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Math Help - Cartesian coordinates

  1. #1
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    Cartesian coordinates

    I am trying to find the equation of the tangent line in cartestian coordinates of the curve given in polar coordinates by

    r=3-2 costheta, at theta=pi/3



    i have that



    from there I am stuck on the rest I know that



    now can someone plz help me solve it
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by karatekid_81 View Post
    I am trying to find the equation of the tangent line in cartestian coordinates of the curve given in polar coordinates by

    r=3-2 costheta, at theta=pi/3



    i have that



    from there I am stuck on the rest I know that



    now can someone plz help me solve it
    here's what you want to do. first, find \frac {dy}{dx}. how? just plug in all the values and solve.

    \frac {dy}{dx} = \frac {r \cos \theta + \frac {dr}{d \theta} \sin \theta}{-r \sin \theta + \frac {dr}{d \theta} \cos \theta}

    we know r and you found \frac {dr}{d \theta}

    \Rightarrow \frac {dy}{dx} = \frac {(3 - 2 \cos \theta) \cos \theta + (2 \sin \theta) \sin \theta}{-(3 - 2 \cos \theta) \sin \theta + (2 \sin \theta) \cos \theta}

    simplify that, and plug in \theta = \frac {\pi}3 when done. or you can just plug it in now and calculate the result, it's up to you

    now, when \theta = \frac {\pi}3, r = 2

    so the point we are concerned with is (r, \theta) = \left( 2, \frac {\pi}3 \right)

    now we must change this coordinate to Cartesian coordinates:

    recall, to go from Polar coordinates (r , \theta) to Cartesian coordinates (x,y), we use the fact that:

    x = r \cos \theta and y = r \sin \theta

    use those equations to find the point (x,y)


    now, finally, use the point slope form to get the equation of the tangent line.

    using m = \frac {dy}{dx} and (x_1,y_1) as the point you found just above, we have by the point slope form:

    y - y_1 = m(x - x_1)

    plug in the values and solve for y and that's your line
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  3. #3
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    these whole trig functions mess with me, what would it look like simplified, i understand from there on but i dont get the whole simplify thing,
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by karatekid_81 View Post
    these whole trig functions mess with me, what would it look like simplified, i understand from there on but i dont get the whole simplify thing,
    as i said. you don't HAVE TO simplify, it would be nice, but it is not necessary. you could just plug in \theta = \frac {\pi}3 into \frac {dy}{dx} = \frac {(3 - 2 \cos \theta) \cos \theta + (2 \sin \theta) \sin \theta}{-(3 - 2 \cos \theta) \sin \theta + (2 \sin \theta) \cos \theta} and work it out.

    however, if you were simplifying, you would do something like this:

    \frac {dy}{dx} = \frac {(3 - 2 \cos \theta) \cos \theta + (2 \sin \theta) \sin \theta}{-(3 - 2 \cos \theta) \sin \theta + (2 \sin \theta) \cos \theta}

    = \frac {3 \cos \theta - 2 \cos^2 \theta + 2 \sin^2 \theta}{-3 \sin \theta + 2 \sin \theta \cos \theta + 2 \sin \theta \cos \theta}

    = \frac {3 \cos \theta - 2 \left( {\color {red} cos^2 \theta - \sin^2 \theta } \right)}{-3 \sin \theta + 2 ({\color {blue} 2 \sin \theta \cos \theta})}

    what is in red is: \cos 2 \theta
    what is in blue is: \sin 2 \theta

    so plug those in, we get:

    \frac {dy}{dx} = \frac {3 \cos \theta - 2 \cos 2 \theta}{2 \sin 2 \theta - 3 \sin \theta}

    now plug in \theta = \frac {\pi}3
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  5. #5
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    thanks I believe I got it.
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