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opus

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- 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##

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

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##