Chapter 5: Problem 9
Determine the amplitude and period of each function. Then graph one period of the function. $$y=3 \sin \frac{1}{2} x$$
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Chapter 5: Problem 9
Determine the amplitude and period of each function. Then graph one period of the function. $$y=3 \sin \frac{1}{2} x$$
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Describe the restriction on the sine function so that it has an inverse function.
For years, mathematicians were challenged by the following problem: What is the area of a region under a curve between two values of \(x ?\) The problem was solved in the seventeenth century with the development of integral calculus. Using calculus, the area of the region under \(y=\frac{1}{x^{2}+1},\) above the \(x\) -axis, and between \(x=a\) and \(x=b\) is \(\tan ^{-1} b-\tan ^{-1} a\). Use this result, shown in the figure, to find the area of the region under \(y=\frac{1}{x^{2}+1}\) above the \(x\) -axis, and between the values of a and b given in Exercises \(97-98\). (GRAPH CANNOT COPY) \(a=0\) and \(b=2\)
Explain what is meant by one radian.
Graph \(y=\tan ^{-1} x\) and its two horizontal asymptotes in a \([-3,3,1]\) by \(\left[-\pi, \pi, \frac{\pi}{2}\right]\) viewing rectangle. Then change the viewing rectangle to \([-50,50,5]\) by \(\left[-\pi, \pi, \frac{\pi}{2}\right] .\) What do you observe?
Your neighborhood movie theater has a 25 -foot-high screen located 8 feet above your eye level. If you sit too close to the screen, your viewing angle is too small, resulting in a distorted picture. By contrast, if you sit too far back, the image is quite small, diminishing the movie's visual impact. If you sit \(x\) feet back from the screen, your viewing angle, \(\theta,\) is given by $$\theta=\tan ^{-1} \frac{33}{x}-\tan ^{-1} \frac{8}{x}$$ (GRAPH CANNOT COPY) Find the viewing angle, in radians, at distances of 5 feet, 10 feet, 15 feet, 20 feet, and 25 feet.
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