Exercise Zone : Laju Perubahan
Table of Contents
Tipe:
No.
Diberikan\dfrac12\sqrt3 rad/s\dfrac13\sqrt3 rad/s\dfrac15\sqrt3 rad/s
\dfrac16\sqrt3 rad/s\dfrac17\sqrt3 rad/s
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
\dfrac{dx}{dt}&=0{,}2\\
&=\dfrac15\end{aligned}\)
\(\begin{aligned} y&=1+2\sin^2x\\ \dfrac{dy}{dx}&=4\sin x\cos x \end{aligned}\)
Laju perubahan y terhadap waktu adalah\dfrac{dy}{dt}
\(\begin{aligned} \dfrac{dy}{dt}&=\dfrac{dy}{dx}\cdot\dfrac{dx}{dt}\\[3.7pt] &=4\sin x\cos x\left(\dfrac15\right)\\[3.7pt] &=\dfrac45\sin x\cos x \end{aligned}\)
\(\begin{aligned} y&=1+2\sin^2x\\ \dfrac{dy}{dx}&=4\sin x\cos x \end{aligned}\)
Laju perubahan y terhadap waktu adalah
\(\begin{aligned} \dfrac{dy}{dt}&=\dfrac{dy}{dx}\cdot\dfrac{dx}{dt}\\[3.7pt] &=4\sin x\cos x\left(\dfrac15\right)\\[3.7pt] &=\dfrac45\sin x\cos x \end{aligned}\)
Saat {x=\dfrac13\pi} ,
\(\begin{aligned} \dfrac{dy}{dt}&=\dfrac45\sin\dfrac13\pi\cos\dfrac13\pi\\[3.7pt] &=\dfrac{\color{red}{\cancel{\color{black}{4}}}}5\left(\dfrac1{\color{red}{\cancel{\color{black}{2}}}}\sqrt3\right)\left(\dfrac1{\color{red}{\cancel{\color{black}{2}}}}\right)\\ &=\boxed{\boxed{\dfrac15\sqrt3}} \end{aligned}\)
\(\begin{aligned} \dfrac{dy}{dt}&=\dfrac45\sin\dfrac13\pi\cos\dfrac13\pi\\[3.7pt] &=\dfrac{\color{red}{\cancel{\color{black}{4}}}}5\left(\dfrac1{\color{red}{\cancel{\color{black}{2}}}}\sqrt3\right)\left(\dfrac1{\color{red}{\cancel{\color{black}{2}}}}\right)\\ &=\boxed{\boxed{\dfrac15\sqrt3}} \end{aligned}\)
Jadi, laju perubahan y terhadap waktu saat {x=\dfrac13\pi} adalah \dfrac15\sqrt3 rad/s.
JAWAB: C
JAWAB: C
No.
Laju perubahan fungsi- 2
- 5
- 6
- 8
- 10
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
f(x)&=\left(x^2-3\right)^2\\
f'(x)&=2\left(x^2-3\right)2x\\
&=4x\left(x^2-3\right)\\
f'(2)&=4(2)\left(2^2-3\right)\\
&=8(4-3)\\
&=8(1)\\
&=\boxed{\boxed{8}}
\end{aligned}\)
Jadi, laju perubahan fungsi f(x) = (x2 − 3)2 pada x = 2 adalah 8.
JAWAB: D
JAWAB: D
No.
Tentukan laju perubahan fungsiALTERNATIF PENYELESAIAN
\(\begin{aligned} f'(x)&=\displaystyle\lim_{h\to0}\dfrac{f(x+h)-f(x)}h\\ &=\displaystyle\lim_{h\to0}\dfrac{3(x+h)^2-4(x+h)-\left(3x^2-4x\right)}h\\ &=\displaystyle\lim_{h\to0}\dfrac{3\left(x^2+2xh+h^2\right)-4x-4h-3x^2+4x}h\\ &=\displaystyle\lim_{h\to0}\dfrac{3x^2+6xh+3h^2-4x-4h-3x^2+4x}h\\ &=\displaystyle\lim_{h\to0}\dfrac{6xh+3h^2-4h}h\\ &=\displaystyle\lim_{h\to0}\dfrac{h(6x+3h-4)}h\\ &=\displaystyle\lim_{h\to0}(6x+3h-4)\\ &=6x+3(0)-4\\ &=6x-4 \end{aligned}\)
Diperoleh
\(\begin{aligned} f'(2)&=6(2)-4\\ &=12-4\\ &=8 \end{aligned}\)
Jadi, laju perubahan fungsi f(x) = 3x2 − 4x di x = 2 adalah 8.
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