- #1

yusukered07

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If A(

*) =***t***t***i**-*t*^{2}**j**+ (**- 1)***t***k**and B(**) = 2***t**t*^{2}**i**+ 6*t***k**, evaluate (*a*) [tex]\int^{2}_{0}A \cdot Bdt ,[/tex] (*b)*[tex]\int^{2}_{0}A \times B dt.[/tex]
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- Thread starter yusukered07
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- #1

yusukered07

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Last edited:

- #2

arildno

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I have no idea!

Can you help me out, by SHOWING WHAT YOU HAVE DONE SO FAR?

Can you help me out, by SHOWING WHAT YOU HAVE DONE SO FAR?

- #3

mathman

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Integral of t = t1. IfA (=t)t i - t, evaluate (b)^{2}j + (t -1) k[tex]\int^{2}_{0} A [/tex]

Integral of t

Integral of (t-1)=t

Net result 2i -(8/3)j

- #4

yusukered07

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Integral of t = t^{2}/2 = 2

Integral of t^{2}= t^{3}/3 = 8/3

Integral of (t-1)=t^{2}/2 - t = 0

Net result 2i -(8/3)j

Sorry for giving a wrong problem....

- #5

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Moderator's note: thread moved from "Calculus & Analysis"

Homework assignments or**any textbook style exercises** are to be posted in the appropriate forum in our https://www.physicsforums.com/forumdisplay.php?f=152" area. This should be done whether the problem is part of one's assigned coursework **or just independent study.**

Start by evaluating**A·B** and *A*x*B*.

Homework assignments or

You need to show an attempt at solving the problem before receiving help!) =tti-t^{2}j+ (- 1)tkand B() = 2tt^{2}i+ 6tk, evaluate (a) [tex]\int^{2}_{0}A \cdot Bdt ,[/tex] (b)[tex]\int^{2}_{0}A \times B dt.[/tex]

Start by evaluating

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- #6

yusukered07

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Moderator's note: thread moved from "Calculus & Analysis"

Homework assignments orany textbook style exercisesare to be posted in the appropriate forum in our https://www.physicsforums.com/forumdisplay.php?f=152" area. This should be done whether the problem is part of one's assigned courseworkor just independent study.

You need to show an attempt at solving the problem before receiving help!

Start by evaluatingandA·BxA.B

Yeah.... That's the process I've made... Then I integrate them with respect to t and evaluate from 0 to 2

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- #7

Dustinsfl

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For the cross product, I obtained -6*t

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