Integration of vector valued functions

[ahmetbaba]

Homework Statement

Evaluate the definite integral.


[[ti + t^2j]] dt from 3 to 0. Sorry I don't know how to add the integration sign here. the parenthesis is referring to the magnitude of the vector.

Homework Equations

The Attempt at a Solution

I have no clue.

 

[LCKurtz]

Homework Statement

Evaluate the definite integral.


[[ti + t^2j]] dt from 3 to 0. Sorry I don't know how to add the integration sign here. the parenthesis is referring to the magnitude of the vector.

Do you mean this:

$$\vec R(t) = ti + t^2j$$

and you want to integrate

$$\int_3^0|\vec R(t)|\, dt\hbox{ ?}$$

Click on the equations to see how to write them. If that is what your question is, what exactly are you stuck on?

 

[ahmetbaba]

$$\int_3^0|\vec R(t)|\, dt\hbox{ ?}$$ the integration is from zero to three. the other way around that I wrote it ( I have to get used to this stuff very soon)

I'm stuck on the second thing you wrote there, and I'm stuck on the whole thing. Our instructor didn't really explain this stuff a lot, and the textbook kind of assumes what we should be doing, so that's not much help also.

So, my main problem is, do we integrate first, or we set up the magnitude, find that and then integrate it. And any help with that would be much appreciated.

 

[LCKurtz]

$$\int_3^0|\vec R(t)|\, dt\hbox{ ?}$$


the integration is from zero to three. the other way around that I wrote it ( I have to get used to this stuff very soon)

I'm stuck on the second thing you wrote there, and I'm stuck on the whole thing. Our instructor didn't really explain this stuff a lot, and the textbook kind of assumes what we should be doing, so that's not much help also.

So, my main problem is, do we integrate first, or we set up the magnitude, find that and then integrate it. And any help with that would be much appreciated.

Do you know how to find the magnitude (length) of a vector? Just write down its length and integrate it.

 

[ahmetbaba]

would it be like this; It comes out of the integral like this

ti^2/2 + t^3/3 + C

and the length of this would be,

(.5^2 + .33^2)^.5 =.599 ?

 

[Mark44]

That's not what I get. What did you get for the magnitude of the vector - |ti + t²j|? I think you skipped that part, and it looks like you just integrated t + t², which is not a vector.

 

[ahmetbaba]

|ti + t^2| isn't this simply the sum of the square root of the coefficients squared. so it would be (2)^.5 = 1.41

 

[HallsofIvy]

|ti + t^2| isn't this simply the sum of the square root of the coefficients squared. so it would be (2)^.5 = 1.41

What do you mean by "it" here? Grammatically, it should refer to |ti+ t^2j| but I think you mean the integral. |ti+ t^2j| is equal to [tex]\sqrt{t^2+ t^4}= t\sqrt{1+ t^2}[/itex]. That should be easy to integrate. Also, you said in your first post that you were integrating "from 3 to 0". Since 3< 0 and your integrand is positive, the integral would be negative. Or are you really integrating from 0 to 3?

Finally, $2^{.5}$ is only approximately equal to 1.41. It would be better to leave it as $$\sqrt{2}$$

 

[Mark44]

|ti + t^2| isn't this simply the sum of the square root of the coefficients squared. so it would be (2)^.5 = 1.41

The coefficients are the coefficients of i and j, not the coefficients of t and t^2. t and t^2 are the coefficients.