Question

Given that the acceleration vector is a(t)=(-4cos(-2t))i + (-4sin(-2t))j + (4t)k, the initial velocity is v(0)=(i + k) , and the initial position vector is r(0)= i + j + k , compute
v(t)=?
r(t)=?

Answer 1

so you know that acceleration is your second derivative of your position/vector "whatchamacall it" ... you can say r(t) correct?

Answer 2

So you must integrate a(t) in which you'll have to integrate each component by itself due to it being a vector

Answer 3

i have integrated already i got v(t)=(-2sin(2t)+1)i +(2cos(2t))j + (2t^2+1)k

Answer 4

but j is wrong

Answer 5

+C

Answer 6

they said j was wrong? let me check to see what is up

Answer 7

the integration of sin will give you a -cos

Answer 8

so before doing any integration pull the -4 out and then make u=-2t

Answer 9

so i got 2cos(-2t)

Answer 10

still wrong

Answer 11

\[-4\int\limits_{}^{}\sin(-2t)dtj\]

let u= -2t
du = -2dt
du/-2=dt

if you pull that -2 out you get

\[2\int\limits_{}^{}\sin(u)du\]

\[2(-\cos(u) + C)\]

Answer 12

did they get -2cos(-2t)

Answer 13

still wrong

Answer 14

what do they have?

Answer 15

its my hw question i cant figure it out

Answer 16

oh well thats not the answer yet lol that's just j... you said that the integration was wrong

Answer 17

after the integration you get + C correct so you need to find your constant vector

Answer 18

C = to v(0)

Answer 19

and it 0 according to v(0)

Answer 20

so after integration of your acceleration vector did you get

\[2\sin(-2t)i-2\cos(-2t)j+2t^2k\]

Answer 21

yea i got that

Answer 22

and only j was wrong when i plug in the v(0)

Answer 23

alright so when you plug in v(0) what did you get before adding like terms

Answer 24

or before adding and subtracting... your

you should get i+k=0-2j+0+c

Answer 25

so then it is -2??

Answer 26

meaning that c=i+2j+k

Answer 27

so it should be +2

Answer 28

k i got it right now

Answer 29

but how did u know to do that or why did u do that??

Answer 30

yeah when you put in v(0) you are basically trying to find c

Answer 31

so for r(t) the i part do i have to do the same thing?

Answer 32

yep

Answer 33

because when you integrate what you have above you'll get another constant vector

Answer 34

so will it bt +3

Answer 35

be not bt..

Answer 36

b/c r(0)= 1,1,1 right??

Answer 37

i'm not sure i haven't integrated that far... these equations aare always sooo long lol hold on

Answer 38

cos(2t)+t+1 i -sin(2t)+2t+1 j + 2/3(t^3)+t+1 k

Answer 39

i got this and only got i wrong

Answer 40

what did you get for your integration without using r(0)

Answer 41

cos(2t)+t i - sin(2t)+2t j + 2/3(t^3) + t k

Answer 42

you mean (-2t) for all your trig ops right/

Answer 43

yea

Answer 44

so if you do r(0)=i+j+k you should get

i+j+k=i+0j+0k+c

minus the i to the other side to get c by itself
should get j+k =c

Answer 45

what r(0) is basically saying is that when your t =0, your answer will be i+j+k

Answer 46

y is j and k 0?

Answer 47

so if you put 0 into the equation that you have above cos(-2t) you get 1

Answer 48

(-sin(-2(0))+0)j = 0

sin(0) is 0

Answer 49

so your c will be j+k

if you put that back into the equation you'll get

\[[\cos(-2t)+t]i+[-\sin(-2t)+2t+1]j+[\frac{2}{3}t^3+t+1]\]

Answer 50

so for i i should put cos(-2t)+t?

Answer 51

k

Answer 52

yep the above should be the equation of the position vector

Answer 53

k i got it right thanks a lot for your help i appreciate it

Answer 54

when solving this problem you go through these steps

1) Integrate to get to the velocity or position vector function (DO NOT FORGET that integrating an indefinite integral produces a +C or a constant vector)
2)Use the given info r(0) of v(0) to find the constant(r(0)=i+j+k, it just means that when t=0, your equation equals i+j+k)
3)once finding c by doing the above put it into your r(t) or v(t)

Answer 55

k i will

Answer 56

yep no problem