December 13th, 2016, 02:15 PM  #1 
Newbie Joined: Dec 2016 From: Hungary Posts: 12 Thanks: 0  Find angles between vectors
WHAT IS THIS =(? SOMEONE HELP ME.. step by step i need to understand pls t0 = 0 equation r(t) =[ln(t+1)*2 root x+4, t square 1] thanks guys Last edited by greg1313; December 13th, 2016 at 04:14 PM. 
December 13th, 2016, 02:23 PM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,531 Thanks: 1390 
$r(t) = (\ln(t+1), 2\sqrt{t+4}, t^2  1)$ I assume that $x$ in there is a typo and should be a $t$ no? $v(t) = \dfrac{d}{dt} r(t)$ $a(t) = \dfrac{d}{dt} v(t)$ $\theta = \arccos\left(\dfrac{v(t) \cdot a(t)}{v(t)a(t)}\right)$ I leave the gory details to you. 
December 13th, 2016, 02:46 PM  #3 
Newbie Joined: Dec 2016 From: Hungary Posts: 12 Thanks: 0 
no idea to be honest, i have no idea about this... i just hope what i sent is at least a good start

December 13th, 2016, 02:50 PM  #4 
Math Team Joined: May 2013 From: The Astral plane Posts: 2,258 Thanks: 929 Math Focus: Wibbly wobbly timeywimey stuff.  
December 13th, 2016, 03:09 PM  #5 
Newbie Joined: Dec 2016 From: Hungary Posts: 12 Thanks: 0 
nop, i do not know anything about it.. i have been studying for other stuff, and then i need to know that now..

December 13th, 2016, 03:09 PM  #6 
Newbie Joined: Dec 2016 From: Hungary Posts: 12 Thanks: 0 
i tried to use web calculators... but it does not show step by step..

December 13th, 2016, 03:20 PM  #7 
Senior Member Joined: Sep 2015 From: USA Posts: 2,531 Thanks: 1390  
December 13th, 2016, 03:22 PM  #8 
Newbie Joined: Dec 2016 From: Hungary Posts: 12 Thanks: 0 
calculus, i know.. but i am not the best O.o!

December 13th, 2016, 03:24 PM  #9  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,258 Thanks: 929 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
This is in vector notation with $\displaystyle r_x = ln(t + 1),~r_y = 2 \sqrt{t + 4},~r_z = t^2  1$, so when we take the time derivative we take the time derivative of all three components. $\displaystyle v(t) = \left ( \frac{1}{t + 1},~(t + 4)^{1/2},~ 2t \right )$ You try a(t). Dan  
December 13th, 2016, 03:42 PM  #10 
Newbie Joined: Dec 2016 From: Hungary Posts: 12 Thanks: 0 
so... a(t) = (1/(t+1)^2 , 1/2(x+4)^3/2 , 2 ) ??? ??


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angles, exercise, find, impossible, vectors 
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