삼각함수의 미분

(a) The object's instantaneous velocity at $t=2$ is given by $h'(2)$. Since $h(t) = 3 + \frac{2\cos(t)}{1.2^t}$, by the quotient rule we can determine that $$\begin{aligned}h'(t) &= \frac{1.2^t \cdot (-2\sin(t)) - 2\cos(t) \cdot 1.2^t \ln(1.2)}{(1.2^t)^2} \\ &= \frac{1.2^t (-2\sin(t) - 2\cos(t) \ln(1.2))}{(1.2^t)^2} \\ &= \frac{-2\sin(t) - 2\cos(t) \ln(1.2)}{1.2^t}\end{aligned}\,.$$ Thus, $h'(2) = \frac{-2\sin(2) - 2\cos(2) \ln(1.2)}{1.2^2} \approx -1.1575$ feet per second is the instanateous (vertical) velocity of the object at $t=2$. (b) To find the object's acceleration at $t=2$, we compute $h''(2)$. First, by the quotient rule, $$\begin{aligned}h''(t) &= \frac{1.2^t(-2\cos(t) + 2\sin(t) \ln(1.2)) - (-2\sin(t) - 2\cos(t) \ln(1.2))1.2^t \ln(1.2)}{(1.2^t)^2} \\ &= \frac{-2\cos(t) + 2\sin(t) \ln(1.2)) + 2\sin(t)\ln(1.2) + 2\cos(t) \ln^2(1.2)}{1.2^t} \\ &= \frac{\cos(t)(-2 + 2\ln^2(1.2)) + 4\ln(1.2)\sin(t))}{1.2^t}\end{aligned}\,.$$ Therefore, the acceleration at $t = 2$ is $h''(2) = \frac{\cos(2)(-2 + 2\ln^2(1.2)) + 4\ln(1.2)\sin(2))}{1.2^2} \approx 1.0193$ feet per second per second. (c) At $t=2$, the object is falling and slowing down. We know the object is falling because its velocity is negative, and slowing down because its negative velocity is increasing (due to $h''(2)$ being positive) and thus getting less negative.