Pls-help-r-2-r-C-where-r-t-is-a-function-and-C-is-a-constant-

Question Number 59361 by Andrew Foxman last updated on 08/May/19

P l s h e l p

r^{2} r^{″} = C w h e r e r (t) i s a f u n c t i o n a n d

C i s a c o n s t a n t

Commented by kaivan.ahmadi last updated on 09/May/19

\frac{d^{2} r}{{d t}^{2}} = \frac{C}{r^{2}} \Rightarrow \frac{d r}{d t} = \int \frac{C}{r^{2}} d t = \frac{C}{r^{2}} t + C_{1} \Rightarrow

r = \frac{{C t}^{2}}{r^{2}} + C_{1} t + C_{2} \Rightarrow {C t}^{2} + (C_{1} t + C_{2}) r^{2} - r^{3} = 0

Commented by mr W last updated on 09/May/19

i t i s w r o n g s i r!

r i n \int \frac{C}{r^{2}} d t i s a f u n c t i o n o f t, i . e . r = r (t),

y o u c a n n o t t r e a t r (t) a s a c o n s t a n t,

\int \frac{C}{r^{2}} d t \neq \frac{C}{r^{2}} t + C_{1}

Commented by maxmathsup by imad last updated on 09/May/19

r i s a f u n c t i o n o f v a r i a b l e t \dots .

Commented by Andrew Foxman last updated on 09/May/19

T h e m i s t e r y o f t h i s e q u a t i o n r e m a i n s \dots

A n y o n e c a n g i v e t h e i r a n s w e r n o w a n d h e r e \dots

I f y o u a r e a h e r o i n m a t h r e s o l v e t h i s e q u a t i o n \dots

Commented by MJS last updated on 10/May/19

I think r (t) can only be given implicitly

Commented by MJS last updated on 10/May/19

this is the answer from a friend who owns a

calculator . I cannot explain it, {it}^{'} s just the

output of that machine :

\frac{C \ln \frac{1}{r (t)}}{\sqrt{B^{3}}} + \frac{r (t) \sqrt{\frac{B r (t) - 2 C}{r (t)}}}{B} - \frac{2 C \ln (\sqrt{\frac{B r (t) - 2 C}{r (t)}} - \sqrt{B})}{\sqrt{B^{3}}} = t + A

with A, B, C \in R

it can be further simplyfied when r (t) > 0 \land b > 0

2 C \ln (\sqrt{B r (t) - 2 C} - \sqrt{B r (t)}) - \sqrt{B r (t)) B r (t) - 2 C} = - \sqrt{B^{3}} (t + A)

Answered by alex041103 last updated on 10/May/19

{L e t}^{'} s t r y r (t) = - {A t}^{α}

\Rightarrow r^{2} = A^{2} t^{2 α}

\Rightarrow r^{″} = - A α (α - 1) t^{α - 2}

\Rightarrow r^{2} r^{″} = - A^{3} α (α - 1) t^{3 α - 2} = C

\Rightarrow {\begin{cases} 3 α - 2 = 0 \\ A^{3} α (1 - α) = C \end{cases}

\Rightarrow α = \frac{2}{3}

\Rightarrow A^{3} \frac{2}{3} (1 - \frac{2}{3}) = \frac{2}{9} A^{3} = C \Rightarrow A^{3} = \frac{9}{2} C

\Rightarrow A = \sqrt[3]{\frac{9 C}{2}}

\Rightarrow O n e s o l u t i o n i s r (t) = - \sqrt[3]{\frac{9 C t^{2}}{2}}

Answered by aleks041103 last updated on 07/Sep/20

O b s e r v e :

r^{″} = \frac{d}{d t} (\frac{d r}{d t}) = \frac{d v}{d t} = \frac{d v}{d r} \frac{d r}{d t} = v \frac{d v}{d r}, w h e r e v = \frac{d r}{d t}

T h e n :

v d v = \frac{C d r}{r^{2}} = - C d (\frac{1}{r}) = - d (\frac{C}{r})

v d v = \frac{1}{2} d (v^{2}) = d (\frac{v^{2}}{2})

\Rightarrow d (\frac{v^{2}}{2}) = - d (\frac{C}{r})

d (\frac{v^{2}}{2} + \frac{C}{r}) = 0 \Rightarrow \frac{v^{2}}{2} = \frac{A}{2} - \frac{C}{r}

w h e r e A i s a c o n s t a n t

\Rightarrow r^{'} = v = \sqrt{A - \frac{2 C}{r}}

\Rightarrow \frac{d r}{\sqrt{A - \frac{2 C}{r}}} = d t

\Rightarrow \int \frac{d r}{\sqrt{A - \frac{2 C}{r}}} = \int d t = t + B, B = c o n s t .

u = \sqrt{r} \Rightarrow r = u^{2} a n d d r = 2 u d u

\int \frac{d r}{\sqrt{A - \frac{2 C}{r}}} = \int \frac{\sqrt{r}}{\sqrt{A r - 2 C}} d r =

= \int \frac{u 2 u d u}{\sqrt{{A u}^{2} - 2 C}} = \frac{1}{\sqrt{A}} \int u \frac{2 u d u}{\sqrt{u^{2} - \frac{2 C}{A}}} =

= \frac{1}{\sqrt{A}} \int u \frac{d (u^{2} - 2 C / A)}{\sqrt{u^{2} - 2 C / A}} =

= \frac{2}{\sqrt{A}} \int u d (\sqrt{u^{2} - \frac{2 C}{A}})

I B P :

\frac{2}{\sqrt{A}} \int u d (\sqrt{u^{2} - \frac{2 C}{A}}) =

= \frac{2}{\sqrt{A}} (u \sqrt{u^{2} - \frac{2 C}{A}} - \int \sqrt{u^{2} - \frac{2 C}{A}} d u) =

= \sqrt{\frac{8 C}{A^{2}}} (u \sqrt{{(\sqrt{\frac{A}{2 C}} u)}^{2} - 1} - \int \sqrt{{(\sqrt{\frac{A}{2 C}} u)}^{2} - 1} d u)

\int \sqrt{{(\sqrt{\frac{A}{2 C}} u)}^{2} - 1} d u = (w = \sqrt{\frac{A}{2 C}} u)

= \sqrt{\frac{2 C}{A}} \int \sqrt{w^{2} - 1} d w

w = s e c θ a n d d w = s e c θ t a n θ d θ

\int \sqrt{w^{2} - 1} d w = \int \frac{{s i n}^{2} θ}{{c o s}^{3} θ} d θ =

= \int \frac{s i n θ}{{c o s}^{3} θ} s i n θ d θ = \frac{1}{2} \int s i n θ d (\frac{1}{{c o s}^{2} θ})

I B P :

\int \sqrt{w^{2} - 1} d w = \frac{1}{2} (\frac{s i n θ}{{c o s}^{2} θ} - \int \frac{d θ}{c o s θ}) =

= \frac{1}{2} (\frac{s i n θ}{{c o s}^{2} θ} - l n (s e c θ + t a n θ)) =

= \frac{1}{2} (t a n θ s e c θ - l n (s e c θ + t a n θ)) =

= \frac{1}{2} (w \sqrt{w^{2} - 1} - l n (w + \sqrt{w^{2} - 1}))

N o w a l l i n t e g r a l s h a v e b e e n c a l c u l a t e d!

W e n e e d t o j u s t g o b a c k a n d p l u g

e v e r y s u b s t i t u t i o n s o w e g e t t h e f i n a l

r e s u l t

\dots