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Mathematics 61 Online
OpenStudy (anonymous):

Someone please help. Suppose a curve C is parameterized by r (t) with a≤t≤b and suppose F ⃗ is a vector field F(t)=r (t)×r ' (t) for a≤t≤b. Find ∫F ⋅dr and explain your answer

OpenStudy (anonymous):

Are we working with general functions here?

OpenStudy (anonymous):

@ChristopherToni yes we are..

OpenStudy (anonymous):

I'll think about this over dinner; I don't think it's going to be too bad. Were there any other assumptions to be made, like, for instance, \(r(t)\) is a parameterization in \(\large \Bbb{R}^3\)? i.e. \(\large r(t) = (x(t),y(t),z(t))\) for \(\large a\leq t\leq b\)?

OpenStudy (anonymous):

The reason why I ask this is because we need to rewrite \(dr\) in the integral formula. Otherwise, I assume we're supposed to show this in \(\large \Bbb{R}^n\) if nothing else is specified or assumed.

OpenStudy (anonymous):

@ChristopherToni r(t)=(x(t),y(t),z(t)) for a≤t≤b this is given. this is the only information provided for this question. Yeah we do need dr in the integral. I am so lost in this problem. It would be such a great help if you can help me on this.

OpenStudy (anonymous):

Thanks for the clarification; it's actually not going to be too bad. I'll help you with this when I get back from dinner (in about 30-45 minutes).

OpenStudy (anonymous):

@ChristopherToni i would really appreciate that. Thanks.

OpenStudy (anonymous):

I think I just realized it's easier than it looks. The first thing to note is that \[\large \,dr = (x^{\prime}(t)\,dt, y^{\prime}(t)\,dt,z^{\prime}(t)\,dt) = r^{\prime}(t)\,dt\]Hence\[\large \int_C F\cdot\,dr = \int_a^b(r(t)\times r^{\prime}(t))\cdot r^{\prime}(t) \,dt\] Now note that \[\large (r(t)\times r^{\prime}(t))\cdot r^{\prime}(t) = \begin{vmatrix}x^{\prime}(t) & y^{\prime}(t) & z^{\prime}(t)\\ x(t) & y(t) & z(t) \\ x^{\prime}(t) & y^{\prime}(t) & z^{\prime}(t)\end{vmatrix} = 0\]since the determinant of a matrix where two of the rows are the same is zero. Therefore, \(\large\displaystyle \int_a^b (r(t)\times r^{\prime}(t))\cdot r^{\prime}(t)\,dt = \int_a^b 0 \,dt = 0\). I hope this makes sense! :-)

OpenStudy (anonymous):

What I used to evaluate \(\large (r(t)\times r^{\prime}(t))\cdot r^{\prime}(t)\) is known as the scalar triple product. See this for more information: http://en.wikipedia.org/wiki/Triple_product#Scalar_triple_product

OpenStudy (anonymous):

@ChristopherToni thanks a lot. That makes perfect sense now. Thanks for your time. I know i am asking for a lot but can you please take a look at this problem? you might give me some information. Let f(x,y,z) be a function of three variables. Suppose that C is an oriented curve lying on the level surface f(x,y,z)=1. Find the line integral ∫∇f⋅dr ⃗ and explain your answer.

OpenStudy (anonymous):

@ChristopherToni Do you have any idea on that problem?

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