anybody any good at differentiating?

timi_h · #1 10-12-2006, 05:59 PM

i want to differentiate X=[2(v^2)/g]sin(a)cos(a) with respect to a so i can prove that 45 degrees is the angle of maximum trajectory but a whole summer off has given me some severe brainrot! its quite embarassing!! anybody a genius at integrals? cmon andre your studying physics!!

ibanezcollector · #2 10-12-2006, 06:25 PM

www.casio.com

LOL sounds scientific..

My brain is beyond rotted for that stuff.

Jem#5150 · #3 10-12-2006, 08:32 PM

i think my brain just sh** it's pants.

..........

yup, there's ooze coming out of my ears, SEE WHAT YOU DID TO ME?!?!?

sorry i'm no help, but are you in college?

Vai7Or7Die · #4 10-12-2006, 08:39 PM

Oh, dude you are a TOTAL newb if you can't figure that one out! Just take the square root 45 and multiply by the cosine of the perpendicular angle of trajectory, add a cross product of the measures of an isoceles triangle's base and leg angles, and divide by 3.0000005456453257897632874320403043. Duh.

jk bro sorry I couldn't help

**jaxadam** · #5 10-12-2006, 08:50 PM

Remember the product rule?

This is a pretty easy fuction to take the derivative of, considering you just have a constant multiplied by the product of two trigonometric functions.

Generic product rule for differentiation is:

f(x) = sin(x)*cos(x)

therefore

f'(x) = d/dx [(sin(x)] * cos(x) + sin(x) * d/dx [cos(x)]

we know that

d/dx [sin(x)] = cos(x)

and d/dx [cos(x)] = -sin(x)

so you essentially have (let c = 2*v^2/g)

f(a) = c*sin(a)*cos(a)

and

f'(a) = c*[cos^2(a) - sin^2(a)]

or f'(a) = 2*v^2/g[cos^2(a)-sin^2(a)]

Now remember your identities here?

cos^2(x) - sin^2(x) = cos(2x)

so know you have a simplified version of the derivative in the form of

f'(a) = 2*v^2/g*cos(2a)

Hope that helps a little! That's about all I can remember right now, but I'm sure there is some significance with the 2cos(2a) part. It's a double angle something or another... I'm gonna go look it up... I'll get back to you when I think about this problem a little more.

**jaxadam** · #6 10-12-2006, 08:54 PM

Quote:

Originally Posted by timi_h

i want to differentiate X=[2(v^2)/g]sin(a)cos(a) with respect to a so i can prove that 45 degrees is the angle of maximum trajectory but a whole summer off has given me some severe brainrot! its quite embarassing!! anybody a genius at integrals? cmon andre your studying physics!!

Hmmm, I just read my post over again, and I skipped lots of steps, so if you got lost, just ask. Plus, this is not integration.

smc · #7 10-12-2006, 08:56 PM

Good work man! I was scratching my head because i differentiated cosx to be sinx and not -sinx...doh, rusty myself.

Just to finish it, if you set the last equation equal to 0 to find the max/min point, 45 is the value of a that will equal 0 (cos90=0).

**jaxadam** · #8 10-12-2006, 09:03 PM

Quote:

Originally Posted by smc

Just to finish it, if you set the last equation equal to 0 to find the max/min point, 45 is the value of a that will equal 0 (cos90=0).

Yup! There you go... I forgot all about the 45 degree angle by the time I was working myself through that one. I'm so used to just dealing with theoretical things, I'm sitting here scratching MY head about the double angle thing still.

Just wait until you get to complex analysis. Trig functions go bye-bye (well, kinda).

Just remember, cos(x) + i*sin(x) = e^ix, and e^i*pi = 1.

It all gets so wonderful. Especially when you get to convolution integrals and residue integration.

StoneLord · #9 10-13-2006, 12:53 AM

You better keep it down, or the guys in the nearby "Shamrock v Ortiz 3" thread will bust in and steal your lunch money and break your pocket protectors.

sorry

Lefty Robb · #10 10-13-2006, 03:44 AM

So easy, see?

timi_h · #11 10-13-2006, 06:49 AM

darn product rule! been over a year since i had to use that hence the forgetting! turns out the bit i had remembered was the top half of the quotient rule so i had a - sign in there which screwed it all up big time!
cheers for the help guys! lets hope this assignment freshens my brain up a bit!!

andy7jem · #12 10-13-2006, 11:36 AM

Quote:

Originally Posted by StoneLord

You better keep it down, or the guys in the nearby "Shamrock v Ortiz 3" thread will bust in and steal your lunch money and break your pocket protectors.

sorry

**jaxadam** · #13 10-13-2006, 12:28 PM

Quote:

Originally Posted by Lefty Robb

So easy, see?

The scary part is, I understand some of this stuff!

Some of that looks like quantum, some looks like E&M.

There doesn't seem to appear to be any interrelationship between these equations, they are all random, because you've got a matrix element representation of one function at the top, but a ket vector notation of a time-dependant waveform in the middle, and what looks like a quantum potential at the bottom, hence the i h-bar, where I'm assuming h-bar in this case is Planck's constant.

Where did you get these?

timi_h · #14 10-13-2006, 12:36 PM

h bar is the modified planck constant isnt it? i seem to recall from somehere (god knows where cos ive never been taught it) that its plancks constant over 2 pi
and theres definitely some electromagnetism in the one with the e^2 on top cos it has 1/4pi epsilon zero aswell as the charge of an electron/proton in the equation

**jaxadam** · #15 10-13-2006, 12:53 PM

Quote:

Originally Posted by timi_h

h bar is the modified planck constant isnt it? i seem to recall from somehere (god knows where cos ive never been taught it) that its plancks constant over 2 pi
and theres definitely some electromagnetism in the one with the e^2 on top cos it has 1/4pi epsilon zero aswell as the charge of an electron/proton in the equation

Yeah, pretty much. It's all the same (kinda). I just always knew either one of them as Planck's constant. They are the same thing, but just different units.

It's like talking about frequency as omega, or f. f is just omega/2*pi.