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 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: 2,426 Thanks: 1312 $r(t) = (\ln(t+1), 2\sqrt{t+4}, t^2 - 1)$ 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. Thanks from greg1313 and topsquark
 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
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Quote:
 Originally Posted by aneka no idea to be honest, i have no idea about this... i just hope what i sent is at least a good start
How about finding v(t) and a(t)? Do you know how to do this part?

-Dan

 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
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Quote:
 Originally Posted by aneka nop, i do not know anything about it.. i have been studying for other stuff, and then i need to know that now..
do you know what a derivative is?

what class is this for?

 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
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Quote:
 Originally Posted by aneka i tried to use web calculators... but it does not show step by step..
$\displaystyle r(t) = (ln(t+1),~2 \sqrt{t + 4},~t^2 - 1)$

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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