Question

Successive differentiation!

Answer 1

If \(\LARGE p^2=a^2 \cos^2 \theta+b^2 \sin^2 \theta\)
Show that :
\[\LARGE p + \frac{d^2p}{d \theta^2}= \frac{a^2b^2}{p^3}\]

Answer 2

I found the first derivative.
\[\Large 2p \frac{dp}{d \theta}= -2a^2 \cos \theta \sin \theta+2 b ^2 \sin \theta \cos \theta\]

Answer 3

\[\Large p \frac{dp}{d \theta}=\sin \theta \cos \theta(b^2-a^2)\]

Answer 4

now what?

Answer 5

use double angle identity so that it becomes cos(2theta) when u differentiate again

Answer 6

yeah I tried that but it almost gives the same result.

Answer 7

Lets differentiate it again.
\[\Large (\frac{dp}{d \theta})^2 + p \frac{d^2p}{d \theta^2}=(b^2-a^2)(\cos^2 \theta-\sin^2 \theta)\]

Answer 8

We can put it as cos 2 theta but still not proceed-able further.

Answer 9

replace dp/dtheta from your 3rd comment

Answer 10

upon p?

Answer 11

s^2c^2 /p^2 (b^2-a^2)^2

see whether you can use, s^2+c^2=1

Answer 12

yes, divided by p

Answer 13

\[\Large (\frac{\sin \theta \cos \theta(b^2-a^2)}{p})^2 + p \frac{d^2p}{d \theta^2}=(b^2-a^2)(\cos^2 \theta-\sin^2 \theta)\]

Answer 14

hmm..that just complicated it...also using p^2 from the question, would not help much

Answer 15

true!

Answer 16

(b^2-a^2)(c^2-s^2)
can be simplified

Answer 17

and p^2 term will come in there...

Answer 18

what do you mean :O

Answer 19

=p^2+b^2c^2+a^2s^2

Answer 20

oh I see..so now?

Answer 21

i am just throwing random ideas at you :P

Answer 22

.......

Answer 23

seems much like an algebraic work...isolate d^2p/dtheta^2
and work on other side

Answer 24

too scared to deal with all those big terms :P

Answer 25

maybe some step that simplifies everything?

Answer 26

follow this thread
http://math.stackexchange.com/questions/907124/if-p2-a2-cos2-thetab2-sin2-theta-show-that-p-p-fraca2b

Answer 27

I don't get it.......

Answer 28

\[\large\begin{align}\\

p^2 &= a^2\cos^2\theta + b^2\sin^2\theta = \frac{1}{2}[a^2+b^2 -(b^2-a^2)\cos2\theta] ~~~\color{Red}{\star}\\~\\
2pp' &= (b^2-a^2)\sin 2\theta \\~\\
p'^2+pp'' &= (b^2-a^2)\cos 2\theta \\~\\
&=a^2+b^2-2p^2~~~ \color{gray}{(~from ~} \color{Red}{\star}~\color{gray}{)} \\~\\


\end{align}\]

Answer 29

\[\begin{align}\\

p'^2 + pp'' &= a^2+b^2-2p^2 \\~\\
p(p+p'') &= a^2 + b^2 - p^2 - p'^2 \\~\\
&=a^2\sin^2\theta + b^2\cos^2\theta - p'^2 \\~\\
&=a^2\sin^2\theta + b^2\cos^2\theta - \dfrac{(b^2-a^2)^2\sin^22\theta}{4p^2} \\~\\
&= \dfrac{4p^2(a^2\sin^2\theta + b^2\cos^2\theta) -(b^2-a^2)^2\sin^22\theta}{4p^2} \\~\\
&= \dfrac{4( a^2\cos^2\theta + b^2\sin^2\theta )(a^2\sin^2\theta + b^2\cos^2\theta) -(b^2-a^2)\sin^22\theta}{4p^2} \\~\\
&= \dfrac{4a^2b^2(\sin^4\theta + \cos^4\theta)+ (a^2+b^2)\sin^22\theta-(b^2-a^2)^2\sin^22\theta}{4p^2} \\~\\
&= \dfrac{4a^2b^2(\sin^4\theta + \cos^4\theta)+2a^2b^2\sin^22\theta}{4p^2} \\~\\
&= \dfrac{4a^2b^2(\sin^2\theta + \cos^2\theta)^2}{4p^2} \\~\\
&= \dfrac{a^2b^2}{p^2} \\~\\

\end{align}\]

Answer 30

that looks lengthy >.<

Answer 31

i'll just..quit this question :P