anonymous
  • anonymous
please help, explanation would be greatly appreciated (picture included
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
baru
  • baru
what we need to know is velocity=rate of change of position with respect to time so, V=dx/dt
anonymous
  • anonymous
so we find the derivatives of both particles

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baru
  • baru
so to find \(v_1 ~and ~v_2\) differentiate \(x_1~and~x_2\) with respect to t
baru
  • baru
yep
anonymous
  • anonymous
yes one moment
anonymous
  • anonymous
so would x=cos2t x'=0?
baru
  • baru
no, \[\frac{d(cosax)}{dx}=-asin(ax)\]
anonymous
  • anonymous
shoot, so -2sin(2t)
baru
  • baru
yep
anonymous
  • anonymous
and the next one we would use the d/dx a^u= (lna)(a^u)(u') ?
baru
  • baru
yes :)
anonymous
  • anonymous
so (lne)(e^((t-3)/2))([quotient rule] ((2)(1)-(t-3)(0))/4)
anonymous
  • anonymous
(1/2)(e^((t-3)/2))
baru
  • baru
you shouldnt use the quotient rule here, you should use it when there are "variables" in the numerator and denominator
anonymous
  • anonymous
isn't t a variable though?
baru
  • baru
\[x=e^{\frac{t-3}{2}}-0.75\]
baru
  • baru
yes, 't' is only in the numerator, the denominator is a constant
baru
  • baru
\[dx/dt=e^{\frac{t-3}{2}}\times \]
baru
  • baru
ahh...ignore that
baru
  • baru
\[dx/dt=e^{\frac{t-3}{2}} \times \frac{d (\frac{t-3}{2})}{dt}\]
anonymous
  • anonymous
that is very complex...
baru
  • baru
no..i've just used your rule
baru
  • baru
ln(a)(a^u)u'
anonymous
  • anonymous
from what i understand then that would be e^((t-3)/2) * (2/t)?
anonymous
  • anonymous
the u' is messing me up, i'm not sure how to find it
baru
  • baru
\[u=\frac{t-3}{2}\\u=\frac{t}{2}-3/2\\u'=\frac{1}{2}\]
anonymous
  • anonymous
so i see we split up the numerator, but i don't understand how we get 1/2 I am so sorry
baru
  • baru
i'm using formulas of differentiation derivative of a constant=0 so the (-3/2) becomes zero and \[\frac{d(ax)}{dx}=a\]
anonymous
  • anonymous
alright so the answer would be e^((t-3)/2)*(1/2)
baru
  • baru
yes
baru
  • baru
now we have v1 and v2
baru
  • baru
the question asks us to find when the two velocities are the same, so v1=v2
baru
  • baru
\[-2\sin2t=\frac{1}{2}e^{\frac{t-3}{2}}\]
baru
  • baru
@CadenN
anonymous
  • anonymous
i dont know how to do that
anonymous
  • anonymous
we multiply both sides by 2
baru
  • baru
this cannot be done using algebra the question says (calculator),j so you should use a calculator
baru
  • baru
u have such a calculator?
anonymous
  • anonymous
i have a graphing calcualtor
baru
  • baru
yes, use that to plot v1 and v2, see where the graphs intersect
anonymous
  • anonymous
do we just start pluging in values from between 0 and 2pi
baru
  • baru
no, graph the functions
anonymous
  • anonymous
THAT IS SO SO SIMPLE
anonymous
  • anonymous
WHY DONT I THINK OF THESE THINGS????
anonymous
  • anonymous
i am SOO sorry
baru
  • baru
haha, i see you have a lot of enthusiasm for mathematics, don't worry, i knew what do only because i've done the several times already :)
anonymous
  • anonymous
thank you so much for helping me i'm so clueless sometimes
anonymous
  • anonymous
my calcualtor shows three points of intersection
baru
  • baru
well, then the answer is 3 :)
anonymous
  • anonymous
Thank you SO much
baru
  • baru
but, does it actually tell you what the points are? because the question says "between t=0 and t=2pi"
anonymous
  • anonymous
well it tells me the intersections on a x- y scale
baru
  • baru
so x=t? in that case u should consider points x<2pi
anonymous
  • anonymous
2pi = 6.283
anonymous
  • anonymous
so x has to be < 6.283 right?
baru
  • baru
yep
anonymous
  • anonymous
all the point have x coordinates below 6
baru
  • baru
then go with three :)
anonymous
  • anonymous
thank you
baru
  • baru
hey, can you check your graphs again, i seem to getting 4 points
baru
  • baru
1 Attachment
anonymous
  • anonymous
ooohhhh this proves i should always zoom in
baru
  • baru
lol
baru
  • baru
https://www.desmos.com/calculator this is the website i used to graph that, you might find it useful later on :)
anonymous
  • anonymous
this will definitely come in handy, thank you @baru

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