Find second derivative of ( r * dr/dt * d^2r/dt^2 )

R

R.K.4.7

New member
Joined
Apr 20, 2017
Messages
21
Please Help me

Find :-
d^2/dt^2 [ r * dr/dt * d^2r/dt^2 ]

i have bolded the changes for an easier review

d/dx (u * v) = ( u * dv/dx ) + ( v * du/dx )
I dont know what happens when there are 3 variable, i tried doing 1st & the 2nd tern as U & V ,
then uv=u, 3rd term = v
but the answer does not match!!

taking the first & second terms
u=r , v=dr/dt
r*d^2r/dt^2 + dr/dt*dr/dt ........eq 1


eq1 * d^2r/dt^2
u = eq1 , v = d^2r/dt^2
[ r*d^2r/dt^2 + dr/dt*dr/dt ] * [ d^3r/dt^3 ] + d(u)/dt * d^2r/dt^2


[ r*d^2r/dt^2 + (dr/dt)^2 ] * [ d^3r/dt^3 ] + d(u)/dt * d^2r/dt^2 .........eq2


now, to find the d(u)/dt,
d(u)/dt = d ( r*d^2r/dt^2 + (dr/dt)^2 ) / dt
d(u)/dt = r*d^3r/dt^3 + dr/dt*d^2r/dt^2 + 2(dr/dt)*(dr/dt)


putting the value of d(u)/dt in......eq2
[ r*d^2r/dt^2 + (dr/dt)^2 ] * [ d^3r/dt^3 ] + ( r*d^3r/dt^3 + dr/dt*d^2r/dt^2 + 2(dr/dt)*(dr/dt) ) * d^2r/dt^2

[ r*d^2r/dt^2 + (dr/dt)^2 ] * [ d^3r/dt^3 ] + ( r*d^3r/dt^3 + dr/dt*d^2r/dt^2 + 2(dr/dt)^2 ) * d^2r/dt^2

im so pissed at this question bro, thi s is just the single derivative, now i have to find the second derivative ! wtf man ! Also the solution given in the book is just done in 1 or 2 lines !! :(
Is the method correct ?
Thanks Guys !
 
Last edited:
R.K.4.7 said:
Find :-
d^2/dt^2 [ r * dr/dt * d^2r/dt^2 ]

i have bolded the changes for an easier review
Click to expand...
"Changes" from what? Changes from what the book gave you?

R.K.4.7 said:
d/dx (u * v) = ( u * dv/dx ) + ( v * du/dx )
I dont know what happens when there are 3 variable...
Click to expand...
What are the "three variables" you're seeing? I'm only seeing r, which is treated as a function of t. (The dr/dt is, of course, the first derivative of r with respect to t.)

R.K.4.7 said:
i tried doing 1st & the 2nd tern as U & V ,
then uv=u, 3rd term = v
but the answer does not match!!
Click to expand...
What "answer" does not match what?

R.K.4.7 said:
taking the first & second terms
u=r , v=dr/dt
r*d^2r/dt^2 + dr/dt*dr/dt ........eq 1
Click to expand...
These are not "terms". They're factors. Also, what are you doing with these factors? How did you get what you posted? By what reasoning?

Instead, try working with the Product Rule and the products they've given you, remembering that the first derivative of a first derivative is a second derivative, the first derivative of a second derivative is a third derivative, and so forth. So the first step would start as:

. . .\(\displaystyle \dfrac{d}{dt}\, \left(r\, \cdot\, \dfrac{dr}{dt}\, \cdot\, \dfrac{d^2r}{dt^2}\right)\)

. . . . .\(\displaystyle =\, \left(\dfrac{d}{dt}\, r\right)\, \cdot\, \left(\dfrac{dr}{dt}\, \cdot\, \dfrac{d^2r}{dt^2}\right) \)

. . . . . . .\(\displaystyle +\, (r)\, \cdot\, \left(\dfrac{d}{dt}\, \dfrac{dr}{dt}\right)\, \cdot\, \left(\dfrac{d^2r}{dt^2}\right)\,\)

. . . . . . . . .\(\displaystyle +\, \left(r\, \cdot\, \dfrac{dr}{dt}\right)\, \cdot\, \left(\dfrac{d}{dt}\, \dfrac{d^2r}{dt^2}\right)\)

. . . . .\(\displaystyle =\, \left(\dfrac{dr}{dt}\right)\, \cdot\, \left(\dfrac{dr}{dt}\, \cdot\, \dfrac{d^2r}{dt^2}\right) \)

. . . . . . .\(\displaystyle +\, (r)\, \cdot\, \left(\dfrac{d^2r}{dt^2}\right)\, \cdot\, \left(\dfrac{d^2r}{dt^2}\right)\,\)

. . . . . . . . .\(\displaystyle +\, \left(r\, \cdot\, \dfrac{dr}{dt}\right)\, \cdot\, \left(\dfrac{d^3r}{dt^3}\right)\)

...and so forth. ;)
 
stapel said:
"Changes" from what? Changes from what the book gave you?


What are the "three variables" you're seeing? I'm only seeing r, which is treated as a function of t. (The dr/dt is, of course, the first derivative of r with respect to t.)


What "answer" does not match what?


These are not "terms". They're factors. Also, what are you doing with these factors? How did you get what you posted? By what reasoning?

Instead, try working with the Product Rule and the products they've given you, remembering that the first derivative of a first derivative is a second derivative, the first derivative of a second derivative is a third derivative, and so forth. So the first step would start as:

. . .\(\displaystyle \dfrac{d}{dt}\, \left(r\, \cdot\, \dfrac{dr}{dt}\, \cdot\, \dfrac{d^2r}{dt^2}\right)\)

. . . . .\(\displaystyle =\, \left(\dfrac{d}{dt}\, r\right)\, \cdot\, \left(\dfrac{dr}{dt}\, \cdot\, \dfrac{d^2r}{dt^2}\right) \)

. . . . . . .\(\displaystyle +\, (r)\, \cdot\, \left(\dfrac{d}{dt}\, \dfrac{dr}{dt}\right)\, \cdot\, \left(\dfrac{d^2r}{dt^2}\right)\,\)

. . . . . . . . .\(\displaystyle +\, \left(r\, \cdot\, \dfrac{dr}{dt}\right)\, \cdot\, \left(\dfrac{d}{dt}\, \dfrac{d^2r}{dt^2}\right)\)

. . . . .\(\displaystyle =\, \left(\dfrac{dr}{dt}\right)\, \cdot\, \left(\dfrac{dr}{dt}\, \cdot\, \dfrac{d^2r}{dt^2}\right) \)

. . . . . . .\(\displaystyle +\, (r)\, \cdot\, \left(\dfrac{d^2r}{dt^2}\right)\, \cdot\, \left(\dfrac{d^2r}{dt^2}\right)\,\)

. . . . . . . . .\(\displaystyle +\, \left(r\, \cdot\, \dfrac{dr}{dt}\right)\, \cdot\, \left(\dfrac{d^3r}{dt^3}\right)\)

...and so forth. ;)
Click to expand...


You make it look so easy man! Now i get it ! Thanks

So then what you did is the first derivative.
Let me try the second derivative...

= ( d^2r/dt^2 ) * ( dr/dt ) * ( d^2r/dt^2 )
+ ( dr/dt ) * ( d^2r/dt^2 ) * ( d^2r/dt^2 )
+ ( dr/dt ) * ( dr/dt ) * ( d^3r/dt^3 )
+
+ ( dr/dt ) * ( d^2r/dt^2 ) * ( d^2r/dt^2 )
+ ( r ) * ( d^3r/dt^3 ) * ( d^2r/dt^2 )
+ ( r ) * ( d^2r/dt^2 ) * ( d^3r/dt^3 )
+
+ ( dr/dt ) * ( dr/dt ) * ( d^3r/dt^3 )
+ ( r ) * ( d^2r/dt^2 ) * ( d^3r/dt^3 )
+ ( r ) * ( dr/dt ) * ( d^4r/dt^4 )

How does it look ?
 
Last edited:
R.K.4.7 said:
You make it look so easy man! Now i get it ! Thanks

So then what you did is the first derivative.
Let me try the second derivative...

= ( d^2r/dt^2 ) * ( dr/dt ) * ( d^2r/dt^2 )
+ ( dr/dt ) * ( d^2r/dt^2 ) * ( d^2r/dt^2 )
+ ( dr/dt ) * ( dr/dt ) * ( d^3r/dt^3 )
+
+ ( dr/dt ) * ( d^2r/dt^2 ) * ( d^2r/dt^2 )
+ ( r ) * ( d^3r/dt^3 ) * ( d^2r/dt^2 )
+ ( r ) * ( d^2r/dt^2 ) * ( d^3r/dt^3 )
+
+ ( dr/dt ) * ( dr/dt ) * ( d^3r/dt^3 )
+ ( r ) * ( d^2r/dt^2 ) * ( d^3r/dt^3 )
+ ( r ) * ( dr/dt ) * ( d^4r/dt^4 )

How does it look ?
Click to expand...
How does this compare with "the answer" (that they gave you?) which you'd referenced earlier? ;)
 
You must log in or register to reply here.
Top