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Question-206571

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Question Number 206571 by mr W last updated on 19/Apr/24
Commented by MathematicalUser2357 last updated on 26/Dec/24
Translation - What is the area S? (This question asks to calculate the area of S)
Answered by cortano21 last updated on 19/Apr/24
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Commented by mr W last updated on 19/Apr/24
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Answered by MM42 last updated on 19/Apr/24
f(x)=2x^2 ;A(a,2a^2 ) ,f′(a)=4a g(x)=2x^2 −8x+16 ; B(b,2b^2 −8b+16) ,g′(b)=4b−8 f′(a)=g′(b)⇒a=b−2 (i) m_(AB) =((2b^2 −8b+16−2a^2 )/(b−a))=4a (ii) (i),(ii)⇒a=1 & b=3 ⇒m_(AB) =4⇒L: y=4x−2 f=g⇒x=2 s_1 =∫_1 ^2 (2x^2 −4x+2)dx=(2/3) s_2 =∫_2 ^3 (2x^2 −8x+16−4x+2)dx=(2/3) ⇒s=(4/3) ✓
$${f}\left({x}\right)=\mathrm{2}{x}^{\mathrm{2}} \:\:\:\:;{A}\left({a},\mathrm{2}{a}^{\mathrm{2}} \right)\:,{f}'\left({a}\right)=\mathrm{4}{a} \\ $$$${g}\left({x}\right)=\mathrm{2}{x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{16}\:\:\:;\:{B}\left({b},\mathrm{2}{b}^{\mathrm{2}} −\mathrm{8}{b}+\mathrm{16}\right)\:\:,{g}'\left({b}\right)=\mathrm{4}{b}−\mathrm{8} \\ $$$${f}'\left({a}\right)={g}'\left({b}\right)\Rightarrow{a}={b}−\mathrm{2}\:\:\:\left({i}\right) \\ $$$${m}_{{AB}} =\frac{\mathrm{2}{b}^{\mathrm{2}} −\mathrm{8}{b}+\mathrm{16}−\mathrm{2}{a}^{\mathrm{2}} }{{b}−{a}}=\mathrm{4}{a}\:\left({ii}\right) \\ $$$$\left({i}\right),\left({ii}\right)\Rightarrow{a}=\mathrm{1}\:\&\:{b}=\mathrm{3} \\ $$$$\Rightarrow{m}_{{AB}} =\mathrm{4}\Rightarrow{L}:\:{y}=\mathrm{4}{x}−\mathrm{2} \\ $$$${f}={g}\Rightarrow{x}=\mathrm{2} \\ $$$${s}_{\mathrm{1}} =\int_{\mathrm{1}} ^{\mathrm{2}} \left(\mathrm{2}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{2}\right){dx}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${s}_{\mathrm{2}} =\int_{\mathrm{2}} ^{\mathrm{3}} \left(\mathrm{2}{x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{16}−\mathrm{4}{x}+\mathrm{2}\right){dx}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\Rightarrow{s}=\frac{\mathrm{4}}{\mathrm{3}}\:\checkmark \\ $$$$ \\ $$
Commented by mr W last updated on 19/Apr/24
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Answered by MathematicalUser2357 last updated on 28/Apr/24
Commented by MathematicalUser2357 last updated on 26/Dec/24
L_A T^E X Version here (Also fixed mistakes) ↓ Q.206571 - What is the area S? min{2x^2 }=0 so start point is (0,0) min{2x^2 −8x+16}=8 so end point is (4,8) Slope: 4, Y-intercept: 0. ⇒ y=4x ← There was a mistake! must be y=4x−2. These two parabolas meet at x=2 So, S=∫_1 ^2 2x^2 dx+∫_2 ^3 (2x^2 −8x+16)dx−∫_1 ^3 (4x−2)dx =[(2/3)x^3 ]_1 ^2 +[(2/3)x^3 −4x^2 +16x]_2 ^3 −[2x^2 −2x]_1 ^3 Note: I′m using built-in Complex number calculator to find area. (to prevent negative number mistake) (((2/3)×2^3 )−((2/3)×1^3 ))+(((2/3)×3^3 −4×3^2 +16×3)−((2/3)×2^3 −4×2^2 +16×2))−((2×3^2 −2×3)−(2×1^2 −2×1)) 1.333333 So S=(4/3).
$$\mathrm{L}_{\mathrm{A}} \mathrm{T}^{\mathrm{E}} \mathrm{X}\:\mathrm{Version}\:\mathrm{here}\:\left(\mathrm{Also}\:\mathrm{fixed}\:\mathrm{mistakes}\right)\:\downarrow \\ $$$$\mathrm{Q}.\mathrm{206571}\:-\:\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{area}\:{S}? \\ $$$$\mathrm{min}\left\{\mathrm{2}{x}^{\mathrm{2}} \right\}=\mathrm{0}\:\mathrm{so}\:\mathrm{start}\:\mathrm{point}\:\mathrm{is}\:\left(\mathrm{0},\mathrm{0}\right) \\ $$$$\mathrm{min}\left\{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{16}\right\}=\mathrm{8}\:\mathrm{so}\:\mathrm{end}\:\mathrm{point}\:\mathrm{is}\:\left(\mathrm{4},\mathrm{8}\right) \\ $$$$\mathrm{Slope}:\:\mathrm{4},\:\mathrm{Y}-\mathrm{intercept}:\:\mathrm{0}.\:\Rightarrow\:\:{y}=\mathrm{4}{x}\:\leftarrow\:\mathrm{There}\:\mathrm{was}\:\mathrm{a}\:\mathrm{mistake}!\:\mathrm{must}\:\mathrm{be}\:{y}=\mathrm{4}{x}−\mathrm{2}.\:\:\mathrm{These}\:\mathrm{two}\:\mathrm{parabolas}\:\mathrm{meet}\:\mathrm{at}\:{x}=\mathrm{2} \\ $$$$\mathrm{So},\:{S}=\int_{\mathrm{1}} ^{\mathrm{2}} \mathrm{2}{x}^{\mathrm{2}} {dx}+\int_{\mathrm{2}} ^{\mathrm{3}} \left(\mathrm{2}{x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{16}\right){dx}−\int_{\mathrm{1}} ^{\mathrm{3}} \left(\mathrm{4}{x}−\mathrm{2}\right){dx} \\ $$$$=\left[\frac{\mathrm{2}}{\mathrm{3}}{x}^{\mathrm{3}} \right]_{\mathrm{1}} ^{\mathrm{2}} +\left[\frac{\mathrm{2}}{\mathrm{3}}{x}^{\mathrm{3}} −\mathrm{4}{x}^{\mathrm{2}} +\mathrm{16}{x}\right]_{\mathrm{2}} ^{\mathrm{3}} −\left[\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}{x}\right]_{\mathrm{1}} ^{\mathrm{3}} \\ $$$$\mathrm{Note}:\:\mathrm{I}'\mathrm{m}\:\mathrm{using}\:\mathrm{built}-\mathrm{in}\:\mathrm{Complex}\:\mathrm{number}\:\mathrm{calculator} \\ $$$$\mathrm{to}\:\mathrm{find}\:\mathrm{area}.\:\left(\mathrm{to}\:\mathrm{prevent}\:\mathrm{negative}\:\mathrm{number}\:\mathrm{mistake}\right) \\ $$$$\left(\left(\frac{\mathrm{2}}{\mathrm{3}}×\mathrm{2}^{\mathrm{3}} \right)−\left(\frac{\mathrm{2}}{\mathrm{3}}×\mathrm{1}^{\mathrm{3}} \right)\right)+\left(\left(\frac{\mathrm{2}}{\mathrm{3}}×\mathrm{3}^{\mathrm{3}} −\mathrm{4}×\mathrm{3}^{\mathrm{2}} +\mathrm{16}×\mathrm{3}\right)−\left(\frac{\mathrm{2}}{\mathrm{3}}×\mathrm{2}^{\mathrm{3}} −\mathrm{4}×\mathrm{2}^{\mathrm{2}} +\mathrm{16}×\mathrm{2}\right)\right)−\left(\left(\mathrm{2}×\mathrm{3}^{\mathrm{2}} −\mathrm{2}×\mathrm{3}\right)−\left(\mathrm{2}×\mathrm{1}^{\mathrm{2}} −\mathrm{2}×\mathrm{1}\right)\right) \\ $$$$\mathrm{1}.\mathrm{333333} \\ $$$$\mathrm{So}\:{S}=\frac{\mathrm{4}}{\mathrm{3}}. \\ $$

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