Kalkulus bantu ya besok jam pertama ini gak paham dari tadi.... Question from @Adeniah1

December 2019 1 72 Report

Kalkulus bantu ya besok jam pertama ini gak paham dari tadi

Adjie564 Saya bantu beberapa nomor aj y mas

4. $\frac{\partial^2{z}}{\partial{x}\partial{y}}$ dari $z=xy\ln{({x}^{2}+{y}^{2})}$

maksud soal itu adalah kita harus mencari turunan dua variabel sekaligus dikarenakan notasinya itu.

Misalkan : u = xy dan v = ln(x²+y²), maka turunan parsial keduanya dapat dicari dengan rumus hasil kali dua fungsi.

$\frac{\partial^2{z}}{\partial{x}\partial{y}}=v\frac{\partial^2{u}}{\partial{x}\partial{y}}+u\frac{\partial^2{v}}{\partial{x}\partial{y}}\\\frac{\partial^2{z}}{\partial{x}\partial{y}}=(y+x)(\ln{({x}^{2}+{y}^{2})})+(xy)\left(\frac{2(x+y)}{{x}^{2}+{y}^{2}}\right)\\\frac{\partial^2{z}}{\partial{x}\partial{y}}=(x+y)\left(\frac{({x}^{2}+{y}^{2})\ln{({x}^{2}+{y}^{2})}+2xy}{{x}^{2}+{y}^{2}}\right)$

6. $\left\{\begin{array}{ccc}u=x{y}^{2}+y{x}^{2}\\x=r\cos{t}\\y=r\sin{t}\end{array}$

Disuruh hitung $\frac{\partial{u}}{\partial{r}}\,dan\,\frac{\partial{u}}{\partial{t}}\,r=2\,t=\frac{\pi}{4}$

Substitusikan terlebih dahulu x dan y nya ke u, sehingga.

$u={r}^{3}\sin^2{t}\cos{t}+{r}^{3}\sin{t}\cos^2{t}\\u={r}^{3}\sin{t}\cos{t}(\sin{t}+\cos{t})=\frac{1}{2}{r}^{3}\sin{2t}(\sin{t}+\cos{t})$

Jadi,

$\frac{\partial{u}}{\partial{r}}=3{r}^{2}\sin{2t}(\sin{t}+\cos{t})\\3{(2)}^{2}\sin{\frac{\pi}{2}}(\sin{\frac{\pi}{4}}+\cos{\frac{\pi}{4}})\\12\sqrt{2}$

$\frac{\partial{u}}{\partial{t}}=2{r}^{3}\cos{2t}(\sin{t}+\cos{t})+{r}^{3}\sin{2t}(\cos{t}-\sin{t})\\2{(2)}^{3}\cos{\frac{\pi}{2}}(\sin{\frac{\pi}{4}}+\cos{\frac{\pi}{4}})+{(2)}^{3}\sin{\frac{pi}{2}}(\cos{\frac{\pi}{4}}-\sin{\frac{\pi}{4}})=8(0)(\sqrt{2})+8(1)(0)=0$

7. PBS dr kurva ${z}^{2}={x}^{2}y-{y}^{2}x+5x$ di titik P(2,3,2).

Rumus yang bisa dipakai adalah :

$(x-x_{0})\dfrac{\partial{\beta}}{\partial{x}}\right|_{(x_{0},y_{0},z_{0})}+(y-y_{0})\dfrac{\partial{\beta}}{\partial{y}}\right|_{(x_{0},y_{0},z_{0})}+(z-z_{0})\dfrac{\partial{\beta}}{\partial{z}}\right|_{(x_{0},y_{0},z_{0})}=0$

misalkan : $\beta={x}^{2}y-{y}^{2}x+5x-{z}^{2}$ , maka :

1) $m_{x}=\dfrac{\partial{\beta}}{\partial{x}}\right|_{(x_{0},y_{0},z_{0})}\\m_{x}=(2xy-{y}^{2}+5)_{(2,3,2)}\\m_{x}=12-9+5=8$

2) $m_{y}=\dfrac{\partial{\beta}}{\partial{y}}\right|_{(x_{0},y_{0},z_{0})}\\m_{y}=({x}^{2}-2xy)_{(2,3,2)}=4-12=-8$

3) $m_{z}=\dfrac{\partial{\beta}}{\partial{z}}\right|_{(x_{0},y_{0},z_{0})}\\m_z=(-2z)_{(2,3,2)}=-4$

Substitusikan hasilnya ke rumus, sehingga :

$\alpha≈(x-2)(8)+(y-3)(-8)+(z-2)(-4)=0\\\alpha≈8x-16-8y+24-4z+8=0\\\alpha≈8x-8y-4z=-16\to\,\alpha≈2x-2y-z=-4$

Semoga membantu. Maaf jikalau nomor 7 kurang tepat.

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Adeniah1 iya gpp.. terimakasih banyak.. btw saya perempuan kak

Adjie564 oh, maaf saya kira laki laki, saya akan nambah sedikit nomornya, yg nmr 7

Adeniah1 oh iya kak silahkan.. sekali lagi terimakasih banyak