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Math Help - Lines Tangent to Curves

  1. #1
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    Lines Tangent to Curves

    Hey, these are some multiple choice questions I can't figure out. Please feel free to answer all or selectively answer some of them. Thank you!

    If m1 is the slope of the curve xy=2 and m2 is the slope of the curve x^2+y^2=3, then the point of intersection of the two curves is:

    a) m1= - m2 b) m1*m2= -1 c) m1= m2 d) m1*m2 = 1

    e) m1*m2 = -2

    ---------------------------------------------------------------------

    The tangent to the curve y^2 -xy +9 = 0 is vertical when:

    a) y=0 b) y= + or - the square root of 3 c) y= .5 d) y= + or - 3

    e) none of these

    ---------------------------------------------------------------------

    The equation of the tangent to the curve y= xsinx at the point (pi/2, pi/2) is

    a) y= x- pi b) y= .5 c) y= pi- x d) y=x+ .5 e) y= x
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  2. #2
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    For #2, we could solve for y and differentiate, but let's do it implicitly.

    y^{2}-xy+9=0..............[1]

    2yy'-xy'-y=0

    Divide by y' and get:

    2y-x-\frac{y}{y'}=0

    Now, take the limit: \lim_{y'\rightarrow{\infty}}\left[2y-x-\frac{y}{y'}\right]=2y-x

    2y-x=0, \;\ y=\frac{x}{2}

    Sub into [1]:

    (\frac{x}{2})^{2}-x(\frac{x}{2})+9=0

    \frac{x^{2}}{4}-\frac{x^{2}}{2}+9=0

    Solve for x and we get x=\pm{6}

    That leaves y=\pm{3}

    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~~~~~~~~~~~~~~~~~~~~~

    Let's do it using 'regular' differentiation and see if we get the same thing.

    Solve y^{2}-xy+9=0 for y and get:

    y=\frac{\sqrt{x^{2}-36}+x}{2}, \;\ y=\frac{x-\sqrt{36-x^{2}}}{2}

    Differentiate: y'=\frac{x}{2\sqrt{x^{2}-36}}+\frac{1}{2}

    Set the denominator to 0 and solve for x:

    2\sqrt{36-x^{2}}=0

    We can see that x=\pm{6}

    Plugging this into y, we see \pm{3}

    I showed you this so you may know how to find a vertical tangent in the future. Study both methods. They may come in handy.
    Last edited by galactus; March 19th 2008 at 12:35 PM.
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