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practice-quiz3-sol

# practice-quiz3-sol - AMS 151(Fall 2009 Joe Mitchell Applied...

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AMS 151 (Fall, 2009) Joe Mitchell Applied Calculus I Practice Problems for Quiz # 3 – Solution Notes 1. Evaluate the following expressions: (a). cos(tan 1 ( 5 9 )) = Draw a right triangle and let θ be one of the two angles that is not π/ 2. Then, θ = tan 1 ( 5 9 ) if the side opposite θ is of length 5 and the side adjacent is of length 9. Then, the hypotenuse is of length 5 2 + 9 2 = 106. Thus, cos(tan 1 ( 5 9 )) = cos θ = 9 106 . (b). tan(sin 1 ( 3 2 )) = Draw a right triangle and let θ be one of the two angles that is not π/ 2. Then, θ = sin 1 ( 3 2 ) if the side opposite θ is of length 3 and the hypotenuse is of length 2. Then, the side adjacent is of length r 2 2 ( 3) 2 = 1. Thus, tan(sin 1 ( 3 2 )) = tan θ = 3. (Note that θ = π/ 3, 60 degrees.) (c). sin( 20 π 3 ) = Note that 20 π 3 = 2 π + 2 π + 2 π + 2 π 3 (i.e., it is 3 revolutions, plus an additional angle 2 π/ 3 (120 degrees). Thus, sin( 20 π 3 ) = sin(2 π/ 3) = 3 2 . (d). sec(cos 1 ( 2 3 )) = Draw a right triangle and let θ be one of the two angles that is not π/ 2. Then, θ = cos 1 ( 2 3 ) if the side adjacent to θ is of length 2 and the hypotenuse is of length 3. Then, the side opposite is of length 3 2 2 2 = 5. Thus, sec(cos 1 ( 2 3 )) = sec θ = 1 / (cos θ ) = 1 / (2 / 3) = 3 / 2. 2. The intensity of radiation, I , oscillates sinusoidally between a low of 100 and a high of 1200. The time between one peak and the next peak is 5 minutes. At time 30 seconds after noon today, the radiation level is measured to be 650, and it is rising. Write the function I ( t ) for intensity as a function of t , the number of seconds after noon today. Since

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practice-quiz3-sol - AMS 151(Fall 2009 Joe Mitchell Applied...

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