1 mark for review a weight suspended by a spring vibrates vertically according to the function d given by…

1 mark for review a weight suspended by a spring vibrates vertically according to the function d given by d(t)=2sin(4π(t + 1/8)), where d(t), in centimeters, is the directed distance of the weight from its central position t seconds after the start of the motion. assume the positive direction is upward. what is the instantaneous rate of change of the weights position, in centimeters per second, at the moment the weight is first 1 centimeter above its central position? a -157.914 b -21.765 c -12 d 0

1 mark for review a weight suspended by a spring vibrates vertically according to the function d given by d(t)=2sin(4π(t + 1/8)), where d(t), in centimeters, is the directed distance of the weight from its central position t seconds after the start of the motion. assume the positive direction is upward. what is the instantaneous rate of change of the weights position, in centimeters per second, at the moment the weight is first 1 centimeter above its central position? a -157.914 b -21.765 c -12 d 0

Answer

Explanation:

Step1: Find the time when $D(t)=1$

Set $D(t)=2\sin(4\pi(t + \frac{1}{8})) = 1$. Then $\sin(4\pi(t+\frac{1}{8}))=\frac{1}{2}$. We know that $\sin\theta=\frac{1}{2}$ when $\theta=\frac{\pi}{6}+2k\pi$ or $\theta=\frac{5\pi}{6}+2k\pi,k\in\mathbb{Z}$. For the first - positive $t$, we consider $4\pi(t+\frac{1}{8})=\frac{5\pi}{6}$. Solving for $t$: [ \begin{align*} 4\pi(t+\frac{1}{8})&=\frac{5\pi}{6}\ t+\frac{1}{8}&=\frac{5}{24}\ t&=\frac{5}{24}-\frac{1}{8}=\frac{5 - 3}{24}=\frac{1}{12} \end{align*} ]

Step2: Differentiate $D(t)$

Using the chain - rule, if $D(t)=2\sin(4\pi(t+\frac{1}{8})) = 2\sin(4\pi t+\frac{\pi}{2})$, and the derivative of $y = \sin(u)$ with respect to $x$ is $y^\prime=\cos(u)\cdot u^\prime$. Here $u = 4\pi t+\frac{\pi}{2}$ and $u^\prime=4\pi$. So $D^\prime(t)=2\cos(4\pi t+\frac{\pi}{2})\cdot4\pi=8\pi\cos(4\pi t+\frac{\pi}{2})$.

Step3: Evaluate $D^\prime(t)$ at $t = \frac{1}{12}$

Substitute $t=\frac{1}{12}$ into $D^\prime(t)$: [ \begin{align*} D^\prime(\frac{1}{12})&=8\pi\cos(4\pi\cdot\frac{1}{12}+\frac{\pi}{2})\ &=8\pi\cos(\frac{\pi}{3}+\frac{\pi}{2})\ &=8\pi\cos(\frac{2\pi + 3\pi}{6})\ &=8\pi\cos(\frac{5\pi}{6})\ &=8\pi\cdot(-\frac{\sqrt{3}}{2})\ &=- 4\sqrt{3}\pi\approx - 21.765 \end{align*} ]

Answer:

B. $-21.765$