# How do you find a possible formula of the following form?

## Jul 22, 2018

$y = \sin t + 2$

#### Explanation:

Both $y = A \sin t + k$ and $y = A \cos t + k$ have the same midline as $y = 0$. The period is $2 \pi$, for both. Amplitude is $\left\mid A \right\mid$ , for both.

Changing $y$ to $y - 2$, the midline becomes $y - 2 = 0$. So, $k = 2$, for both.

The minimum amplitude $= 0$, in the midline $y = 2$. At $t = 0$, $y = A \sin t + 2 = 2$. OK.

It is not so for .$y = A \cos t + 2 = 3$, with amplitude, the maximum $A$. So change $t$ to $\left(t \pm \frac{\pi}{2}\right)$. Then

$y = A \cos \left(t \pm \frac{\pi}{2}\right) + 2 = 2$, at $t = 0 \Rightarrow$ the amplitude is $0$.

See graphs, with $A = 1$.

Graph of $y = \sin t + 2$:

graph{(y-sin x -2)(y-2) = 0}

Graph of $y = \cos \left(t - \frac{\pi}{2}\right) + 2$:

graph{(y - cos ( x - pi/2 )-2)(y-2) = 0}

Graph of $y = \cos \left(t + \frac{\pi}{2}\right) + 2$:

graph{(y - cos ( x + pi/2 )-2)(y-2) = 0}

Observe that the first two are the same and the third graph is different.