Contoh Soal 9 Turunan Fungsi Trigonometri (Bagian 2)

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

$\begin{array}{l}\\ 42.& \text{Sebuah benda duhubungkan dengan}\\ & \text{pegas dan bergerak sepanjang sumbu}\\ & \text{X dengan formula persamaan}:\\ & x=\sin 2t+\sqrt{3}\cos 2t\\ & \text{Jarak terjauh dari titik}O\text{yang dapat}\\ & \text{dicapai oleh benda tersebut adalah}....\\ & \begin{array}{l}\\ \text{a}.& 1\\ \text{b}.& 2\\ \text{c}.& 3\\ \text{d}.& 4\\ \text{e}.& 5\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \text{Diketahui gerak benda yang bergerak}\\ & \text{mengikuti formula}:\\ & x=\sin 2t+\sqrt{3}\cos 2t\\ & \text{Jarak terjauh dicapai saat}{x}^{\prime }={\displaystyle \frac{dx}{dt}=0}\\ & x^{\prime }=2\cos 2t-2\sqrt{3}\sin 2t=0\\ & \iff 2\cos 2t=2\sqrt{3}\sin 2t\\ & \iff {\displaystyle \frac{\sin 2t}{\cos 2t}={\displaystyle \frac{1}{3}\sqrt{3}}}\\ & \iff \tan 2t=\tan {30}^{\circ }\\ & \iff 2t={30}^{\circ }+k{.180}^{\circ }\\ & \iff t={15}^{\circ }+k{.90}^{\circ }\{\begin{array}{ll}k=0,& t={15}^{\circ }\\ k=1,& t={105}^{\circ }\\ k=2,& t={195}^{\circ }\\ k=3,& t={285}^{\circ }\\ k=4,& t={375}^{\circ }\\ & \text{dst}\end{array}\\ & \text{Ambil}t={15}^{\circ },\text{maka nilai}\\ & x-\text{nya adalah}:\\ & x=\sin 2t+\sqrt{3}\cos 2t\\ & \iff x=\sin 2({15}^{\circ })+\sqrt{3}\cos 2({15}^{\circ })\\ & \iff x={\displaystyle \frac{1}{2}+\sqrt{3}\left({\displaystyle \frac{1}{2}\sqrt{3}}\right)}\\ & \iff x={\displaystyle \frac{1}{2}+\frac{3}{2}=2}\end{array}\end{array}$

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

$.\begin{array}{rl}& \text{Misalkan koordinat titik}P\left({\displaystyle \frac{2\pi }{3},k}\right)\\ & \text{maka},\\ & x_p={\displaystyle \frac{2\pi }{3},y_p=k=\sin {\displaystyle \frac{2\pi }{3}={\displaystyle \frac{1}{2}\sqrt{3}}}}\\ & \text{Persamaan garis singgung di titik P}:\\ & y=m_{x_p}(x-x_p)+y_p\\ & \{\begin{array}{ll}\left({\displaystyle \frac{2\pi }{3},k}\right)& =\left({\displaystyle \frac{2\pi }{3},\frac{1}{2}\sqrt{3}}\right)\\ m_{x_p}={\displaystyle \frac{dy}{dx}}& =y^{\prime }=\cos x\\ m_{x_p}& =\cos \left({\displaystyle \frac{2\pi }{3}}\right)=-\frac{1}{2}\end{array}\\ & \text{Sehingga persamaan garis singgungnya}\\ & y=(-{\displaystyle \frac{1}{2}})(x-{\displaystyle \frac{2\pi }{3}})+{\displaystyle \frac{1}{2}\sqrt{3}}\\ & \iff 2y=-x+{\displaystyle \frac{2\pi }{3}+{\displaystyle \frac{1}{2}\sqrt{3}}}\\ & \bullet \text{memotong sumbu-X, maka}{y}_A=0\\ & 2y_A=-x_A+{\displaystyle \frac{2\pi }{3}+\sqrt{3}}\\ & 0=-x_A+{\displaystyle \frac{2\pi }{3}+\sqrt{3}}\\ & x_A={\displaystyle \frac{2\pi }{3}+\sqrt{3}}\\ & \bullet \text{memotong sumbu-Y, maka}{x}_B=0\\ & 2y_B=-x_B+{\displaystyle \frac{2\pi }{3}+\sqrt{3}}\\ & 2y_B=0+{\displaystyle \frac{2\pi }{3}+\sqrt{3}}\\ & y_B={\displaystyle \frac{\pi }{3}+{\displaystyle \frac{1}{2}\sqrt{3}}}\\ & \bullet \text{Luas}\mathrm{△}AOB=\left[AOB\right]={\displaystyle \frac{x_A.y_B}{2}}\\ & ={\displaystyle \frac{\left({\displaystyle \frac{2\pi }{3}+\sqrt{3}}\right).\left({\displaystyle \frac{\pi }{3}+{\displaystyle \frac{1}{2}\sqrt{3}}}\right)}{2}}\\ & ={\displaystyle \frac{1}{6}(2\pi +3\sqrt{3}).{\displaystyle \frac{1}{6}(2\pi +3\sqrt{3})}}\\ & ={\displaystyle \frac{1}{36}{(2\pi +3\sqrt{3})}^2}\end{array}$

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

$\begin{array}{ll}.& \\ & \begin{array}{ll}\text{a}.& {\displaystyle \frac{\pi }{3}}\\ \text{b}.& {\displaystyle \frac{\pi }{4}}\\ \text{c}.& {\displaystyle \frac{\pi }{5}}\\ \text{d}.& {\displaystyle \frac{\pi }{6}}\\ \text{e}.& {\displaystyle \frac{\pi }{8}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& \text{Supaya memuat dapat maksimum}\\ & \text{maka luas penampang haruslah}\\ & \text{MAKSIMUM, yaitu}\end{array}\end{array}$

gambar 1

gambar 2

$.\begin{array}{rl}& \text{Luas penampang}=\text{Luas Trapesium}\\ & \text{dengan}\{\begin{array}{ll}t& =3\sin \theta \\ n& =3\cos \theta \end{array}\\ & \text{Luas Penampang}={\displaystyle \frac{1}{2}(\sum \text{sisi sejajar})\times t}\\ & \iff L={\displaystyle \frac{1}{2}(6+2n)\times t}\\ & \iff L=(3+n)\times t\\ & \iff L=(3+3\cos \theta )\times 3\sin \theta \\ & \iff L=9\sin \theta +9\sin \theta \cos \theta \\ & \iff L=9\sin \theta +{\displaystyle \frac{9}{2}\sin 2\theta }\\ & \text{Suapa luas penampang}\text{MAKSIMUM}\\ & \text{maka}{L}^{\prime }={\displaystyle \frac{dL}{d\theta }=0}\\ & \iff L^{\prime }=9\cos \theta +9\cos 2\theta =0\\ & \iff 9\cos \theta +9\cos 2\theta =0\\ & \iff 9\cos \theta +9(2{\cos }^2\theta -1)=0\\ & \iff 2{\cos }^2\theta +\cos \theta -1=0\\ & \iff (\cos \theta +1)(2\cos \theta -1)=0\\ & \iff \cos \theta =-1\text{atau}2\cos \theta =1\\ & \iff \cos \theta =-1\text{atau}\cos \theta ={\displaystyle \frac{1}{2}}\\ & \iff \cos \theta =\cos \pi \text{atau}\cos \theta =\cos {\displaystyle \frac{\pi }{3}}\\ & \iff \theta =\pi \text{atau}\theta ={\displaystyle \frac{\pi }{3}}\end{array}$

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

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