Graphs Of Sine And Cosine Homework

Problems 1-4, Determine the amplitude and period of each function without graphing. y = 5 sin(πx) a = 5 period: 2 y = -4 cos(3/2 x) a = 4 period: 4π/3 y = 2 sin(3x - π) + 4 a = 2 period: 2π/3 y = 7 cos(4x + π) - 3 a = 7 period: π/2 Problems 5-8, Use transformations to graph each equation from (-2π, 2π) y = 2 sin x a = 2 int-max-int-min-int y = -sin(1/2 x) a = -1 int-min-int-max-int period: 4π y = 2 cos(x) - 1 a = 1 d = -1 max-int-min-int-max y = -3 sin(x) + 2 a = 3 d = 2 int-min-int-max-int Problems 9-12. Graph each equation. Equation Characteristics Graph (y=sin(x-\frac{\pi}{4})+1) Amplitude: 1 [graph of a sine function shifted right by π/4 and up by 1] Period: 2π Phase Shift: right π/4 Midline: y=1 int-max-int-min-int (y=-cos[2(x-\frac{\pi}{2})]) Amplitude: 1 [graph of a cosine function reflected over x-axis, shifted right by π/2, and with a period of π] Period: π Phase Shift: right π/2 Midline: y=0 min-int-max-int-min (y=2~sin\frac{1}{2}(\theta-\pi)+2) Amplitude: 2 [graph of a sine function stretched vertically by 2, shifted right by π, and up by 2, with a period of 4π] Period: 4π Phase Shift: right π Midline: y=2 int-max-int-min-int (y=-3~cos\frac{\pi}{3}(\theta+1)-2) Amplitude: 3 [graph of a cosine function reflected over x-axis, stretched vertically by 3, shifted left by 1, and down by 2, with a period of 6] Period: 6 Phase Shift: left 1 Midline: y=-2 min-int-max-int-min (y=2~cos(\frac{\pi}{2}x)) B (y=-2~sin(\pi x)) D (y=-2~cos(2x)) A (y=2~sin(\frac{1}{2}x)) C [\frac{2\pi}{\theta}=\frac{5\pi}{2}] [A=3] [D=0] [y=3~cos(\frac{4}{5}x)] [B=\frac{4}{5}] [\frac{2\pi}{\theta}=\frac{4\pi}{3}] [A=3] [p=-2] [B=\frac{3}{2}] [y=\underline{-3}sin(\frac{3}{2}x)-2] The voltage V produced by an ac (alternating current) generator is (V = 120sin(120πt)) A. What is the amplitude? 120 B. What is the period? [\frac{1}{60}] C. Ohm's Law states (V = IR,) where R is the resistance and I is the current. If a resistance of (R = 5) ohms is present, what is the current I? [V = IR] [120sin(120πt) = I(5)] [I = 24sin(120πt)] D. What is the amplitude and period of the current I? (A = 24) period: [\frac{1}{60}] One day all 330 million people in the United States climb up on tables and jump off (t = 0). The shock as we hit the Earth's surface starts the entire Earth vibrating in such a way that its surface first moves down from its normal position, and then moves up an equal distance above its normal position. Assuming that the amplitude is 50 feet, the displacement (distance above and below the normal position) of the surface will be a sinusoidal function of time with a period of 54 minutes.   A. Sketch the graph showing displacement versus time. B. Write an equation expressing displacement in terms of time elapsed since the people jumped. [d(t) = -50sin(\frac{π}{27}t)]