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Calculating Vx and Vy given V, X, Y

haruspex

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I tried to do that and I ended up with an equation of the sort of du/dt * C = 0, it was a veeeeery long equation in the beginning but all sines/cosines etc got cancelled in one way or another.
Where C is v/(sqrt(c+2)) if I recall correctly (I dont have the papers infront of me).
Then you made mistakes, but if you don't post your working I have no idea where.
Try just differentiating your first equation in post #23 with respect to time and post what you get.
 
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vx = du/dt * cos(c * u(t)) - u(t) * c * du/dt * sin(c * u(t))
vy = du/dt * sin(c * u(t)) + u(t) * c * du/dt * cos(c * u(t))
vz = du/dt
Then when I square them I am using the (a+-b)^2 formula to unwrap them for vx and vy
vx^2 = (du/dt)^2 * cos^2(c * u(t)) - 2*du/dt*cos(c * u(t))*u(t)*c*du/dt*sin(c * u(t)) + u(t)^2 * c^2 * (du/dt)^2 * sin^2(c * u(t))
vy^2 = (du/dt)^2 * sin^2(c * u(t)) + 2*du/dt*sin(c * u(t))*u(t)*c*du/dt*cos(c * u(t)) + u(t)^2 * c^2 * (du/dt)^2 * cos^2(c * u(t))
vz^2 = (du/dt)^2
The -2 and +2 terms in vx and vy cancel
v^2 = (du/dt)^2 * cos^2(c*u(t)) + u(t)^2 * c^2 * (du/dt)^2 * sin^2(c*u(t)) + (du/dt)^2*sin^2(c*u(t)) + u(t)^2*c^2*(du/dt)^2*cos^2(c*u(t)) + (du/dt)^2
All terms have (du/dt)^2, so im pulling it out:
v^2 = (du/dt)^2 * (cos^2(c*u(t)) + u(t)^2 * c^2 * sin^2(c*u(t)) + sin^2(c*u(t)) + u(t)^2*c^2*cos^2(c*u(t)) + 1)
cos^2(a) + sin^2(a) = 1, so we are down a sin and a cos now:
v^2 = (du/dt)^2 * (u(t)^2 * c^2 * sin^2(c * u(t)) + u(t)^2 * c^2 * cos^2(c * u(t)) + 2)
Take out u(t) and c and we have another sin^2 cos^2 identity:
v^2 = (du/dt)^2 * u(t)^2 * c^2 *(3)
(du/dt)^2 * u(t)^2 - v^2/(3*c^2) = 0
v^2/(3*c^2) = C
(du/dt)^2 * u(t)^2 - C = 0

Ok, a different result this time, can you double check if this is correct? And if it is, how do I actually use it? I've solved differential equations in my math classes in the past, but they never explained what they actually are and how to use them. If I remember correctly DEs give solutions in the form of functions, so from here I will find u(t) and use that back in my original equations?
 

haruspex

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To make it more readable I have converted your work to Latex.
##v_x = \dot u \cos(c u) - u c \dot u \sin(c u)##
##v_y = \dot u \sin(c u) + u c \dot u \cos(c u)##
##v_z = \dot u##
Then when I square them I am using the ##(a+-b)^2## formula to unwrap them for ##v_x## and ##v_y##
##v_x^2 = \dot u^2 \cos^2(c u) - 2 \dot u \cos(c u) u c \dot u \sin(c u) + u^2 c^2 \dot u^2 \sin^2(c u)##
##v_y^2 = \dot u^2 \sin^2(c u) + 2 \dot u \sin(c u) u c \dot u \cos(c u) + u^2 c^2 \dot u^2 \cos^2(c u)##
##v_z^2 = \dot u^2##
The -2 and +2 terms in ##v_x## and ##v_y## cancel
##v^2 = \dot u^2 \cos^2(c u) + u^2 c^2 \dot u^2 \sin^2(c u) + \dot u^2 \sin^2(c u) + u^2 c^2 \dot u^2 \cos^2(c u) + \dot u^2##
All terms have ##\dot u^2##, so im pulling it out:
##v^2 = \dot u^2 (\cos^2(c u) + u^2 c^2 \sin^2(c u) + \sin^2(c u) + u^2 c^2 \cos^2(c u) + 1)##
##\cos^2(a) + \sin^2(a) = 1##, so we are down a ##\sin## and a ## \cos## now:
##v^2 = \dot u^2 (u^2 c^2 \sin^2(c u) + u^2 c^2 \cos^2(c u) + 2)##
Take out u and c and we have another ##\sin^2 +\cos^2## identity:
##v^2 = \dot u^2 u^2 c^2 (3)##
Fine until that last step. The last term was 2, not ##2u^2 c^2 ##
 
22
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Okay, still, what do I do with it?
To simplify it a bit:
v^2 = (du/dt)^2 * u(t)^2 * c^2 *(3)
v = du/dt * u(t) * c * sqrt(3)
du/dt*u(t) - v/(c*sqrt(3)) = 0

P.S. How do I activate LaTeX?
 

haruspex

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Okay, still, what do I do with it?
To simplify it a bit:
v^2 = (du/dt)^2 * u(t)^2 * c^2 *(3)
v = du/dt * u(t) * c * sqrt(3)
du/dt*u(t) - v/(c*sqrt(3)) = 0

P.S. How do I activate LaTeX?
No, you misunderstand. I am saying you made a mistake. The last line of what I quoted in post #28 is wrong.
 

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