if-a-2-2ab-3b-2-1-prove-that-a-3b-3-d-2-b-da-2-2-a-2-2ab-3b-2-0-

Question Number 115380 by mathdave last updated on 25/Sep/20

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a^{2} + 2 a b + 3 b^{2} = 1, p r o v e t h a t

{(a + 3 b)}^{3} \frac{d^{2} b}{{d a}^{2}} + 2 (a^{2} + 2 a b + 3 b^{2}) = 0

Answered by Dwaipayan Shikari last updated on 25/Sep/20

a^{2} + 2 ab + {3 b}^{2} = 1

2 a + 2 b + 2 a \frac{db}{da} + 6 b \frac{db}{da} = 0

\frac{db}{da} = - \frac{a + b}{a + 3 b} \to (a)

2 + 2 \frac{db}{da} + 2 a \frac{d^{2} b}{{da}^{2}} + 2 \frac{db}{da} + 6 b \frac{d^{2} b}{{da}^{2}} + 6 {(\frac{db}{da})}^{2} = 0

2 + 4 \frac{db}{da} + 2 (a + 3 b) \frac{d^{2} b}{{da}^{2}} + 6 (\frac{db}{da}) = 0

1 - \frac{2 a + 2 b}{a + 3 b} + (a + 3 b) . \frac{d^{2} b}{{da}^{2}} + 3 {(\frac{a + b}{a + 3 b})}^{2} = 0

\frac{d^{2} b}{{da}^{2}} (a + 3 b) = - 3 {(\frac{a + b}{a + 3 b})}^{2} + \frac{2 (a + b)}{a + 3 b} - 1

\frac{d^{2} b}{{da}^{2}} {(a + 3 b)}^{3} = - {3 a}^{2} - 6 ab - {3 b}^{2} + 2 (a + 3 b) (a + b) - a^{2} - 6 ab - {9 b}^{2}

\frac{d^{2} b}{{da}^{2}} {(a + 3 b)}^{3} = - {4 a}^{2} - 12 ab - {12 b}^{2} + {2 a}^{2} + 6 ab + 2 ab + {6 b}^{2}

\frac{d^{2} b}{{da}^{2}} {(a + 3 b)}^{3} = - 2 (a^{2} + 2 ab + {3 b}^{2})

\frac{d^{2} b}{{da}^{2}} {(a + 3 b)}^{3} + 2 (a^{2} + 2 ab + {3 b}^{2}) = 0

Answered by 1549442205PVT last updated on 25/Sep/20

a^{2} + 2 a b + 3 b^{2} = 1 \Leftrightarrow {3 b}^{2} + 2 ab + a^{2} - 1 = 0

this is quadratic eqn . w . r . t b with

Δ^{'} = a^{2} - {3 a}^{2} + 3 = 3 - {2 a}^{2}

\Rightarrow b = \frac{- a \pm \sqrt{3 - {2 a}^{2}}}{3} (1)

\Rightarrow \frac{db}{da} = (- 1 \mp \frac{2 a}{\sqrt{3 - {2 a}^{2}}}) / 3

\Rightarrow \frac{d^{2} b}{{da}^{2}} = \mp (\frac{2 \sqrt{3 - {2 a}^{2}} + \frac{{4 a}^{2}}{\sqrt{3 - {2 a}^{2}}}}{3 (3 - {2 a}^{2})})

= \mp (\frac{2}{(3 - {2 a}^{2}) \sqrt{3 - {2 a}^{2}}}) (2)

From (1) we have 3 b + a = \pm \sqrt{3 - {2 a}^{2}}

\Rightarrow {(3 b + a)}^{3} = \pm (3 - {2 a}^{2}) \sqrt{3 - {2 a}^{2}} (3)

From (2) (3) we get

{(3 b + a)}^{3} \frac{d^{2} b}{{da}^{2}} = - 2

\Rightarrow {(3 b + a)}^{3} \frac{d^{2} b}{{da}^{2}} + 2 (a^{2} + 2 ab + {3 b}^{2}) = 0

(Q . E . D)