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 March 21st, 2019, 03:26 PM #1 Senior Member   Joined: Oct 2015 From: Greece Posts: 137 Thanks: 8 How to integrate a vector function? I have this $\displaystyle \vec{v} = v_0 \cdot \vec{i} - gt \cdot \vec{j}$ and I want to calculate this: $\displaystyle \Delta{x} = \int_{t_o}^{t} \vec{v} \cdot dt$ But I never saw an integral that has a vector inside it: $\displaystyle \Delta{x} = \int_{t_o}^{t} \vec{v} \cdot dt \Leftrightarrow \Delta{x} = \int_{t_o}^{t} (v_0 \cdot \vec{i} - gt \cdot \vec{j}) \cdot dt$ Someone told me that I just need to integrate the magnitude of the vector $\displaystyle \Delta{x} = \int_{t_o}^{t} ||\vec{v}|| \cdot dt \Leftrightarrow \Delta{x} = \int_{t_o}^{t} \sqrt{v_0^2 + g^2t^2} \cdot dt$ But why is that?
 March 21st, 2019, 03:27 PM #2 Senior Member   Joined: Aug 2012 Posts: 2,409 Thanks: 753 We are mathematicians and we do not put arrow hats on our vectors! If we say, "Let $v$ be a vector," that's good enough. And remember, the scalars may be regarded as vectors themselves. So there!! Just wanted to get that off my chest.
 March 21st, 2019, 03:36 PM #3 Global Moderator   Joined: May 2007 Posts: 6,834 Thanks: 733 If the integral was that of the magnitude you would have | | bracketing the vector. If you integrate the components of the vector you will get a vector for the answer.
 March 21st, 2019, 03:39 PM #4 Senior Member   Joined: Oct 2015 From: Greece Posts: 137 Thanks: 8 Guys, I didn't understand lol
March 21st, 2019, 04:42 PM   #5
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Quote:
 Originally Posted by babaliaris Guys, I didn't understand lol
I'm using hats! Thpppppttt!

If you have the integral:
$\displaystyle \int (a(x) \hat{i} + b(x) \hat{j} )~dx$

then you integrate component by component:
$\displaystyle \int (a(x) \hat{i} + b(x) \hat{j} )~dx = \int a(x) \hat{i}~dx + \int b(x) \hat{j}~dx = \left ( \int a(x)~dx \right ) ~ \hat{i} + \left ( \int b(x)~dx \right ) ~ \hat{j}$

The unit vectors act just like constants. (If they don't depend on x, that is. Be cautious when using cylindrical or spherical polar coordinates.)

Notice that your integral in the OP will be a vector.

Note: Please do not use the notation $\displaystyle \vec{v} \cdot dt$. It makes it look like you are using a dot product, which of course you can't here, as dt is a not a vector.

-Dan

Last edited by topsquark; March 21st, 2019 at 04:45 PM.

March 21st, 2019, 05:05 PM   #6
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Quote:
 Originally Posted by topsquark I'm using hats! Thpppppttt! If you have the integral: $\displaystyle \int (a(x) \hat{i} + b(x) \hat{j} )~dx$ then you integrate component by component: $\displaystyle \int (a(x) \hat{i} + b(x) \hat{j} )~dx = \int a(x) \hat{i}~dx + \int b(x) \hat{j}~dx = \left ( \int a(x)~dx \right ) ~ \hat{i} + \left ( \int b(x)~dx \right ) ~ \hat{j}$ The unit vectors act just like constants. (If they don't depend on x, that is. Be cautious when using cylindrical or spherical polar coordinates.) Notice that your integral in the OP will be a vector. Note: Please do not use the notation $\displaystyle \vec{v} \cdot dt$. It makes it look like you are using a dot product, which of course you can't here, as dt is a not a vector. -Dan
So, the integration of the magnitude of v in my OP, is going to give a function that gives only a magnitude too?

 March 21st, 2019, 05:15 PM #7 Senior Member   Joined: Oct 2015 From: Greece Posts: 137 Thanks: 8 $\displaystyle \Delta{x} = i\int_{t_0}^{t} v_0dt - j\int_{t_0}^{t} gt dt = 0i - g(t-t_0)j \\ \\ \Delta{x} = - g(t-t_0)j$ Is this correct? Last edited by babaliaris; March 21st, 2019 at 05:19 PM.
March 21st, 2019, 05:23 PM   #8
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Quote:
 Originally Posted by babaliaris $\displaystyle \Delta{x} = i\int_{t_0}^{t} v_0dt - j\int_{t_0}^{t} gt dt = 0i - g(t-t_0)j \\ \\ \Delta{x} = - g(t-t_0)j$ Is this correct?
I completely destroyed that one lol

$\displaystyle \Delta{x} = i\int_{t_0}^{t} v_0dt - j\int_{t_0}^{t} gt dt = v_0(t-t_0)i - \frac{1}{2}g(t^2-t_0^2)j$

Last edited by babaliaris; March 21st, 2019 at 05:45 PM.

March 21st, 2019, 08:12 PM   #9
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Quote:
 Originally Posted by babaliaris So, the integration of the magnitude of v in my OP, is going to give a function that gives only a magnitude too?
Yes.

Quote:
 Originally Posted by babaliaris I completely destroyed that one lol $\displaystyle \Delta{x} = i\int_{t_0}^{t} v_0dt - j\int_{t_0}^{t} gt dt = v_0(t-t_0)i - \frac{1}{2}g(t^2-t_0^2)j$
That's correct. So the main problem remaining in the thread is are you talking about a distance $\displaystyle \Delta x$ or are you talking about a displacement $\displaystyle \vec{ \Delta x}$? The first integral you posted is a displacement, since it's a vector, and the second (with the magnitude) is a distance.

-Dan

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