Question

Please help me with this question if you can! its for my homework and I would like guidance.

Answer 1

Since d represents the ball's distance from the rest position, the position of 2 inches above the rest position is described by d=2.

To find the values of t for which the ball is located at d=2, replace d=2 and solve for t:

`\begin{gathered} d=-4\cos ((\pi)/(3)t) \\ \Rightarrow2=-4\cos ((\pi)/(3)t) \\ \Rightarrow(2)/(-4)=(-4\cos ((\pi)/(3)t))/(-4) \\ \Rightarrow-(1)/(2)=\cos ((\pi)/(3)t) \\ \Rightarrow\cos ((\pi)/(3)t)=-(1)/(2) \end{gathered}`

Take the inverse cosine of both members of the equation:

`\begin{gathered} \Rightarrow\cos ^(-1)(\cos ((\pi)/(3)t))=\cos ^(-1)(-(1)/(2)) \\ \Rightarrow(\pi)/(3)t=\pm(2)/(3)\pi+2\pi k \end{gathered}`

Where k is any integer number.

Multiply both sides by 3 and divide both sides by π:

`\begin{gathered} (3)/(\pi)*(\pi)/(3)t=(3)/(\pi)*(\pm(2)/(3)\pi+2\pi k) \\ \Rightarrow t=\pm2+6k \end{gathered}`

Therefore, all the values of t for which the ball is 2 inches above its rest position are described by the expression:

`\begin{gathered} t=6k\pm2 \\ k\in\Z \end{gathered}`