Question

please help, explanation would be greatly appreciated (picture included

Answer 1

what we need to know is
velocity=rate of change of position with respect to time

so, V=dx/dt

Answer 2

so we find the derivatives of both particles

Answer 3

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

Answer 4

yep

Answer 5

yes one moment

Answer 6

so would x=cos2t x'=0?

Answer 7

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

Answer 8

shoot, so -2sin(2t)

Answer 9

yep

Answer 10

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

Answer 11

yes :)

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

isn't t a variable though?

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

ahh...ignore that

Answer 20

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

Answer 21

that is very complex...

Answer 22

no..i've just used your rule

Answer 23

ln(a)(a^u)u'

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

i'm using formulas of differentiation
derivative of a constant=0 so the (-3/2) becomes zero

and
\[\frac{d(ax)}{dx}=a\]

Answer 29

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

Answer 30

yes

Answer 31

now we have v1 and v2

Answer 32

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

Answer 33

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

Answer 34

@CadenN

Answer 35

i dont know how to do that

Answer 36

we multiply both sides by 2

Answer 37

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

Answer 38

u have such a calculator?

Answer 39

i have a graphing calcualtor

Answer 40

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

Answer 41

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

Answer 42

no, graph the functions

Answer 43

THAT IS SO SO SIMPLE

Answer 44

WHY DONT I THINK OF THESE THINGS????

Answer 45

i am SOO sorry

Answer 46

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 :)

Answer 47

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

Answer 48

my calcualtor shows three points of intersection

Answer 49

well, then the answer is 3 :)

Answer 50

Thank you SO much

Answer 51

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

Answer 52

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

Answer 53

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

Answer 54

2pi = 6.283

Answer 55

so x has to be < 6.283 right?

Answer 56

yep

Answer 57

all the point have x coordinates below 6

Answer 58

then go with three :)

Answer 59

thank you

Answer 60

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

Answer 61

ooohhhh this proves i should always zoom in

Answer 62

lol

Answer 63

https://www.desmos.com/calculator

this is the website i used to graph that, you might find it useful later on :)

Answer 64

this will definitely come in handy, thank you @baru