Vectors, path of a particle

**a8281333** · October 27th 2010, 12:23 PM

vector r of a moving particle has the equation with respect to the origin of:

r(t) = (r[0]/t^2[0])(t^2i - 2tt[0]j + (t^2 + 2tt[0])k) where r[0] and t[0] are constants

show that the particle goes through the point P(4,-4,8)r[0]. at what time does it do this?

what I did is i subbed (4,-4,8) into the equation but it didnt make anything clear to me, anyone any advice on how to tackle this problem?

Anyone know how id go about this, i need to find this part to complete parts b) and c) of which i know how to do which is the frustrating part!

i got the following:

r(4,-4,8)= (r[0]/t[0])*(16i/t[0] + 8j + (64/t[0] +16)k)?

**skeeter** · October 27th 2010, 01:41 PM

let's use $a = r_0$ and $b = t_0$ as the given constants to minimize the confusion of the notation.

if this is $r(t)$ ...

$\displaystyle r(t) = \frac{a}{b^2}\left[t^2 \vec{i} - 2bt \vec{j} + (t^2 + 2bt) \vec{k} \right]$

show that the particle goes through the point P(4,-4,8)r[0].

$\displaystyle \frac{a}{b^2}t^2 = 4a<br />$

$t^2 - 4b^2 = 0$

$(t - 2b)(t + 2b) = 0$

$\displaystyle -\frac{a}{b^2}(2bt) = -4a$

$t = 2b$

$\displaystyle \frac{a}{b^2}(t^2 + 2bt) = 8a$

$t^2 + 2bt - 8b^2 = 0$

$(t + 4b)(t - 2b) = 0$

common solution for all three equations is $t = 2b$

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