If-y-f-x-ax-2-bx-c-and-at-some-x-say-x-p-0-p-ydx-y-p-y-p-y-p-p-then-find-p-

Question Number 44543 by ajfour last updated on 01/Oct/18

I f y = f (x) = {a x}^{2} + b x + c

a n d a t s o m e x, s a y x = p

\int_{0}^{p} y d x = y (p) = y^{'} (p) = y^{″} (p) = p,

t h e n f i n d p .

Commented by MrW3 last updated on 01/Oct/18

y^{'} (x) = 2 a x + b

y^{″} (x) = 2 a

2 a = p \Rightarrow a = \frac{p}{2}

2 \times \frac{p}{2} \times p + b = p \Rightarrow b = p (1 - p)

\frac{p}{2} \times p^{2} + p (1 - p) \times p + c = p

\Rightarrow c = p - p^{2} + \frac{p^{3}}{2}

\int_{0}^{p} y d x = \frac{a}{3} \times p^{3} + \frac{b}{2} \times p^{2} + c p = p

\frac{a}{3} \times p^{2} + \frac{b}{2} \times p + c = 1

\frac{p}{6} \times p^{2} + \frac{p (1 - p)}{2} \times p + p - p^{2} + \frac{p^{3}}{2} = 1

\frac{p^{3}}{6} + \frac{p^{2}}{2} - \frac{p^{3}}{2} + p - p^{2} + \frac{p^{3}}{2} = 1

\frac{p^{3}}{6} - \frac{p^{2}}{2} + p = 1

p^{3} - 3 p^{2} + 6 p - 6 = 0

\Rightarrow p \approx 1.6

Commented by MrW3 last updated on 01/Oct/18

t h a n k s f o r c h e c k i n g s i r!

Commented by ajfour last updated on 01/Oct/18

y e s s i r, i t i s c o r r e c t t h e r e, i d i n t

n o t i c e p r o p e r l y, p l e a s e p a r d o n .

T h a n k y o u .

Answered by ajfour last updated on 01/Oct/18

p = 2 a = 2 a p + b = {a p}^{2} + b p + c

= \frac{{a p}^{3}}{3} + \frac{{b p}^{2}}{2} + c p

\Rightarrow {a p}^{2} + b p + c = p . . (i)

\Rightarrow \frac{{a p}^{2}}{3} + \frac{b p}{2} + c = 1 . . (i i)

(i) - (i i) g i v e s

\frac{2 {a p}^{2}}{3} + \frac{b p}{2} = p - 1, a n d w e a l r e a d y

h a v e 2 a p + b = p

w i t h 2 a = p, b = p - p^{2}

S o \frac{p^{3}}{3} + \frac{p}{2} (p - p^{2}) = p - 1

\Rightarrow \frac{p^{3}}{6} - \frac{p^{2}}{2} + p - 1 = 0

o r p^{3} - 3 p^{2} + 6 p - 6 = 0

l e t p = t + 1

\Rightarrow t^{3} + 3 t + 1 - 6 t - 3 + 6 t = 0

\Rightarrow t^{3} + 3 t - 2 = 0

l e t t = u + v

\Rightarrow u^{3} + v^{3} + 3 (u + v) (u v + 1) = 2

f u r t h e r l e t u v = - 1

t h e n u^{3} + v^{3} = 2

u^{3}, v^{3} a r e r o o t s o f q u a d r a t i c

z^{2} - 2 z - 1 = 0

z = 1 \pm \sqrt{2}

t = u + v = {(\sqrt{2} + 1)}^{1 / 3} - {(\sqrt{2} - 1)}^{1 / 3}

p = t + 1 \approx 1.59607164

Commented by MrW3 last updated on 01/Oct/18

I k n e w {y o u}^{'} l l g i v e t h e e x a c t s o l u t i o n,

t h a n k s f o r t h e p e r f e c t r e s u l t!

Commented by ajfour last updated on 01/Oct/18

t h a n k s m r W S i r (f o r {C a r d a n o}^{'} s) .

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