Contoh Soal 9 Turunan Fungsi Trigonometri (Bagian 2)

$\begin{array}{ll}\\ 41.&\textrm{Sebuah mesin diprogram untuk dapat}\\ &\textrm{begerak tiap waktu mengikuti posisi}\\ &x=2\cos 3t\: \: \textrm{dan}\: \: y=2\cos 2t \: \: \textrm{di mana}\\ &x,y\: \: \textrm{dalam}\: \: cm\: ,\: \textrm{dan}\: \: t\: \: \textrm{dalam detik}\\ &\textrm{Jika kecepatakan dirumuskan dengan}\\ &v=\sqrt{\left ( v_{x} \right )^{2}+\left ( v_{y} \right )^{2}},\: \textrm{maka nilai}\: \: v\\ &\textrm{saat}\: \: t=30\: detik\: \textrm{adalah}\: ...\: cm/detik\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&4\sqrt{3}\\ \textrm{b}.&2\sqrt{11}\\ \textrm{c}.&2\sqrt{10}\\ \textrm{d}.&6\\ \textrm{e}.&4\sqrt{2} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}&\textrm{Diketahui Kecepatan gerak mesin}\\ &\begin{cases} x=2\cos 3x &\Rightarrow \displaystyle \frac{dx}{dt}=-6\sin 3t \\ y=2\cos 2x &\Rightarrow \displaystyle \frac{dy}{dt}=-4\sin 2t \end{cases}\\ &\color{black}\textrm{Maka kecepatan mesin saat}\: \: t=30\\ &\: \: \color{red}v=\sqrt{\left ( v_{x} \right )^{2}+\left ( v_{y} \right )^{2}}\\ &\: \: \color{black}v=\sqrt{\left ( -6\sin 3t \right )^{2}+\left ( -4\sin 2t \right )^{2}}\\ &\quad \color{black}=\sqrt{\left ( -6\sin 3(30) \right )^{2}+\left ( -4\sin 2(30) \right )^{2}}\\ &\quad \color{black}=\sqrt{\left ( -6(1) \right )^{2}+\left ( -4\left ( \displaystyle \frac{1}{2}\sqrt{3} \right ) \right )^{2}}\\ &\quad \color{black}=\sqrt{36+12}=\sqrt{48}=\sqrt{16.3}\\ &\quad \color{red}=4\sqrt{3} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 42.&\textrm{Sebuah benda duhubungkan dengan}\\ &\textrm{pegas dan bergerak sepanjang sumbu}\\ &\textrm{X dengan formula persamaan}:\\ &\qquad x=\sin 2t+\sqrt{3}\cos 2t\\ &\textrm{Jarak terjauh dari titik}\: \: O\: \: \textrm{yang dapat}\\ &\textrm{dicapai oleh benda tersebut adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&1\\ \color{red}\textrm{b}.&2\\ \textrm{c}.&3\\ \textrm{d}.&4\\ \textrm{e}.&5 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}&\textrm{Diketahui gerak benda yang bergerak}\\ &\textrm{mengikuti formula}:\\ &\qquad \color{red}x=\sin 2t+\sqrt{3}\cos 2t\\ &\color{black}\textrm{Jarak terjauh dicapai saat}\: \: x'=\displaystyle \frac{dx}{dt}=0\\ &\: \: \color{red}x'=2\cos 2t-2\sqrt{3}\sin 2t=0\\ &\: \: \color{black}\Leftrightarrow \: 2\cos 2t=2\sqrt{3}\sin 2t\\ &\: \: \color{black}\Leftrightarrow \: \displaystyle \frac{\sin 2t}{\cos 2t}=\displaystyle \frac{1}{3}\sqrt{3}\\ &\: \: \color{black}\Leftrightarrow \: \tan 2t=\tan 30^{\circ}\\ &\: \: \color{black}\Leftrightarrow \: 2t=30^{\circ}+k.180^{\circ}\\ &\: \: \color{black}\Leftrightarrow \: t=15^{\circ}+k.90^{\circ}\begin{cases} k=0, &t=15^{\circ} \\ k=1, &t=105^{\circ} \\ k=2, &t=195^{\circ} \\ k=3, &t=285^{\circ}\\ k=4, &t=375^{\circ}\\ &\textrm{dst} \end{cases}\\ &\textrm{Ambil}\: \: t=15^{\circ},\: \textrm{maka nilai}\\ &x-\textrm{nya adalah}:\\ &\quad \color{red}x=\sin 2t+\sqrt{3}\cos 2t\\ &\quad \Leftrightarrow \: \color{black}x=\sin 2(15^{\circ})+\sqrt{3}\cos 2(15^{\circ})\\ &\quad \Leftrightarrow \: \color{black}x=\displaystyle \frac{1}{2}+\sqrt{3}\left ( \displaystyle \frac{1}{2}\sqrt{3} \right )\\ &\quad \Leftrightarrow \: \color{red}x=\displaystyle \frac{1}{2}+\frac{3}{2}=2 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 43.&\textrm{Pada kurva}\: \: y=\sin x\: \: \: \textrm{dibuat}\\ &\textrm{garis singgung melalui titik}\: \: \left ( \displaystyle \frac{2\pi }{3},k \right )\\ &\textrm{garis singgung tersebut memotong}\\ &\textrm{sumbu-X di A dan sumbu-Y di B}.\\ &\textrm{Luas}\: \: \triangle AOB\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{\left ( 3\pi +2\sqrt{3} \right )^{2}}{36}\\ \color{red}\textrm{b}.&\displaystyle \frac{\left ( 3\pi +3\sqrt{3} \right )^{2}}{36}\\ \textrm{c}.&\displaystyle \frac{\left ( 3\pi +2\sqrt{3} \right )^{2}}{16}\\ \textrm{d}.&\displaystyle \frac{\left ( 3\pi +2\sqrt{3} \right )^{2}}{18}\\ \textrm{e}.&\displaystyle \frac{\left ( 3\pi +3\sqrt{3} \right )^{2}}{18} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\textrm{Perhatikan ilustrasi berikut} \end{array}$

$.\: \qquad\color{blue}\begin{aligned}&\textrm{Misalkan koordinat titik}\: \: P\left ( \displaystyle \frac{2\pi }{3},k \right )\\ & \color{black}\textrm{maka},\\ &x_{p}=\displaystyle \frac{2\pi }{3},\: y_{p}=k=\sin \displaystyle \frac{2\pi }{3}=\displaystyle \frac{1}{2}\sqrt{3}\\ &\textrm{Persamaan garis singgung di titik P}:\\ &\color{red}y=m_{x_{p}}\left ( x-x_{p} \right )+y_{p}\\ &\color{black}\begin{cases} \left ( \displaystyle \frac{2\pi }{3},k \right ) &=\color{red}\left ( \displaystyle \frac{2\pi }{3},\frac{1}{2}\sqrt{3} \right ) \\ m_{x_{p}}=\displaystyle \frac{dy}{dx} &=y'=\cos x\\ \qquad m_{x_{p}}&=\cos \left ( \displaystyle \frac{2\pi }{3} \right )=\color{red}-\frac{1}{2} \end{cases}\\ &\textrm{Sehingga persamaan garis singgungnya}\\ &\color{red}y=\left ( -\displaystyle \frac{1}{2} \right )\left ( x-\displaystyle \frac{2\pi }{3} \right )+\displaystyle \frac{1}{2}\sqrt{3}\\ &\Leftrightarrow \color{red}2y=-x+\displaystyle \frac{2\pi }{3}+\displaystyle \frac{1}{2}\sqrt{3}\\ &\bullet\quad \textrm{memotong sumbu-X, maka}\: \: y_{A}=0\\ &\qquad \color{red}2y_{A}=-x_{A}+\displaystyle \frac{2\pi }{3}+\sqrt{3}\\ &\qquad \color{black}0=-x_{A}+\displaystyle \frac{2\pi }{3}+\sqrt{3}\\ &\qquad \color{black}x_{A}=\displaystyle \frac{2\pi }{3}+\sqrt{3}\\ &\bullet\quad \textrm{memotong sumbu-Y, maka}\: \: x_{B}=0\\ &\qquad \color{red}2y_{B}=-x_{B}+\displaystyle \frac{2\pi }{3}+\sqrt{3}\\ &\qquad \color{black}2y_{B}=0+\displaystyle \frac{2\pi }{3}+\sqrt{3}\\ &\qquad \color{black}y_{B}=\displaystyle \frac{\pi }{3}+\displaystyle \frac{1}{2}\sqrt{3}\\ &\bullet\quad \color{blue}\textrm{Luas}\: \: \triangle AOB=\left [ AOB \right ]=\color{red}\displaystyle \frac{x_{A}.y_{B}}{2}\\ &\qquad \color{black}=\displaystyle \frac{\left ( \displaystyle \frac{2\pi }{3}+\sqrt{3} \right ).\left ( \displaystyle \frac{\pi }{3}+\displaystyle \frac{1}{2}\sqrt{3} \right )}{2}\\ &\qquad \color{black}=\displaystyle \frac{1}{6}\left ( 2\pi +3\sqrt{3} \right ).\displaystyle \frac{1}{6}\left ( 2\pi +3\sqrt{3} \right )\\ &\qquad \color{red}=\displaystyle \frac{1}{36}\left ( 2\pi +3\sqrt{3} \right )^{2} \end{aligned}$

$\begin{array}{ll}\\ 44.&\textrm{Sebuah wadah penampung air hujan}\\ &\textrm{memiliki ukuran sisi samping 3 m dan}\\ &\textrm{sisi horisontal juga 3 m. Sisi samping}\\ &\textrm{membentuk sudut}\: \: \theta \left ( 0\leq \theta \leq \displaystyle \frac{\pi }{2} \right )\\ &\textrm{dengan garis vertikal (lihat gambar)}\\ &\textrm{Nilai}\: \: \theta \: \: \textrm{supaya wadah dapat menampung}\\ &\textrm{air hujan maksimum adalah}\: ....\\ \end{array}$

$\begin{array}{l} .\: \qquad&\\ &\begin{array}{llll} \color{red}\textrm{a}.&\displaystyle \frac{\pi }{3}\\ \textrm{b}.&\displaystyle \frac{\pi }{4}\\ \textrm{c}.&\displaystyle \frac{\pi }{5}\\ \textrm{d}.&\displaystyle \frac{\pi }{6}\\ \textrm{e}.&\displaystyle \frac{\pi }{8} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}&\textrm{Supaya memuat dapat maksimum}\\ &\textrm{maka luas penampang haruslah}\\ &\color{black}\textrm{MAKSIMUM, yaitu} \end{aligned} \end{array}$

gambar 1

gambar 2

$.\: \qquad\color{blue}\begin{aligned}&\textrm{Luas penampang}=\textrm{Luas Trapesium}\\ &\color{red}\textrm{dengan}\color{black}\begin{cases} t &=3\sin \theta \\ n &=3\cos \theta \end{cases}\\ &\textrm{Luas Penampang}=\color{red}\displaystyle \frac{1}{2}\left ( \sum \textrm{sisi sejajar} \right )\times t\\ &\Leftrightarrow \: L=\displaystyle \frac{1}{2}\left ( 6+2n \right )\times t\\ &\Leftrightarrow \: L=\left ( 3+n \right )\times t\\ &\Leftrightarrow \: L=\left ( 3+3\cos \theta \right )\times 3\sin \theta \\ &\Leftrightarrow \: L=9\sin \theta +9\sin \theta \cos \theta \\ &\Leftrightarrow \: L=9\sin \theta +\displaystyle \frac{9}{2}\sin 2\theta \\ &\color{black}\textrm{Suapa luas penampang}\: \: \color{red}\textrm{MAKSIMUM}\\ &\color{black}\textrm{maka}\: \: L'=\displaystyle \frac{dL}{d\theta }=0\\ &\Leftrightarrow \: L'=9\cos \theta +9\cos 2\theta =0\\ &\Leftrightarrow \: 9\cos \theta +9\cos 2\theta =0\\ &\Leftrightarrow \: 9\cos \theta +9\left ( 2\cos ^{2}\theta -1 \right ) =0\\ &\Leftrightarrow \: 2\cos^{2} \theta +\cos \theta -1=0\\ &\Leftrightarrow \: \left (\cos \theta +1 \right )\left ( 2\cos \theta -1 \right )=0\\ &\Leftrightarrow \: \cos \theta =-1\: \: \color{black}\textrm{atau}\: \: \color{blue}2\cos \theta =1\\ &\Leftrightarrow \: \cos \theta =-1\: \: \color{black}\textrm{atau}\: \: \color{blue}\cos \theta =\displaystyle \frac{1}{2}\\ &\Leftrightarrow \: \cos \theta =\cos \pi \: \: \color{black}\textrm{atau}\: \: \color{blue}\cos \theta =\cos \displaystyle \frac{\pi }{3}\\ &\Leftrightarrow \: \theta =\pi \: \: \color{black}\textrm{atau}\: \: \color{red}\theta =\displaystyle \frac{\pi }{3} \end{aligned}$

$\begin{array}{ll}\\ 45.&\textrm{Seseorang melempar bola dari atap}\\ &\textrm{sebuah rumah. Ketinggian bola saat}\\ &t\: (detik)\: \: \textrm{dinyatakan dengan persamaan}\\ &h(t)=5+\cos ^{2}\pi t.\: \: \textrm{Kecepatan bola}\\ &\textrm{ditentukan dengan formula}\: \: v=\displaystyle \frac{dh}{dt}\\ &\textrm{Besar kecepatan bola saat}\: \: t=0,25\\ &\textrm{detik adalah}\: ....\\ &\begin{array}{llll} \textrm{a}.&0\\ \color{red}\textrm{b}.&\pi \\ \textrm{c}.&2\pi \\ \textrm{d}.&3\pi \\ \textrm{e}.&4\pi \end{array}\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}&\textrm{Diketahui}\: \: h(t)=5+\cos ^{2}\pi t.\: \: \color{red}\textrm{maka}\\ &v=\displaystyle \frac{dh}{dt}=2\cos \pi t\left ( -\sin \pi t \right ).(\pi )\\ &\Leftrightarrow v=-\pi \sin 2\pi t\\ &\color{black}\textrm{Saat}\: \: t=0,25=\displaystyle \frac{1}{4},\: \: \color{red}\textrm{maka}\\ &\textrm{besar kecepatannya adalah}:\\ &\Leftrightarrow \: \color{black}v=-\pi \sin 2\pi \left ( \displaystyle \frac{1}{4} \right )\\ &\Leftrightarrow \: \quad =-\pi \sin \displaystyle \frac{\pi }{2}\\ &\Leftrightarrow \: \quad =\color{red}-\pi\\ &\textrm{Tanda negatif menunjukkan}\\ &\color{black}\textrm{arah kecepatan ke bawah}\\ &\color{black}\textrm{Karena kecepatan merupakan salah}\\ &\textrm{satu besaran}\: \color{red}\textrm{VEKTOR} \end{aligned} \end{array}$

DAFTAR PUSTAKA

1. Kanginan, M., Nurdiansyah, H., & Akhmad G. 2016. Matematika untuk Siswa SMA/MA Kelas XII Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: YRAMA WIDYA
2. Noormandiri. 2017. Matematika Jilid 3 untuk SMA/MA Kelas XII Kelompok Peminatan MAtematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA
3. Sembiring, S., Zulkifli, M., Marsito, & Rusdi, I. 2016. Matematika untuk Siswa SMA/MA Kelas XII Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: SEWU