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# Vector Calculus: Indefinite Integral of a Vector

Posted on 2011-09-20
Medium Priority
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r''(t) = -t^(-2) i, t > 0
r'(1) = i + j
r(1) = j

Find r(t).

My question is, how do I find C when I take the indefinite integrals?
So far i have:
r'(t) = t^(-1)i + C ==> C = j
&
r(t) = ln(t)i + tj + C ==> C = 0k but I'm not sure if it's right, or how to check if I am right. Any help is (as always) much appreciated.
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Question by:Zenoture
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LVL 84

Assisted Solution

ozo earned 664 total points
ID: 36571694
adjust C to make r(1) = j
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Author Comment

ID: 36571752
Why?
If r(t) = ln(t)i + tj + C, then
r(1) = j = ln(1)i + (1)j + C
j = 0i + j + C
j = j + C
0 = C

But that just means we can ignore C doesn't it?
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LVL 84

Expert Comment

ID: 36571770
Do you usually ignore parameters when their value is 0?
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Author Comment

ID: 36571774
Yes
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Assisted Solution

BigRat earned 668 total points
ID: 36572875
All I can say is tsk,tsk. When one plays with differential equations, each integration introduces a new constant. I always call then A,B,C,... so as to avoid confusion.

So the first integration introduces A which turns out to be 1. And the second integration introduces B which turns out to be 0.

Furthermore if we are dealing with three dimensions i,j and k, then the first integration introduces A and B, and the second C and D. This keeps things clean and avoids confusion.
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LVL 18

Accepted Solution

deighton earned 668 total points
ID: 36573133
r'(t) = (1/t) i + Ai + Bj

r'(1) = i + j
hence A = 0 B = 1

so

r'(t) = (1/t) i  + j

r(t) = log(t) i + tj + Ci + Dj

r(1) = j

so C and D are zero

r(t) = log(t)i + tj

0

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