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December 13th, 2016, 02:15 PM   #1
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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.
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December 13th, 2016, 02:23 PM   #2
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$r(t) = (\ln(t+1), 2\sqrt{t+4}, t^2 - 1)$

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.
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December 13th, 2016, 02:46 PM   #3
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no idea to be honest, i have no idea about this... i just hope what i sent is at least a good start
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December 13th, 2016, 02:50 PM   #4
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Originally Posted by aneka View Post
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
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December 13th, 2016, 03:09 PM   #5
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nop, i do not know anything about it.. i have been studying for other stuff, and then i need to know that now..
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December 13th, 2016, 03:09 PM   #6
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i tried to use web calculators... but it does not show step by step..
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December 13th, 2016, 03:20 PM   #7
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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?
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December 13th, 2016, 03:22 PM   #8
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calculus, i know.. but i am not the best O.o!
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December 13th, 2016, 03:24 PM   #9
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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
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December 13th, 2016, 03:42 PM   #10
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so... a(t) = (-1/(t+1)^2 , -1/2(x+4)^3/2 , 2 ) ??? ??
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