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anybody any good at differentiating?

3K views 21 replies 13 participants last post by  caprile 
#1 ·
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!! :p
 
#4 ·
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.
:lol:
jk bro sorry I couldn't help
 
#5 ·
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.
 
#7 ·
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).
 
#8 ·
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.
 
#13 ·
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?
 
#11 ·
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!!
 
#14 · (Edited)
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
 
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