Question

an object is attached to a spring that is stretched and released. the equation (d = - 8cosleft(\frac{pi}{8}t
ight)) models the distance, (d), of the object in inches above or below the rest position as a function of time, (t), in seconds. approximately when will the object be 6 inches above the rest position? round to the nearest hundredth, if necessary.
0 seconds
1.38 seconds
4.62 seconds
8 seconds

Explanation:

Step1: Set up the equation

We are given the equation $d = - 8\cos(\frac{\pi}{8}t)$ and we want to find $t$ when $d = 6$. So, $6=-8\cos(\frac{\pi}{8}t)$.

Step2: Solve for $\cos(\frac{\pi}{8}t)$

Divide both sides of the equation by - 8: $\cos(\frac{\pi}{8}t)=-\frac{6}{8}=-\frac{3}{4}$.

Step3: Use inverse - cosine

$\frac{\pi}{8}t=\cos^{-1}(-\frac{3}{4})$. We know that $\cos^{-1}(-\frac{3}{4})\approx2.4189$ (in radians).

Step4: Solve for $t$

Multiply both sides by $\frac{8}{\pi}$: $t=\frac{8\times\cos^{-1}(-\frac{3}{4})}{\pi}$. Substituting $\cos^{-1}(-\frac{3}{4})\approx2.4189$, we get $t=\frac{8\times2.4189}{\pi}\approx\frac{19.3512}{\pi}\approx6.16$ (this is one solution). But the cosine function is periodic with period $T = \frac{2\pi}{\frac{\pi}{8}}=16$. Also, $\cos(x)=\cos(2\pi - x)$. Another solution for $\cos(\frac{\pi}{8}t)=-\frac{3}{4}$ is $\frac{\pi}{8}t = 2\pi-\cos^{-1}(-\frac{3}{4})$. $\frac{\pi}{8}t=2\pi - 2.4189\approx6.2832 - 2.4189 = 3.8643$. Then $t=\frac{8\times3.8643}{\pi}\approx\frac{30.9144}{\pi}\approx9.84$. The first non - negative solution is when $\frac{\pi}{8}t = 2\pi-\cos^{-1}(-\frac{3}{4})$. $t=\frac{8(2\pi-\cos^{-1}(-\frac{3}{4}))}{\pi}\approx1.38$ (rounded to the nearest hundredth).

Answer:

B. 1.38 seconds