Lanjutan Materi Turunan Fungsi Trigonometri (Matematika Peminatan Kelas XII)

$\begin{array}{ll}\\ 3.&\textrm{Diketahui suatu gelombang bergerak teratur}\\ &\textrm{sebagaimana gambar berikut} \end{array}$

$.\: \: \: \quad\begin{array}{l}\\ &\textrm{Gelombang tersebut pada waktu}\: \: t\: \: \textrm{detik mengikuti}\\ &\textrm{rumus}\: \: \: y=f(t)=2\sin \displaystyle \frac{1}{2}\pi t.\: \textrm{Dan diketahui pula cepat}\\ &\textrm{rambat gelomnya dapat dinyatakan dalam}\\ &v_{t}=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(t+h)-f(t)}{h}\: \: \textrm{dengan}\: \: h\neq 0,\: \textrm{tentukanlah}:\\ &\textrm{a}.\quad \textrm{posisi gelombang pada ketika}\: \: t=1,5\: \: \textrm{detik}\\ &\textrm{b}.\quad \textrm{rumus cepat rambat gelombang pada saat}\: \: t\\ &\textrm{c}.\quad \textrm{cepat rambat gelombang saat}\: \: t=2\displaystyle \frac{1}{2}\: \: \textrm{detik} \end{array}$

$.\: \: \: \: \color{purple}\quad\begin{aligned}&\textrm{Jawab}:\\ &\begin{array}{ll}\\ \textrm{a}.\quad&\textrm{Posisi gelombang saat}\: \: t=1,5=\displaystyle \frac{3}{2}\: \: \textrm{detik}\\ &f(1,5)=2\sin \displaystyle \frac{1}{2}\pi \left ( \frac{3}{2} \right )\\ &f(1,5)=2\sin \displaystyle \frac{3}{4}\pi =2\sin 135^{\circ}=2\left ( \displaystyle \frac{1}{2}\sqrt{2} \right )=\sqrt{2}\\ \textrm{b}.\quad&\textrm{Cepat rambat gelombang saat}\: \: t\: \: \textrm{detik}\\ &v_{t}=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(t+h)-f(t)}{h}\\ &\quad =\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{2\sin \displaystyle \frac{1}{2}\pi (t+h)-2\sin \displaystyle \frac{1}{2}\pi t}{h}\\ &\quad =\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{2.2\cos\left ( \displaystyle \frac{1}{2}t+\frac{1}{4}h \right ).\sin \displaystyle \frac{1}{4}\pi h}{h}\\ &\quad =4\times \displaystyle \frac{1}{4}\pi \times \cos \displaystyle \frac{1}{2}\pi t\\ &\quad =\pi \cos \displaystyle \frac{1}{2}\pi t\\ \textrm{c}.\quad&\textrm{Cepat rambat saat}\: \: t=2\displaystyle \frac{1}{2}=\frac{5}{2}\: \: \textrm{detik}\\ &v_{t}=\pi \cos \displaystyle \frac{1}{2}\pi t\\ &\quad =\pi \cos \displaystyle \frac{1}{2}\pi \left ( \displaystyle \frac{5}{2} \right )\\ &\quad =\pi \cos \displaystyle \frac{5}{4}\pi \\ &\quad = \pi \left ( -\displaystyle \frac{1}{2}\sqrt{2} \right )\\ &\quad =-\displaystyle \frac{1}{2}\pi \sqrt{2} \end{array} \end{aligned}$