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.


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

$S o m e p e o p l e m a y h a v e n o t i c e d t h a t i$ $u s u a l l y c a l c u l a t e a r e a s c o n c e r n i n g$ $p a r a b o l a d i r e c t l y, w i t h o u t a p p l y i n g$ $c o m p l i c a t e d i n t e g r a l c a l c u l u s .$ $H e r e i a m g i v i n g y o u t h e b a c k g r o u d .$ $A c t u a l l y y o u k n o w a l l t h e s e t h i n g s a n d$ $y o u a r e a b l e t o p r o v e t h e m . M a y b e$ $y o u j u s t f o r g e t t o a p p l y t h e m .$
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

$Wow sir, it really helped me . God bless you more sir$ $for this .$ $I will love to see more of this once in a while .$ $I appreciate your time and effort sir$
Commented by Learner-123 last updated on 12/Apr/20
Wonderful!
Commented by Prithwish Sen 1 last updated on 13/Apr/20

$Great sir . Thanks for this and thanks for almost$ $everything . Thank you sir .$
Commented by Prithwish Sen 1 last updated on 13/Apr/20

$Sir how could one be good at geometry ? Is there$ $any good book / s to follow ? If there then please$ $name them . Or else give your valuable sugesstion .$
Commented by mr W last updated on 13/Apr/20

$s o r r y, i {c a n}^{'} t t e l l y o u t h e t r u t h, b e c a u s e$ $t h e t r u t h i s t h a t i r e a l l y k n o w n o b o o k,$ $a l l w h a t i k n o w a b o u t g e o m e t r y c o m e s$ $f r o m m y s c h o o l t i m e, l o n g l o n g t i m e$ $a g o . . .$ $b u t s e r i o u s l y, i h a v e l o v e g e o m e t r y$ $i n t h e s c h o o l a n d i w a s e v e n b e t t e r t h a n a l l$ $t h e t e a c h e r s t h e r e . a n d t h i n g s y o u l o v e$ $y o u {w o n}^{'} t f o r g e t! j u s t l o v e i t, t h i s$ $i s t h e o n l y s u g g e s t i o n i c a n g i v e y o u .$
Commented by Prithwish Sen 1 last updated on 13/Apr/20

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

$yes my man prof w i could see you were$ $good than your teacher . God has really$ $bless you with all necessary knowledge$ $in mathematics and physics and may$ $the good God continue to add more$ $because you have been a blessing to all$ $of us here on this noble platform . what$ $i like about you most is your detailed$ $explanation to questions and different$ $approaches to questions and above all$ $you can solve every question . am even$ $shot of words$
Commented by mr W last updated on 13/Apr/20

$n o t a l l t r u e, b u t t h a n k s f o r y o u r k i n d n e s s!$ $a n d g o d b l e s s y o u t o o!$
Commented by mr W last updated on 13/Apr/20
شكرا
Commented by otchereabdullai@gmail.com last updated on 13/Apr/20

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

$Thanks sir . how is A_{1} = \frac{1}{3} ab and A_{2} = \frac{2}{3} ab and for the second$ $parabola . The A_{1} = \frac{1}{3} ab is for both sides ? (+ and - sides ?)$ $or we do A_{1} for positive side and A_{1} for negative side .$
Commented by mr W last updated on 17/Apr/20

$p a r a b o l a : y = {k x}^{2}$ $A_{1} = \int_{0}^{a} y d x = \int_{0}^{a} {k x}^{2} d x = \frac{{k a}^{3}}{3} = \frac{a \times {k a}^{2}}{3} = \frac{a b}{3}$ $A_{2} = a b - A_{1} = \frac{2 a b}{3}$ $w e a r e t r e a t i n g h e r e a r e a, w h i c h i s$ $a p o s i t i v e q u a n t i t y .$ $a a n d b a r e l e n g t h o f l i n e s e g m e n t s,$ $t h e y a r e a l s o p o s i t i v e q u a n t i t y . t h e y$ $a r e n o t a l w a y s t h e s a m e a s c o o r d i n a t e s!$ $t h e f o r m u l a i s c e r t a i n l y a l s o f o r$ $t h e r i g h t s i d e o f t h e p a r a b o l a .$
Commented by I want to learn more last updated on 17/Apr/20

$I understand now sir$
Commented by mr W last updated on 17/Apr/20

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

$Thanks sir$
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

$s e e a l s o Q 89781$
Commented by mr W last updated on 06/May/20

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

$s a y y = {a x}^{2}$ $y_{1} = {a x}_{1}^{2}$ $y_{2} = {a x}_{2}^{2}$ $\Rightarrow \frac{y_{2}}{y_{1}} = \frac{{a x}_{2}^{2}}{{a x}_{1}^{2}} = {(\frac{x_{2}}{x_{1}})}^{2}$
	Answered by mr W last updated on 12/Apr/20

	$E x a m p l e 1$ $(s e e Q 88616)$
	Commented by mr W last updated on 12/Apr/20

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

	$w i d t h = a = 4 - 0 = 4$ $y_{P} = \frac{9 + 1}{2} = 5$ $x_{Q} = \frac{0 + 4}{2} = 2$ $y_{Q} = {(2 - 3)}^{2} = 1$ $h e i g h t b = y_{P} - y_{Q} = 5 - 1 = 4$ $a r e a o f s h a d e d r e g i o n$ $A = \frac{2}{3} a b = \frac{2}{3} \times 4 \times 4 = \frac{32}{3}$
	Answered by mr W last updated on 12/Apr/20

	$E x a m p l e 2$ $(s e e Q 88606)$
	Commented by mr W last updated on 12/Apr/20

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

	$a r e a o f s h a d e d r e g i o n i s \frac{9}{2} . f i n d t = ?$ $w i d t h a = t$ $a t m i d p o i n t o f w i d t h, i . e . a t x = \frac{t}{2},$ $y_{p a r a b o l a} = 4 \times \frac{t}{2} - {(\frac{t}{2})}^{2} = 2 t - \frac{t^{2}}{4}$ $y_{c h o r d} = \frac{1}{2} y (t) = \frac{1}{2} (4 t - t^{2}) = 2 t - \frac{t^{2}}{2}$ $h e i g h t b = y_{p a r a b o l a} - y_{c h o r d}$ $= 2 t - \frac{t^{2}}{4} - (2 t - \frac{t^{2}}{2}) = \frac{t^{2}}{4}$ $a r e a o f s h a d e d r e g i o n$ $A = \frac{2}{3} a b = \frac{2}{3} \times t \times \frac{t^{2}}{4} = \frac{t^{3}}{6}$ $\frac{t^{3}}{6} = \frac{9}{2}$ $t^{3} = 27$ $\Rightarrow t = 3$
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