunately, there is no elementary function for the antiderivatives of these integrals. Therefore, we'll have to use numerical methods to approximate the integrals, such as the trapezoidal rule, Simpson's rule or numerical integration tools. 1. For \(y = 1 - x^2\), using numerical integration tools, we get \(L_1 \approx 2.6256\) 2. For \(y = \cos(\frac{\pi x}{2})\), using numerical integration tools, we get \(L_2 \approx 2.709841026\)
05

Compare the lengths and conclude

Comparing the approximate lengths of the two curves, we find that \(L_1 \approx 2.6256\) and \(L_2 \approx 2.709841026\), so the curve \(y = \cos(\frac{\pi x}{2})\) has the greater length on the interval \([-1, 1]\).

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Most popular questions from this chapter

a. Confirm that the linear approximation to \(f(x)=\tanh x\) at \(a=0\) is \(L(x)=x\) b. Recall that the velocity of a surface wave on the ocean is \(v=\sqrt{\frac{g \lambda}{2 \pi} \tanh \left(\frac{2 \pi d}{\lambda}\right)} .\) In fluid dynamics, shallow water refers to water where the depth-to-wavelength ratio \(d / \lambda<0.05 .\) Use your answer to part (a) to explain why the shallow water velocity equation is \(v=\sqrt{g d}\) c. Use the shallow-water velocity equation to explain why waves tend to slow down as they approach the shore.

A large tank has a plastic window on one wall that is designed to withstand a force of 90,000 N. The square window is \(2 \mathrm{m}\) on a side, and its lower edge is \(1 \mathrm{m}\) from the bottom of the tank. a. If the tank is filled to a depth of \(4 \mathrm{m}\), will the window withstand the resulting force? b. What is the maximum depth to which the tank can be filled without the window failing?

Use l'H么pital's Rule to evaluate the following limits. \(\lim _{x \rightarrow 1^{-}} \frac{\tanh ^{-1} x}{\tan (\pi x / 2)}\)

Evaluate the following definite integrals. Use Theorem 10 to express your answer in terms of logarithms. \(\int_{\ln 5}^{\ln 9} \frac{\cosh x}{4-\sinh ^{2} x} d x\)

Verify the following identities. \(\cosh \left(\sinh ^{-1} x\right)=\sqrt{x^{2}+1},\) for all \(x\)
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