please help, explanation would be greatly appreciated (picture included

- anonymous

please help, explanation would be greatly appreciated (picture included

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- anonymous

##### 1 Attachment

- baru

what we need to know is
velocity=rate of change of position with respect to time
so, V=dx/dt

- anonymous

so we find the derivatives of both particles

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## More answers

- baru

so to find \(v_1 ~and ~v_2\)
differentiate \(x_1~and~x_2\) with respect to t

- baru

yep

- anonymous

yes one moment

- anonymous

so would x=cos2t x'=0?

- baru

no,
\[\frac{d(cosax)}{dx}=-asin(ax)\]

- anonymous

shoot, so -2sin(2t)

- baru

yep

- anonymous

and the next one we would use the d/dx a^u= (lna)(a^u)(u') ?

- baru

yes :)

- anonymous

so (lne)(e^((t-3)/2))([quotient rule] ((2)(1)-(t-3)(0))/4)

- anonymous

(1/2)(e^((t-3)/2))

- baru

you shouldnt use the quotient rule here, you should use it when there are "variables" in the numerator and denominator

- anonymous

isn't t a variable though?

- baru

\[x=e^{\frac{t-3}{2}}-0.75\]

- baru

yes, 't' is only in the numerator, the denominator is a constant

- baru

\[dx/dt=e^{\frac{t-3}{2}}\times \]

- baru

ahh...ignore that

- baru

\[dx/dt=e^{\frac{t-3}{2}} \times \frac{d (\frac{t-3}{2})}{dt}\]

- anonymous

that is very complex...

- baru

no..i've just used your rule

- baru

ln(a)(a^u)u'

- anonymous

from what i understand then that would be e^((t-3)/2) * (2/t)?

- anonymous

the u' is messing me up, i'm not sure how to find it

- baru

\[u=\frac{t-3}{2}\\u=\frac{t}{2}-3/2\\u'=\frac{1}{2}\]

- anonymous

so i see we split up the numerator, but i don't understand how we get 1/2 I am so sorry

- 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

alright so the answer would be e^((t-3)/2)*(1/2)

- baru

yes

- baru

now we have v1 and v2

- baru

the question asks us to find when the two velocities are the same,
so v1=v2

- baru

\[-2\sin2t=\frac{1}{2}e^{\frac{t-3}{2}}\]

- baru

@CadenN

- anonymous

i dont know how to do that

- anonymous

we multiply both sides by 2

- baru

this cannot be done using algebra
the question says (calculator),j so you should use a calculator

- baru

u have such a calculator?

- anonymous

i have a graphing calcualtor

- baru

yes, use that to plot v1 and v2, see where the graphs intersect

- anonymous

do we just start pluging in values from between 0 and 2pi

- baru

no, graph the functions

- anonymous

THAT IS SO SO SIMPLE

- anonymous

WHY DONT I THINK OF THESE THINGS????

- anonymous

i am SOO sorry

- 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

thank you so much for helping me i'm so clueless sometimes

- anonymous

my calcualtor shows three points of intersection

- baru

well, then the answer is 3 :)

- anonymous

Thank you SO much

- baru

but, does it actually tell you what the points are?
because the question says "between t=0 and t=2pi"

- anonymous

well it tells me the intersections on a x- y scale

- baru

so x=t?
in that case u should consider points x<2pi

- anonymous

2pi = 6.283

- anonymous

so x has to be < 6.283 right?

- baru

yep

- anonymous

all the point have x coordinates below 6

- baru

then go with three :)

- anonymous

thank you

- baru

hey, can you check your graphs again, i seem to getting 4 points

- baru

##### 1 Attachment

- anonymous

ooohhhh this proves i should always zoom in

- baru

lol

- baru

https://www.desmos.com/calculator
this is the website i used to graph that, you might find it useful later on :)

- anonymous

this will definitely come in handy, thank you @baru

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