Calculate the distance $h(t)=1+e^{-t}\cos(t)$ that drops A weight is hanging from a spring, and oscillates with heights modeled by the function $h(t)=1+e^{-t}\cos(t)$ at time $t\geq 0.$ Does it travel an infinite distance (in other words, does it converge or diverge) as $t\to\infty$ and also, what is the total distance that it drops ?
Note: the distance dropped indicates only the downward motion of the function, not when the function increases.
I have tried to plot this and it seems from the graph that it only drops to $\frac{3\pi}{4}$ and increases from that point on. I got a final answer of $\frac{1}{1-e^{-\pi}}.$ For the other question of whether the weight travels infinitely, the answer is a simple no.
 A: To solve this problem, we find the time intervals where the height increases and when it decreases. The height increases on the intervals (where $k\in \mathbb{N}\cup \{0\}$)
$$[3\pi/4+2k\pi,7\pi/4+2k\pi],$$
and decreases on
$$[0,3\pi/4],\quad [7\pi/4+2k\pi,11\pi/4+2k\pi].$$
This can be seen by noting that $h'(t)=-e^{-t}(\sin(t)+\cos(t))=-\sqrt{2}e^{-t}\sin(t+\pi/4)$. The distance travelled going downwards is therefore given by the sum
$$\scriptsize (2-(1+e^{-3\pi/4}\cos(3\pi/4)))+\sum_{k=0}^{\infty} (1+e^{-(7/4+2k)\pi}\cos((7/4+2k)\pi))-(1+e^{-(11/4+2k)\pi}\cos((11/4+2k)\pi)),$$
which simplifies to
$$1+\frac{e^{-3\pi/4}}{\sqrt{2}}+\frac{1}{\sqrt{2}}\sum_{k=0}^{\infty} (e^{-(7/4+2k)\pi}+e^{-(11/4+2k)\pi})$$
which by the geometric series (which converges since $|e^{-2\pi}|<1$) is given by
$$1+\frac{e^{-3\pi/4}}{\sqrt{2}}+\frac{e^{-3\pi/4}}{\sqrt{2}(e^{\pi}-1)}=1+\frac{e^{\pi/4}}{\sqrt{2}(e^{\pi}-1)}\approx 1.0700467.$$
You can similarly find the distance going upwards by computing the sum
$$\sum_{k=0}^{\infty} (1+e^{-(7/4+2k)\pi}\cos((7/4+2k)\pi))-(1+e^{-(3/4+2k)\pi}\cos((3/4+2k)\pi)),$$
which by a similar method is given by
$$\frac{e^{\pi/4}}{\sqrt{2}(e^{\pi}-1)}\approx 0.0700467.$$
Alternatively, you can use continuity of $h$ and that the difference between the starting height and the height as $t\to \infty$ is $1$ to conclude the same result.
Therefore the total distance travelled by the weight is finite.
A: We have $\cos(t)=\frac{1}{2}\cdot(e^{it}+e^{-it}),$ so rewrite to
$$1+e^{-t}\cdot\frac{1}{2}\cdot(e^{it}+e^{-it})=1+\frac{1}{2}\cdot(e^{t(i-1)}+e^{t(-i-1)})$$
Now integrate over t to get $t+1/2(\frac{e^{t(i-1)}}{i-1}-\frac{e^{t(-i-1)}}{i+1})$ from $0$ to infinity to get infinity and thus it travels an infinite distance.
