K Kabeio k. Joined Jan 11, 2005 Messages 557 Gender Male HSC 2005 Oct 3, 2005 #1 Hw do you find the second derivative of the following: 10 sin pi(t)/2 Thanks,
haboozin Do you uhh.. Yahoo? Joined Aug 3, 2004 Messages 707 Gender Male HSC 2005 Oct 3, 2005 #2 Kabeio said: Hw do you find the second derivative of the following: 10 sin pi(t)/2 Thanks, Click to expand... f' = 5pi cos pit/2 f'' = -5/2pi^2 sinpit/2 f''(t) = -pi/2^2f(x)
Kabeio said: Hw do you find the second derivative of the following: 10 sin pi(t)/2 Thanks, Click to expand... f' = 5pi cos pit/2 f'' = -5/2pi^2 sinpit/2 f''(t) = -pi/2^2f(x)
K Kabeio k. Joined Jan 11, 2005 Messages 557 Gender Male HSC 2005 Oct 3, 2005 #3 i dont get how you go from f'' to f''(t) ?
rnitya_25 Abhishek's Rani.. Joined Mar 19, 2005 Messages 1,577 Location Mars Gender Female HSC 2005 Oct 3, 2005 #4 because i don't think your (t) was meant to mean that you need to subtitute t in....the 2nd line is right i think
because i don't think your (t) was meant to mean that you need to subtitute t in....the 2nd line is right i think
S speed2 Member Joined Sep 28, 2004 Messages 209 Location ? Gender Male HSC 2005 Oct 3, 2005 #5 f(t) = 10 sin [П(t)/2] f'(t) = 5П cos [П(t)/2] f''(t) = -5/2.П^2.{(1/10).10sin[П(t)/2] since f(t)=10 sin [П(t)/2] f''(t) = -5/2.П^2.{(1/10).f(t)] f''(t) = -П/4^2f(t) Last edited: Oct 3, 2005
f(t) = 10 sin [П(t)/2] f'(t) = 5П cos [П(t)/2] f''(t) = -5/2.П^2.{(1/10).10sin[П(t)/2] since f(t)=10 sin [П(t)/2] f''(t) = -5/2.П^2.{(1/10).f(t)] f''(t) = -П/4^2f(t)
