Question

Find r(t) given the following:

r''(t)=cos(2t) i + sin(2t) j
r'(0)=2i-1/2j
r(0)=3/4i+j

I know that I must integrate r''(t) twice, but is the r'(0) the constant value or what?

Answer 1

yes. integrating once gives:

r'(t) = (1/2)(sin(2t)i - cos(2t)j) + C

taking r'(0) = 2i - (1/2)j = (1/2)(sin(0)i - cos(0)j) + C

C = 2i.

therefore r'(t) = (1/2)(sin(2t)i - cos(2t)j) + 2i

doing this again and solving for C again will find r(t)

Answer 2

is it not the constant value, but it is used to find the constant value

Answer 3

Ok cool! Thanks!

Answer 4

Although I'm still not sure why is C=2i when cos(0)=1

So wouldn't r'(0)=0i+2i-(1/2)j-(1/2)j?

Answer 5

check again ^_^

C = 0i + 2i - (1/2)j + (1/2)J

Answer 6

I'm still not sure I get it. If you simply add the given r'(0) to the found r'(0), then it is sin0-cos0, so it is (1/2)(0-1), so therefore it is (-1/2)+(-1/2)

Answer 7

2i - (1/2)j = (1/2)(sin(0)i - cos(0)j) + C

keep C on the right and move everything to the left.

2i - (1/2)j = (1/2)(0 - j) + C
2i - (1/2)j - (-1/2)j = C

Answer 8

Awesome! I didn't understand exactly what that operation was until you wrote it out. Thanks again!