# Initial Value w/ Vector-Valued Function

• opus
In summary, the position vector of a particle at time t is given by <1, 6, 0>, where <C_1, C_2, C_3> are the three unknown constants.
opus
Gold Member
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##

Here is my handwritten work if that would help too.

#### Attachments

• Screen Shot 2019-04-09 at 9.32.13 PM.png
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opus said:
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?

opus said:
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?

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!

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

opus said:
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.

LCKurtz said:
I don't see it either. This will let @Greg Bernhardt know.
I don’t have near future plans on bringing it back

Ok just wanted to make sure. Thanks all!

## What is an initial value with a vector-valued function?

An initial value with a vector-valued function is a specific point on the graph of the function that is used as a starting point for calculations or analysis. It is usually denoted as (t0, x0) where t0 is the initial time and x0 is the initial value of the vector function at that time.

## Why is an initial value important in vector-valued functions?

An initial value is important because it allows us to determine the behavior of the vector function at a specific point. It serves as a reference point for understanding the overall behavior of the function and can help us make predictions about its future values.

## How do you find the initial value of a vector-valued function?

The initial value of a vector-valued function can be found by evaluating the function at the initial time t0. This will give us the initial value of the vector function, denoted as x0. Alternatively, the initial value can also be given as a separate set of coordinates (t0, x0).

## Can the initial value of a vector-valued function change?

Yes, the initial value of a vector-valued function can change if the initial time t0 is changed. This will result in a different point on the graph of the function being used as the starting point for calculations or analysis.

## What is the significance of the initial value in differential equations?

In differential equations, the initial value is used to determine the particular solution of the equation. It serves as the starting point for solving the equation and helps us find the behavior of the function at a specific point in time.

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