Question-134878

Question Number 134878 by bemath last updated on 08/Mar/21

Answered by EDWIN88 last updated on 08/Mar/21

B(θ) = (1/2).100.sin θ = 50 sin θ (i) radius of semi circle is r = 10.sin ((1/2)θ) so the area of semi circle A(θ)=(1/2)π(100.sin^2 ((1/2)θ)) A(θ) = 50π sin^2 ((1/2)θ) lim_(θ→0^+ ) ((50π sin^2 ((1/2)θ))/(50 sin θ)) = lim_(θ→0^+ ) ((π sin ((1/2)θ))/(2cos ((1/2)θ))) = 0

$$\mathrm{B}\left(\theta\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{100}.\mathrm{sin}\:\theta\:=\:\mathrm{50}\:\mathrm{sin}\:\theta\: \\ $$$$\left(\mathrm{i}\right)\:\mathrm{radius}\:\mathrm{of}\:\mathrm{semi}\:\mathrm{circle}\:\mathrm{is}\:\mathrm{r}\:=\:\mathrm{10}.\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\theta\right) \\ $$$$\mathrm{so}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{semi}\:\mathrm{circle}\:\mathrm{A}\left(\theta\right)=\frac{\mathrm{1}}{\mathrm{2}}\pi\left(\mathrm{100}.\mathrm{sin}\:^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\theta\right)\right) \\ $$$$\mathrm{A}\left(\theta\right)\:=\:\mathrm{50}\pi\:\mathrm{sin}\:^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\theta\right) \\ $$$$\underset{\theta\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\frac{\mathrm{50}\pi\:\mathrm{sin}\:^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\theta\right)}{\mathrm{50}\:\mathrm{sin}\:\theta}\:=\:\underset{\theta\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\frac{\pi\:\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\theta\right)}{\mathrm{2cos}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\theta\right)}\:=\:\mathrm{0} \\ $$

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Question-134878

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