Exercise Zone : Laju Perubahan

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Berikut ini adalah kumpulan soal mengenai Laju Perubahan. Jika ingin bertanya soal, silahkan gabung ke grup Telegram, Signal, Discord, atau WhatsApp.

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StandarSBMPTNHOTS

No. 1

Diberikan y = 1 + 2 sin² x untuk 0 ≤ x ≤ π dan nilai x bertambah pada laju 0,2 rad/s. Laju perubahan y terhadap waktu saat {x=\dfrac13\pi} adalah

1. \dfrac12\sqrt3 rad/s
2. \dfrac13\sqrt3 rad/s
3. \dfrac15\sqrt3 rad/s

4. \dfrac16\sqrt3 rad/s
5. \dfrac17\sqrt3 rad/s

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ALTERNATIF PENYELESAIAN

$\begin{array}{rl}{\displaystyle \frac{dx}{dt}}& =0,2\\ & ={\displaystyle \frac{1}{5}}\end{array}$

$\begin{array}{rl}y& =1+2{\sin }^{2}x\\ {\displaystyle \frac{dy}{dx}}& =4\sin x\cos x\end{array}$

Laju perubahan y terhadap waktu adalah \dfrac{dy}{dt}

$\begin{array}{rl}{\displaystyle \frac{dy}{dt}}& ={\displaystyle \frac{dy}{dx}}\cdot {\displaystyle \frac{dx}{dt}}\\ & =4\sin x\cos x\left({\displaystyle \frac{1}{5}}\right)\\ & ={\displaystyle \frac{4}{5}}\sin x\cos x\end{array}$

Saat {x=\dfrac13\pi},

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$\begin{array}{rl}{\displaystyle \frac{dy}{dt}}& ={\displaystyle \frac{4}{5}}\sin {\displaystyle \frac{1}{3}}\pi \cos {\displaystyle \frac{1}{3}}\pi \\ & ={\displaystyle \frac{\text{color}red\text{cancel}\text{color}black4}{5}}\left({\displaystyle \frac{1}{\text{color}red\text{cancel}\text{color}black2}}\sqrt{3}\right)\left({\displaystyle \frac{1}{\text{color}red\text{cancel}\text{color}black2}}\right)\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{1}{5}}\sqrt{3}}}}}\end{array}$

Jadi, laju perubahan y terhadap waktu saat {x=\dfrac13\pi} adalah \dfrac15\sqrt3 rad/s.
JAWAB: C

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No. 2

Laju perubahan fungsi f(x) = (x² − 3)² pada x = 2 adalah

1. 2
2. 5
3. 6

4. 8
5. 10

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ALTERNATIF PENYELESAIAN

$\begin{array}{rl}f(x)& ={(x^2-3)}^2\\ f^{\prime }(x)& =2(x^2-3)2x\\ & =4x(x^2-3)\\ f^{\prime }(2)& =4(2)(2^2-3)\\ & =8(4-3)\\ & =8(1)\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 8}}}}\end{array}$

Jadi, laju perubahan fungsi f(x) = (x² − 3)² pada x = 2 adalah 8.
JAWAB: D

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No. 3

Tentukan laju perubahan fungsi f(x) = 3x² − 4x di x = 2.

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ALTERNATIF PENYELESAIAN

f(x) = 3x² − 4x maka f(x + h) = 3(x + h)² − 4(x + h) sehingga laju perubahannya:
$\begin{array}{rl}{f}^{\prime }(x)& ={\displaystyle \underset{h\to 0}{lim}{\displaystyle \frac{f(x+h)-f(x)}{h}}}\\ & ={\displaystyle \underset{h\to 0}{lim}{\displaystyle \frac{3(x+h{)}^2-4(x+h)-(3x^2-4x)}{h}}}\\ & ={\displaystyle \underset{h\to 0}{lim}{\displaystyle \frac{3(x^2+2xh+h^2)-4x-4h-3x^2+4x}{h}}}\\ & ={\displaystyle \underset{h\to 0}{lim}{\displaystyle \frac{3x^2+6xh+3h^2-4x-4h-3x^2+4x}{h}}}\\ & ={\displaystyle \underset{h\to 0}{lim}{\displaystyle \frac{6xh+3h^2-4h}{h}}}\\ & ={\displaystyle \underset{h\to 0}{lim}{\displaystyle \frac{h(6x+3h-4)}{h}}}\\ & ={\displaystyle \underset{h\to 0}{lim}(6x+3h-4)}\\ & =6x+3(0)-4\\ & =6x-4\end{array}$
Diperoleh f'(x) = 6x − 4. Dengan demikian:
$\begin{array}{rl}{f}^{\prime }(2)& =6(2)-4\\ & =12-4\\ & =8\end{array}$

Jadi, laju perubahan fungsi f(x) = 3x² − 4x di x = 2 adalah 8.

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