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Question Number 88758 by mr W last updated on 12/Apr/20

Some people may have noticed that i usually calculate areas concerning parabola directly, without applying complicated integral calculus. Here i am giving you the backgroud. Actually you know all these things and you are able to prove them. Maybe you just forget to apply them.

$${Some}\:{people}\:{may}\:{have}\:{noticed}\:{that}\:{i} \\ $$$${usually}\:{calculate}\:{areas}\:{concerning} \\ $$$${parabola}\:{directly},\:{without}\:{applying} \\ $$$${complicated}\:{integral}\:{calculus}. \\ $$$${Here}\:{i}\:{am}\:{giving}\:{you}\:{the}\:{backgroud}.\: \\ $$$${Actually}\:{you}\:{know}\:{all}\:{these}\:{things}\:{and} \\ $$$${you}\:{are}\:{able}\:{to}\:{prove}\:{them}.\:{Maybe} \\ $$$${you}\:{just}\:{forget}\:{to}\:{apply}\:{them}. \\ $$

Commented by mr W last updated on 12/Apr/20

Commented by mr W last updated on 12/Apr/20

Commented by TawaTawa1 last updated on 12/Apr/20

Wow sir, it really helped me. God bless you more sir for this. I will love to see more of this once in a while. I appreciate your time and effort sir

$$\mathrm{Wow}\:\mathrm{sir},\:\mathrm{it}\:\mathrm{really}\:\mathrm{helped}\:\mathrm{me}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{more}\:\mathrm{sir} \\ $$$$\mathrm{for}\:\mathrm{this}. \\ $$$$\mathrm{I}\:\mathrm{will}\:\mathrm{love}\:\mathrm{to}\:\mathrm{see}\:\mathrm{more}\:\mathrm{of}\:\mathrm{this}\:\mathrm{once}\:\mathrm{in}\:\mathrm{a}\:\mathrm{while}. \\ $$$$\mathrm{I}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{time}\:\mathrm{and}\:\mathrm{effort}\:\mathrm{sir} \\ $$$$ \\ $$

Commented by Learner-123 last updated on 12/Apr/20

Wonderful!

Commented by Prithwish Sen 1 last updated on 13/Apr/20

Great sir. Thanks for this and thanks for almost everything. Thank you sir.

$$\mathrm{Great}\:\mathrm{sir}.\:\mathrm{Thanks}\:\mathrm{for}\:\mathrm{this}\:\mathrm{and}\:\mathrm{thanks}\:\mathrm{for}\:\mathrm{almost} \\ $$$$\mathrm{everything}.\:\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Commented by Prithwish Sen 1 last updated on 13/Apr/20

Sir how could one be good at geometry ? Is there any good book/s to follow ? If there then please name them. Or else give your valuable sugesstion.

$$\mathrm{Sir}\:\mathrm{how}\:\mathrm{could}\:\mathrm{one}\:\mathrm{be}\:\mathrm{good}\:\mathrm{at}\:\mathrm{geometry}\:?\:\mathrm{Is}\:\mathrm{there}\: \\ $$$$\mathrm{any}\:\mathrm{good}\:\mathrm{book}/\mathrm{s}\:\mathrm{to}\:\mathrm{follow}\:?\:\mathrm{If}\:\mathrm{there}\:\mathrm{then}\:\mathrm{please} \\ $$$$\mathrm{name}\:\mathrm{them}.\:\mathrm{Or}\:\mathrm{else}\:\mathrm{give}\:\mathrm{your}\:\mathrm{valuable}\:\mathrm{sugesstion}. \\ $$

Commented by mr W last updated on 13/Apr/20

sorry, i can′t tell you the truth, because the truth is that i really know no book, all what i know about geometry comes from my school time, long long time ago... but seriously, i have love geometry in the school and i was even better than all the teachers there. and things you love you won′t forget! just love it, this is the only suggestion i can give you.

$${sorry},\:{i}\:{can}'{t}\:{tell}\:{you}\:{the}\:{truth},\:{because} \\ $$$${the}\:{truth}\:{is}\:{that}\:{i}\:{really}\:{know}\:{no}\:{book},\: \\ $$$${all}\:{what}\:{i}\:{know}\:{about}\:{geometry}\:{comes} \\ $$$${from}\:{my}\:{school}\:{time},\:{long}\:{long}\:{time} \\ $$$${ago}... \\ $$$${but}\:{seriously},\:{i}\:{have}\:{love}\:{geometry} \\ $$$${in}\:{the}\:{school}\:{and}\:{i}\:{was}\:{even}\:{better}\:{than}\:{all} \\ $$$${the}\:{teachers}\:{there}.\:{and}\:{things}\:{you}\:{love} \\ $$$${you}\:{won}'{t}\:{forget}!\:{just}\:{love}\:{it},\:{this} \\ $$$${is}\:{the}\:{only}\:{suggestion}\:{i}\:{can}\:{give}\:{you}. \\ $$

Commented by Prithwish Sen 1 last updated on 13/Apr/20

Thanks sir.

$$\mathrm{Thanks}\:\mathrm{sir}.\: \\ $$

Commented by otchereabdullai@gmail.com last updated on 13/Apr/20

yes my man prof w i could see you were good than your teacher . God has really bless you with all necessary knowledge in mathematics and physics and may the good God continue to add more because you have been a blessing to all of us here on this noble platform. what i like about you most is your detailed explanation to questions and different approaches to questions and above all you can solve every question . am even shot of words

$$\mathrm{yes}\:\mathrm{my}\:\mathrm{man}\:\mathrm{prof}\:\mathrm{w}\:\mathrm{i}\:\mathrm{could}\:\mathrm{see}\:\mathrm{you}\:\mathrm{were} \\ $$$$\mathrm{good}\:\mathrm{than}\:\mathrm{your}\:\mathrm{teacher}\:.\:\mathrm{God}\:\mathrm{has}\:\mathrm{really} \\ $$$$\mathrm{bless}\:\mathrm{you}\:\mathrm{with}\:\mathrm{all}\:\mathrm{necessary}\:\mathrm{knowledge} \\ $$$$\mathrm{in}\:\mathrm{mathematics}\:\mathrm{and}\:\mathrm{physics}\:\mathrm{and}\:\mathrm{may} \\ $$$$\mathrm{the}\:\mathrm{good}\:\mathrm{God}\:\mathrm{continue}\:\mathrm{to}\:\mathrm{add}\:\mathrm{more}\: \\ $$$$\mathrm{because}\:\mathrm{you}\:\mathrm{have}\:\mathrm{been}\:\mathrm{a}\:\mathrm{blessing}\:\mathrm{to}\:\mathrm{all} \\ $$$$\mathrm{of}\:\mathrm{us}\:\mathrm{here}\:\mathrm{on}\:\mathrm{this}\:\mathrm{noble}\:\mathrm{platform}.\:\mathrm{what} \\ $$$$\mathrm{i}\:\mathrm{like}\:\mathrm{about}\:\mathrm{you}\:\mathrm{most}\:\mathrm{is}\:\mathrm{your}\:\mathrm{detailed}\: \\ $$$$\mathrm{explanation}\:\mathrm{to}\:\mathrm{questions}\:\mathrm{and}\:\mathrm{different} \\ $$$$\mathrm{approaches}\:\mathrm{to}\:\mathrm{questions}\:\mathrm{and}\:\mathrm{above}\:\mathrm{all} \\ $$$$\mathrm{you}\:\mathrm{can}\:\mathrm{solve}\:\mathrm{every}\:\mathrm{question}\:.\:\mathrm{am}\:\mathrm{even} \\ $$$$\mathrm{shot}\:\mathrm{of}\:\mathrm{words} \\ $$

Commented by mr W last updated on 13/Apr/20

not all true, but thanks for your kindness! and god bless you too!

$${not}\:{all}\:{true},\:{but}\:{thanks}\:{for}\:{your}\:{kindness}! \\ $$$${and}\:{god}\:{bless}\:{you}\:{too}! \\ $$

Commented by mr W last updated on 13/Apr/20

شكرا

Commented by otchereabdullai@gmail.com last updated on 13/Apr/20

Amen

$$\mathrm{Amen} \\ $$

Commented by I want to learn more last updated on 17/Apr/20

Thanks sir. how is A_1 = (1/3)ab and A_2 = (2/3)ab and for the second parabola. The A_1 = (1/3)ab is for both sides? (+ and − sides?) or we do A_1 for positive side and A_1 for negative side.

$$\mathrm{Thanks}\:\mathrm{sir}.\:\:\mathrm{how}\:\mathrm{is}\:\:\:\:\mathrm{A}_{\mathrm{1}} \:\:=\:\:\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ab}\:\:\mathrm{and}\:\:\mathrm{A}_{\mathrm{2}} \:\:=\:\:\frac{\mathrm{2}}{\mathrm{3}}\mathrm{ab}\:\:\:\mathrm{and}\:\mathrm{for}\:\mathrm{the}\:\mathrm{second} \\ $$$$\mathrm{parabola}.\:\mathrm{The}\:\mathrm{A}_{\mathrm{1}} \:=\:\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ab}\:\mathrm{is}\:\mathrm{for}\:\mathrm{both}\:\mathrm{sides}?\:\left(+\:\:\mathrm{and}\:−\:\mathrm{sides}?\right) \\ $$$$\mathrm{or}\:\mathrm{we}\:\mathrm{do}\:\mathrm{A}_{\mathrm{1}} \:\mathrm{for}\:\mathrm{positive}\:\mathrm{side}\:\mathrm{and}\:\mathrm{A}_{\mathrm{1}} \:\mathrm{for}\:\mathrm{negative}\:\mathrm{side}. \\ $$

Commented by mr W last updated on 17/Apr/20

parabola: y=kx^2 A_1 =∫_0 ^( a) ydx=∫_0 ^( a) kx^2 dx=((ka^3 )/3)=((a×ka^2 )/3)=((ab)/3) A_2 =ab−A_1 =((2ab)/3) we are treating here area, which is a positive quantity. a and b are length of line segments, they are also positive quantity. they are not always the same as coordinates! the formula is certainly also for the right side of the parabola.

$${parabola}:\:{y}={kx}^{\mathrm{2}} \\ $$$${A}_{\mathrm{1}} =\int_{\mathrm{0}} ^{\:{a}} {ydx}=\int_{\mathrm{0}} ^{\:{a}} {kx}^{\mathrm{2}} {dx}=\frac{{ka}^{\mathrm{3}} }{\mathrm{3}}=\frac{{a}×{ka}^{\mathrm{2}} }{\mathrm{3}}=\frac{{ab}}{\mathrm{3}} \\ $$$${A}_{\mathrm{2}} ={ab}−{A}_{\mathrm{1}} =\frac{\mathrm{2}{ab}}{\mathrm{3}} \\ $$$$ \\ $$$${we}\:{are}\:{treating}\:{here}\:{area},\:{which}\:{is} \\ $$$${a}\:{positive}\:{quantity}. \\ $$$${a}\:{and}\:{b}\:{are}\:{length}\:{of}\:{line}\:{segments}, \\ $$$${they}\:{are}\:{also}\:{positive}\:{quantity}.\:{they} \\ $$$${are}\:{not}\:{always}\:{the}\:{same}\:{as}\:{coordinates}!\: \\ $$$${the}\:{formula}\:{is}\:{certainly}\:{also}\:{for} \\ $$$${the}\:{right}\:{side}\:{of}\:{the}\:{parabola}. \\ $$

Commented by I want to learn more last updated on 17/Apr/20

I understand now sir

$$\mathrm{I}\:\mathrm{understand}\:\mathrm{now}\:\mathrm{sir} \\ $$

Commented by mr W last updated on 17/Apr/20

Commented by I want to learn more last updated on 17/Apr/20

Thanks sir

$$\mathrm{Thanks}\:\mathrm{sir} \\ $$

Commented by mr W last updated on 19/Apr/20

Commented by mr W last updated on 19/Apr/20

Commented by mr W last updated on 19/Apr/20

see also Q89781

$${see}\:{also}\:{Q}\mathrm{89781} \\ $$

Commented by mr W last updated on 06/May/20

Commented by mr W last updated on 06/May/20

say y=ax^2 y_1 =ax_1 ^2 y_2 =ax_2 ^2 ⇒(y_2 /y_1 )=((ax_2 ^2 )/(ax_1 ^2 ))=((x_2 /x_1 ))^2

$${say}\:{y}={ax}^{\mathrm{2}} \\ $$$${y}_{\mathrm{1}} ={ax}_{\mathrm{1}} ^{\mathrm{2}} \\ $$$${y}_{\mathrm{2}} ={ax}_{\mathrm{2}} ^{\mathrm{2}} \\ $$$$\Rightarrow\frac{{y}_{\mathrm{2}} }{{y}_{\mathrm{1}} }=\frac{{ax}_{\mathrm{2}} ^{\mathrm{2}} }{{ax}_{\mathrm{1}} ^{\mathrm{2}} }=\left(\frac{{x}_{\mathrm{2}} }{{x}_{\mathrm{1}} }\right)^{\mathrm{2}} \\ $$

Answered by mr W last updated on 12/Apr/20

Example 1 (see Q88616)

$${Example}\:\mathrm{1} \\ $$$$\left({see}\:{Q}\mathrm{88616}\right) \\ $$

Commented by mr W last updated on 12/Apr/20

Commented by mr W last updated on 12/Apr/20

width=a=4−0=4 y_P =((9+1)/2)=5 x_Q =((0+4)/2)=2 y_Q =(2−3)^2 =1 height b=y_P −y_Q =5−1=4 area of shaded region A=(2/3)ab=(2/3)×4×4=((32)/3)

$${width}={a}=\mathrm{4}−\mathrm{0}=\mathrm{4} \\ $$$${y}_{{P}} =\frac{\mathrm{9}+\mathrm{1}}{\mathrm{2}}=\mathrm{5} \\ $$$${x}_{{Q}} =\frac{\mathrm{0}+\mathrm{4}}{\mathrm{2}}=\mathrm{2} \\ $$$${y}_{{Q}} =\left(\mathrm{2}−\mathrm{3}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$$ \\ $$$${height}\:{b}={y}_{{P}} −{y}_{{Q}} =\mathrm{5}−\mathrm{1}=\mathrm{4} \\ $$$$ \\ $$$${area}\:{of}\:{shaded}\:{region} \\ $$$${A}=\frac{\mathrm{2}}{\mathrm{3}}{ab}=\frac{\mathrm{2}}{\mathrm{3}}×\mathrm{4}×\mathrm{4}=\frac{\mathrm{32}}{\mathrm{3}} \\ $$

Answered by mr W last updated on 12/Apr/20

Example 2 (see Q88606)

$${Example}\:\mathrm{2} \\ $$$$\left({see}\:{Q}\mathrm{88606}\right) \\ $$

Commented by mr W last updated on 12/Apr/20

Commented by mr W last updated on 12/Apr/20

area of shaded region is (9/2). find t=? width a=t at midpoint of width, i.e. at x=(t/2), y_(parabola) =4×(t/2)−((t/2))^2 =2t−(t^2 /4) y_(chord) =(1/2)y(t)=(1/2)(4t−t^2 )=2t−(t^2 /2) height b=y_(parabola) −y_(chord) =2t−(t^2 /4)−(2t−(t^2 /2))=(t^2 /4) area of shaded region A=(2/3)ab=(2/3)×t×(t^2 /4)=(t^3 /6) (t^3 /6)=(9/2) t^3 =27 ⇒t=3

$${area}\:{of}\:{shaded}\:{region}\:{is}\:\frac{\mathrm{9}}{\mathrm{2}}.\:{find}\:{t}=? \\ $$$${width}\:{a}={t} \\ $$$${at}\:{midpoint}\:{of}\:{width},\:{i}.{e}.\:{at}\:{x}=\frac{{t}}{\mathrm{2}}, \\ $$$${y}_{{parabola}} =\mathrm{4}×\frac{{t}}{\mathrm{2}}−\left(\frac{{t}}{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{2}{t}−\frac{{t}^{\mathrm{2}} }{\mathrm{4}} \\ $$$${y}_{{chord}} =\frac{\mathrm{1}}{\mathrm{2}}{y}\left({t}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{4}{t}−{t}^{\mathrm{2}} \right)=\mathrm{2}{t}−\frac{{t}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$ \\ $$$${height}\:{b}={y}_{{parabola}} −{y}_{{chord}} \\ $$$$=\mathrm{2}{t}−\frac{{t}^{\mathrm{2}} }{\mathrm{4}}−\left(\mathrm{2}{t}−\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\right)=\frac{{t}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$ \\ $$$${area}\:{of}\:{shaded}\:{region} \\ $$$${A}=\frac{\mathrm{2}}{\mathrm{3}}{ab}=\frac{\mathrm{2}}{\mathrm{3}}×{t}×\frac{{t}^{\mathrm{2}} }{\mathrm{4}}=\frac{{t}^{\mathrm{3}} }{\mathrm{6}} \\ $$$$\frac{{t}^{\mathrm{3}} }{\mathrm{6}}=\frac{\mathrm{9}}{\mathrm{2}} \\ $$$${t}^{\mathrm{3}} =\mathrm{27} \\ $$$$\Rightarrow{t}=\mathrm{3} \\ $$

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