Question-43386

Question Number 43386 by ajfour last updated on 10/Sep/18

Commented by ajfour last updated on 10/Sep/18

I f t h e t w o d i f f e r e n t l y s h a d e d

a r e a s a r e e q u a l, f i n d s l o p e m o f

l i n e, i n t e r m s o f a .

Commented by Tawa1 last updated on 10/Sep/18

Sir, please how did you draw the smoth curve using lekh diagram ?

Commented by ajfour last updated on 10/Sep/18

q u i c k l y m o v e d f i n g e r l i k e a

p a r a b o l a (o n e b r a n c h l o n g, o t h e r

s h o r t), (l e k h r e c o g n i s e d i t),

c o p i e d i t, f l i p p e d t h e c o p y

v e r i c a l t h e n h o r i z o n t a l a n d

j o i n e d t h e b r a n c h e n d s o f

o r i g i n a l a n d c o p i e d o n e, t h e n

g r o u p e d t h e m, t h e n r e s i z e d i t .

Commented by Tawa1 last updated on 10/Sep/18

God bless you sir

Answered by MrW3 last updated on 10/Sep/18

y = x (x^{2} - a^{2})

\frac{d y}{d x} = 3 x^{2} - a^{2}

a t x = 0 : \tan θ_{0} = - a^{2}

\Rightarrow r \sin θ = r \cos θ (r^{2} \cos^{2} θ - a^{2})

\Rightarrow r^{2} = \frac{\tan θ + a^{2}}{\cos^{2} θ}

A = \int_{θ_{0}}^{2 π} \frac{r^{2} d θ}{2} = \frac{1}{2} \int_{θ_{0}}^{2 π} \frac{\tan θ + a^{2}}{\cos^{2} θ} d θ

= \frac{1}{2} \int_{θ_{0}}^{2 π} (\tan θ + a^{2}) d (\tan θ)

= \frac{1}{2} {[\frac{\tan^{2} θ}{2} + a^{2} \tan θ]}_{θ_{0}}^{2 π}

= \frac{1}{2} [- \frac{a^{4}}{2} + a^{4}] = \frac{a^{4}}{4}

l e t \tan θ_{1} = m

A_{1} = \frac{1}{2} {[\frac{\tan^{2} θ}{2} + a^{2} \tan θ]}_{θ_{1}}^{2 π}

= \frac{1}{2} [- \frac{\tan^{2} θ_{1}}{2} - a^{2} \tan θ_{1}]

= - \frac{1}{2} [\frac{m^{2}}{2} + a^{2} m] = \frac{A}{2} = \frac{a^{4}}{8}

\Rightarrow 2 m^{2} + 4 a^{2} m + a^{4} = 0

\Rightarrow m = \frac{- 4 a^{2} \pm \sqrt{16 a^{4} - 8 a^{4}}}{4}

= {\begin{cases} - \frac{2 + \sqrt{2}}{2} a^{2} \\ - \frac{2 - \sqrt{2}}{2} a^{2} \end{cases}

s i n c e ∣ θ_{1} ∣<∣ θ_{0} ∣\Rightarrow∣ m ∣=∣ \tan θ_{1} ∣<∣ \tan θ_{0} ∣= a^{2}

t h e s u i t a b l e s o l u t i o n i s :

m = - \frac{2 - \sqrt{2}}{2} a^{2} \approx - 0.293 a^{2}

Commented by ajfour last updated on 10/Sep/18

T h a n k s S i r, i g o t t h e s a m e a n s w e r .

Answered by ajfour last updated on 10/Sep/18

I n t e r s e c t i o n p o i n t s o f l i n e

a n d c u r v e :

y = m x = x (x^{2} - a^{2})

\Rightarrow O (0, 0) a n d P (\sqrt{a^{2} + m}, y_{P})

A = - \int_{0}^{a} (x^{3} - a^{2} x) d x = - (\frac{a^{4}}{4} - \frac{a^{4}}{2})

\Rightarrow A = \frac{a^{4}}{4}

\int_{0}^{\sqrt{a^{2} + m}} (m x - x^{3} + a^{2} x) d x = \frac{a^{4}}{8} (= \frac{A}{2})

\Rightarrow \frac{m (a^{2} + m)}{2} - \frac{{(a^{2} + m)}^{2}}{4} + \frac{a^{2} (a^{2} + m)}{2} = \frac{a^{4}}{8}

\Rightarrow {(a^{2} + m)}^{2} = \frac{a^{4}}{2}

\Rightarrow m = (\frac{1}{\sqrt{2}} - 1) a^{2}

s a m e a s m = - (\frac{2 - \sqrt{2}}{2}) a^{2} .

[\frac{d y}{d x} ∣_{x = 0} = - a^{2}; s o w e r e j e c t

m = - (1 + \frac{1}{\sqrt{2}}) a^{2}] .

Answered by tanmay.chaudhury50@gmail.com last updated on 10/Sep/18

s o l v i n g m x = x (x^{2} - a^{2})

x {m - (x^{2} - a^{2}} = 0

x = 0

x^{2} - a^{2} = m

x = \pm \sqrt{a^{2} + m}

y = m x i n t e r s e c t t h e c u r v e a t {- \sqrt{a^{2} + m}, 0, \sqrt{a^{2} + m}}

t h e c u r v e y = x (x^{2} - a^{2}) i n t e r s e c t x a x i s a t {- a, 0, a}

n o w a r e a o f d o t s h a d e d + l i n e s h a d e d =

\int_{0}^{a} x (x^{2} - a^{2}) d x

= \int_{0}^{a} x^{3} - x^{} a^{2} d x

=∣ \frac{x^{4}}{4} - \frac{x^{2}}{2} \times a^{2} ∣_{0}^{a}

= \frac{a^{4}}{4} - \frac{a^{4}}{2}

t a k i n g t h e m o d o f v a l u e o f a r e a = \frac{a^{4}}{4}

s o d o t s h a d e d = \frac{a^{4}}{8}

l i n e s h a d e d = \frac{a^{4}}{8}

a r e a l i n e s h a d e d

\int_{0}^{\sqrt{a^{2} + m}} m x d x + \int_{\sqrt{a^{2} + m}}^{a} x (x^{2} - a^{2}) d x

=∣ \frac{{m x}^{2}}{2} ∣_{0}^{\sqrt{a^{2} + m}} + ∣ \frac{x^{4}}{4} - \frac{x^{2} a^{2}}{2} ∣_{\sqrt{a^{2} + m}}^{a}

= \frac{m (a^{2} + m)}{2} + {(\frac{a^{4}}{4} - \frac{a^{4}}{2}) - (\frac{a^{4} + 2 a^{2} m + m^{2}}{4} - \frac{a^{2} (a^{2} + m)}{2}}

= \frac{a^{2} m + m^{2}}{2} + {(\frac{- a^{4}}{4}) - (\frac{a^{4} + 2 a^{2} m + m^{2} - 2 a^{4} - 2 a^{2} m}{4})}

= \frac{a^{2} m + m^{2}}{2} + {\frac{- a^{4} - a^{4} - 2 a^{2} m - m^{2} + 2 a^{4} + 2 a^{2} m}{4}}

= \frac{a^{2} m + m^{2}}{2} - \frac{m^{2}}{4}

= \frac{2 a^{2} m + 2 m^{2} - m^{2}}{4}

= \frac{2 a^{2} m + m^{2}}{4}

\frac{2 a^{2} m + m^{2}}{4} = \frac{a^{4}}{8}

4 a^{2} m + 2 m^{2} - a^{4} = 0

2 m^{2} + 4 a^{2} m - a^{4} = 0

m = \frac{- 4 a^{2} \pm \sqrt{16 a^{4} + 8 a^{4}}}{4}

= \frac{- 4 a^{2} \pm 2 \sqrt{6} a^{2}}{4} = \frac{- 2 a^{2} \pm \sqrt{6} a^{2}}{2}

a s p e r f i g u r e m i s n e g d t i v e

m = \frac{- a^{2} (2 + \sqrt{6})}{2}

p l s c h e c k

Commented by ajfour last updated on 10/Sep/18

p l e a s e c h e c k, s o m e l i t t l e e r r o r, S i r .

Commented by tanmay.chaudhury50@gmail.com last updated on 10/Sep/18

i h a v e d o n e

\int_{0}^{\sqrt{a^{2} + m}} m x + \int_{\sqrt{a^{2} + m}}^{a} x (x^{2} - a^{2}) d x t o f i n d t h e l i n e s h a d e d

a r e a

s u m o f t h e v a l u e o f t h i s t w o i n t r e g a l = \frac{a^{4}}{8}

b u t y o u h a v e d o n e

\int_{0}^{\sqrt{a^{2} + m}} m x - x (x^{2} - a^{2}) d x = \frac{a^{4}}{8}

p l s s a y w h i c h a r e a b e l o n g s t o t h i s i n t r e g a l

Commented by ajfour last updated on 10/Sep/18

t h e d o t s h a d e d a r e a i s e q u a l t o

w h a t i h a v e i n t e g r a t e d .

Commented by tanmay.chaudhury50@gmail.com last updated on 10/Sep/18

i d o n o t a g r e e \dots \int_{0}^{\sqrt{a^{2} + m}} m x d x i s t h e a r e a o f t r i a n g l e

f o r m e d b y y = m x a n d x a x i s a n d s t l i n e x = \sqrt{a^{2} + m}

\int_{0}^{\sqrt{a^{2} + m}} x (x^{2} - a^{2}) d x

i s t h e a r e a f o r m e d b y x a x i s a n d c u r v e x (x^{2} - a^{2}) a n d s t

l i n e x = \sqrt{a^{2} + m} [

Commented by tanmay.chaudhury50@gmail.com last updated on 10/Sep/18

p l s f i n d l i n e s h a d e d a r e a \dots

Commented by ajfour last updated on 10/Sep/18

h o w w i l l y o u f i n d a r e a o f

{(x - h)}^{2} + {(y - k)}^{2} = r^{2}, S i r ?

Commented by tanmay.chaudhury50@gmail.com last updated on 10/Sep/18

y o u f i n d e r r o r i n m y c a l c u l a t i o n \dots l o c a t e i t

Commented by ajfour last updated on 10/Sep/18

\int_{0}^{\sqrt{a^{2} + m}} m x d x + \int_{\sqrt{a^{2} + m}}^{a} x (x^{2} - a^{2}) d x = - \frac{a^{4}}{8}

(a r e a i s b e l o w x - a x i s s o - v e)

t h e n y o u s h a l l g e t

4 a^{2} m + 2 m^{2} + a^{4} = 0

\Rightarrow 2 m^{2} + 4 a^{2} m + a^{4} = 0

m = \frac{- 4 a^{2} \pm \sqrt{16 a^{4} - 8 a^{4}}}{4}

o r m = (- 1 + \frac{1}{\sqrt{2}}) a^{2}

(n o t a c c e p t i n g t h e o t h e r r o o t) .

Commented by tanmay.chaudhury50@gmail.com last updated on 11/Sep/18

o k t h a n k y o u \dots

Answered by MJS last updated on 10/Sep/18

p : y = x^{3} - a^{2} x

l : y = m x

p \cap l

x^{3} - (m + a^{2}) x = 0 \Rightarrow x = \sqrt{m + a^{2}}

dotted area : D = \underset{0}{\int^{\sqrt{m + a^{2}}}} (x^{3} - a^{2} x) d x - m \underset{0}{\int^{\sqrt{m + a^{2}}}} x d x

shaded area : S = m \underset{0}{\int^{\sqrt{m + a^{2}}}} x d x + \underset{\sqrt{m + a^{2}}}{\int^{a}} (x^{3} - a^{2} x) d x

D = \frac{1}{4} (m^{2} - a^{4}) - \frac{1}{2} m (m + a^{2}) = - \frac{1}{4} {(m + a^{2})}^{2}

S = \frac{1}{2} m (m + a^{2}) - \frac{1}{4} m^{2} = \frac{1}{4} m (m + 2 a^{2})

S - D = 0

m^{2} + 2 a^{2} m + \frac{a^{4}}{2} = 0

m = (- 1 + \frac{\sqrt{2}}{2}) a^{2} [the 2^{nd} solution {doesn}^{'} t

give a real value for \sqrt{m + a^{2}}]

Facebook Tweet Pin