Chapter 6: Problem 2
Suppose the velocity of an object moving along a line is positive. Are displacement and distance traveled equal? Explain.
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Chapter 6: Problem 2
Suppose the velocity of an object moving along a line is positive. Are displacement and distance traveled equal? Explain.
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A tsunami is an ocean wave often caused by earthquakes on the ocean floor; these waves typically have long wavelengths, ranging between \(150\) to \(1000 \mathrm{km} .\) Imagine a tsunami traveling across the Pacific Ocean, which is the deepest ocean in the world, with an average depth of about \(4000 \mathrm{m}.\) Explain why the shallow-water velocity equation (Exercise 71 ) applies to tsunamis even though the actual depth of the water is large. What does the shallow-water equation say about the speed of a tsunami in the Pacific Ocean (use \(d=4000 \mathrm{m}\) )?
Differentiate \(\ln x\) for \(x>0\) and differentiate \(\ln (-x)\) for \(x<0\) to conclude that \(\frac{d}{d x}(\ln |x|)=\frac{1}{x}\).
Consider the functions \(f(x)=a \sin 2 x\) and \(g(x)=(\sin x) / a,\) where \(a>0\) is a real number. a. Graph the two functions on the interval \([0, \pi / 2],\) for \(a=\frac{1}{2}, 1\) and 2. b. Show that the curves have an intersection point \(x^{*}\) (other than \(x=0)\) on \([0, \pi / 2]\) that satisfies \(\cos x^{*}=1 /\left(2 a^{2}\right),\) provided \(a>1 / \sqrt{2}\) c. Find the area of the region between the two curves on \(\left[0, x^{*}\right]\) when \(a=1\) d. Show that as \(a \rightarrow 1 / \sqrt{2}^{+}\). the area of the region between the two curves on \(\left[0, x^{*}\right]\) approaches zero.
Find the volume of the solid of revolution. Sketch the region in question. The region bounded by \(y=e^{x}, y=0, x=0,\) and \(x=2\) revolved about the \(x\) -axis
a. The definition of the inverse hyperbolic cosine is \(y=\cosh ^{-1} x \Leftrightarrow x=\cosh y,\) for \(x \geq 1,0 \leq y<\infty.\) Use implicit differentiation to show that \(\frac{d}{d x}\left(\cosh ^{-1} x\right)=\) \(1 / \sqrt{x^{2}-1}.\) b. Differentiate \(\sinh ^{-1} x=\ln (x+\sqrt{x^{2}+1})\) to show that \(\frac{d}{d x}\left(\sinh ^{-1} x\right)=1 / \sqrt{x^{2}+1}.\)
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