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Initial Value w/ Vector-Valued Function

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opus

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Problem Statement
Solve the initial value problem for the vector function ##\vec r(t)##.
Relevant Equations
##\frac{d^2\vec r}{dt^2} = -\cos(t)\hat i - 12\sin(2t)\hat j - 9.8\hat k## ;
##\vec r(0) = 10\hat j + 150\hat k##
##\vec r'(0) = \hat i + 6\hat j##
From ##\vec r''(t)## we integrate to get
$$\vec r'(t) = \left(-\sin(t)+C_1\right)\hat i + \left(6\cos(2t)+C_2\right)\hat j - \left(9.8t+C_3\right)\hat k$$

Solving for the C constants using ##\vec r'(0) = 1\hat i + 6\hat j + 0\hat k##,
##\vec r'(0) = <C_1, C_2, C_3>##
##=<1, 6, 0>##

So we now have $$\vec r'(t) = \left(-\sin(t)+1\right)\hat i + \left(6\cos(2t)+6\right)\hat j - \left(9.8t\right)\hat k$$

Using the same process one more time,

##\vec r(t) = \left(\cos(t) + t + C_4\right)\hat i + \left(3\sin(2t) + 6t + C_5\right)\hat j - \left(4.9t^2 +C_6\right)\hat k##

Solving for the constants like before, and using the initial values given,
##\vec r(0) = < C_4, C_5, C_6 >##
##= <0, 10, 150>##

And we can now state the position vector as:
$$\vec r(t) = \left(\cos(t)+t\right)\hat i + \left(3sin(2t)+6t+10\right)\hat j + \left(-4.9t^2+150\right)\hat k$$

Would someone mind pointing me to where I have made a mistake? This doesn't match the given solution key which is:
##\vec r(t) = \left(\cos(t) + t-1\right)\hat i + \left(3\sin(2t)+10\right)\hat j + \left(150-4.9t^2\right)\hat k##
 

LCKurtz

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From ##\vec r''(t)## we integrate to get
$$\vec r'(t) = \left(-\sin(t)+C_1\right)\hat i + \left(6\cos(2t)+C_2\right)\hat j - \left(9.8t+C_3\right)\hat k$$

Solving for the C constants using ##\vec r'(0) = 1\hat i + 6\hat j + 0\hat k##,
##\vec r'(0) = <C_1, C_2, C_3>##
##=<1, 6, 0>##

So we now have $$\vec r'(t) = \left(-\sin(t)+1\right)\hat i + \left(6\cos(2t)+6\right)\hat j - \left(9.8t\right)\hat k$$
##C_2## should be ##0##. I didn't check further.
 

opus

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Thank you for the reply. What's the reasoning?
We have ##\vec r'(0) = 1\hat i + 6\hat j + 0\hat k##
So wouldn't it follow that ##\vec r'(0) = <C_1, C_2, C_3> = <1, 6, 0>## and therefore C2 would be 6?
 

LCKurtz

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Thank you for the reply. What's the reasoning?
We have ##\vec r'(0) = 1\hat i + 6\hat j + 0\hat k##
So wouldn't it follow that ##\vec r'(0) = <C_1, C_2, C_3> = <1, 6, 0>## and therefore C2 would be 6?
Put ##t=0## in your ##\vec r'(t)## and see what you get.
 

opus

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I get ##\vec r'(0) = 1\hat i + 12\hat j + 0\hat k## which does not match the given initial conditions. I wonder if my integration was wrong?
 

LCKurtz

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It is because your value for ##C_2## should be zero as I originally told you. Your solution is$$
\vec r'(t) = \langle -\sin t, 6\cos(2t), -9.8t \rangle + \langle C_1,C_2, C_3 \rangle $$so when you put ##t=0## your initial condition becomes$$
\vec r'(0) = \langle 0, 6, 0 \rangle + \langle C_1,C_2, C_3 \rangle = \langle 1,6,0\rangle $$giving ##C_1=1,~C_2=0,~C_3=0##.
 

opus

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Oh ok I see what you mean. Sorry I was a little confused there. From the book and class I don't yet really grasp what's going on with the initial conditions with these, but they way you wrote it there by adding the constant vector separately makes more sense.
Thanks!
 

opus

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Side question: Is there a "mark solved" button in this new PF layout? Can't seem to find it.
 

opus

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Ok just wanted to make sure. Thanks all!
 

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