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Math Help - Urgent help please...

  1. #1
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    Urgent help please...

    Let vector a <2,-3,4> and vector b <-3,4,7>


    Compute 3a - 2b


    and


    Use calculus to determine the values of
    θ where r = cosθ has a slope of 0.



    for the first one i just didnt know if to take the magnitude then multiply or multiply each number then get the magnitude...

    and still dont exactly know how to do 2
    Last edited by ThePerfectHacker; March 28th 2009 at 02:51 PM.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by tiga killa View Post
    for the first one i just didnt know if to take the magnitude then multiply or multiply each number then get the magnitude...
    no.

    let c be a scalar, and \vec a = \left< a_1, a_2 \right> be a 2-dimensional vector, then c \vec a = c \left< a_1, a_2 \right> = \left< c a_1, ca_2 \right>

    Let \vec b = \left< b_1, b_2 \right> be another vector. we define \vec a + \vec b as follows:

    \vec a + \vec b = \left< a_1, a_2 \right> + \left< b_1, b_2 \right> = \left< a_1 + b_1, a_2 + b_2 \right>

    that is, we just add corresponding components


    those are some basic definitions you must know. now can you get anywhere?
    and still dont exactly know how to do 2
    you do not know how to differentiate cosine?

    you haven't answered the question of whether we should be thinking of polor coordinates here or not.
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  3. #3
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    I get the first one..


    and yes the 2nd one is using polar coordinates
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by tiga killa View Post
    I get the first one..
    great! what was your answer?

    and yes the 2nd one is using polar coordinates
    ok, the "hard" way is to use the formula \frac {dy}{dx} = \frac {\frac {dr}{d \theta} \sin \theta + r \cos \theta}{\frac {dr}{d \theta} \cos \theta - r \sin \theta}  , then set the expression to 0 and solve. you can see how it is derived here


    Alternatively, begin with r = \cos \theta

    multiply through by r, to get

    r^2 = r \cos \theta

    now we know r^2 = x^2 + y^2 and x = r \cos \theta in polar coordinates. thus we have

    x^2 + y^2 = x

    now you can differentiate implicitly to find \frac {dy}{dx}, then set it equal to zero to find x and y, and hence the points you seek
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