이계도함수와 오목·볼록

The question asks for $s''(1)$. We start our differentiation using the quotient rule:

$$\begin{aligned} s'(t)&=\frac{e^t(t^2+1)-e^t(2t)}{(t^2+1)^2}\\ &=\frac{e^t(t^2-2t+1)}{(t^2+1)^2} \\ \text{Using the quotient rule again,} \\ s''(t)&=\frac{(t^2+1)^2 \frac{\mathrm{d}}{\mathrm{d}t}\{e^t(t^2-2t+1)\}-e^t(t^2-2t+1) \frac{\mathrm{d}}{\mathrm{d}t}\{(t^2+1)^2\}}{(t^2+1)^4}\\ &=\frac{(t^2+1)^2\cdot\left[ e^t(2t-2)+e^t(t^2-2t+1)\right]- e^t(t^2-2t+1) \cdot 2(t^2+1)(2t)}{(t^2+1)^4}\\ &=\frac{e^t(t^2+1)^2(t^2-1)-4te^t(t-1)^2(t^2+1)}{(t^2+1)^4}\\ s''(1)&=0 \end{aligned}$$

The velocity of the particle is given by $h'(t)=\sin t$. Note $0<1<\pi$, so $h'(1)>0$–the particle is rising (moving in the positive direction, in this case “up"). The acceleration of the particle is $h''(t)=\cos t$. Since $0<1<\frac{\pi}{2}$, $h''(t)>0$, so $h'(t)$ is increasing: the particle is moving up, and it's doing so at an increasing rate. So, the particle is speeding up.

For this problem, remember that velocity has a sign indicating direction, while speed does not.

The velocity of the particle is given by $h'(t)=3t^2-2t-5$. At $t=1$, the velocity of the particle is $-4$, so the particle is moving downwards with a speed of 4 units per second. The acceleration of the particle is $h''(t)=6t-2$, so when $t=1$, the acceleration is (positive) $4$ units per second per second. That means the velocity (currently $-4$ units per second) is becoming a bigger number–since the velocity is negative, a bigger number is closer to zero, so the speed of the particle is getting smaller. (For instance, a velocity of $-3$ represents a slower motion than a velocity of $-4$.) So, the particle is slowing down at $t=1$.