A weight suspended by a spring vibrates vertically according to the function D D given by D(t)=2sin(4π(t+18)) D ( t ) = 2 sin ( 4 π ( t + 1 8 ) ) , where D(t) D ( t ) , in centimeters, is the directed distance of the weight from its central position t t seconds after the start of the motion. Assume the positive direction is upward. What is the instantaneous rate of change of the weight’s position, in centimeters per second, at the moment the weight is first 1 centimeter above its central position?

Answer:21.75 cm/sStep-by-step explanation:1 cm above central position means D = 1, so we plug in 1 into D(t) and find the time and which this occurs.[tex]D(t)=2Sin (4\pi(t+18))\\1=2Sin (4\pi(t+18))\\\frac{1}{2}=Sin(4\pi t +72\pi)\\4\pi t + 72 \pi = \frac{\pi}{6}\\4\pi t = \frac{\pi}{6}-72\pi\\t=\frac{\frac{\pi}{6}-72\pi}{4\pi}[/tex]This is the time at which this occurs.To find instantaneous rate of change, we differentiate D(t) and plug in this t we found. Remembering that d/dt (Sin t) = Cos t[tex]D(t)=2Sin (4\pi(t+18))\\D(t)=2Sin(4\pi t + 72\pi)\\D'(t)=2Cos(4\pi t + 72\pi)(4\pi)[/tex]Now putting t:[tex]D'(t)=2Cos(4\pi t + 72\pi)(4\pi)\\D'(\frac{\frac{\pi}{6}-72\pi}{4\pi})=2Cos(4\pi (\frac{\frac{\pi}{6}-72\pi}{4\pi}) + 72\pi)(4\pi)\\=8\pi Cos(\frac{\pi}{6})\\=21.75[/tex]Thus, the instantaneous rate would be around 21.75 cm/s