Initial Value w/ Vector-Valued Function

[opus]

Homework 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:
$\vec r(t) = \left(\cos(t) + t-1\right)\hat i + \left(3\sin(2t)+10\right)\hat j + \left(150-4.9t^2\right)\hat k$

 

[opus]

Here is my handwritten work if that would help too.

 

[LCKurtz]

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]

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]

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]

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]

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]

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]

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

 

[LCKurtz]

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

I don't see it either. This will let @Greg Bernhardt know.

 

[Greg Bernhardt]

I don't see it either. This will let @Greg Bernhardt know.

I don’t have near future plans on bringing it back

 

[opus]

Ok just wanted to make sure. Thanks all!