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Math Help - BC calc problems

  1. #1
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    BC calc problems

    1973 -
    Given the curve x + xy +2y^2 = 6, find the coordinates of all other points on this curve with slop equal to the slope at (2,1)

    1970 -
    the function f is defined for all x not equal to 0, in the interval -1<x<1 by f(x)=\frac{x-\sin 2x}{\sin x}

    a) how should f(0) be defined in order that f be continuous for all x in the interval -1<x<1?
    b) With f(0) defined as in part a), use the definition of the derivative to determine whether f'(0) exists. Show your work.


    For 1973, ive been able to find the derivative as \frac{dy}{dx} = \frac{-y-1}{x+4y}. However, ive tried solving it with the original equation but cannot get an answer

    For 1970, I've got f(0) = -1, but using the definition of the derivitive gets really really messy and i cant get the h to cancel out.

    all help appreciated!!
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  2. #2
    hpe
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    Quote Originally Posted by dizzystudent
    1973 -
    Given the curve x + xy +2y^2 = 6, find the coordinates of all other points on this curve with slop equal to the slope at (2,1)
    That's the correct formula for the derivative. Set x = 2, y = 1 to find the slope. Then set the derivative equal to this slope to get an equation for the unknown point(s). It will be linear, so you can solve it for x or y (you choose). The other equation is the original equation x + xy +2y^2 = 6. When you eliminate one of the two unknowns, you will be left with a quadratic equation for the other one. One solution corresponds to the point (2,1) which you know already, the other one is new.
    1970 -
    the function f is defined for all x not equal to 0, in the interval -1<x<1 by f(x)=\frac{x-\sin 2x}{\sin x}
    \lim_{x \to 0} f(x) = -1 is correct. Mathcad tells me the limit of the difference quotient is 0. Are you allowed to use L'Hopital's rule? Otherwise the only thing you can do is use \lim_{x \to 0} \frac{\sin(x)}{x} = 1.
    Last edited by hpe; November 12th 2005 at 05:30 PM.
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  3. #3
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    thanks for helping me with those... but i still need help with part B on my second question.
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  4. #4
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    Quote Originally Posted by hpe
    \lim_{x \to 0} \frac{\sin(x)}{x} = 0.
    It's 1. I see you fixed it though.
    Last edited by Jameson; November 13th 2005 at 12:29 PM.
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    Quote Originally Posted by dizzystudent
    thanks for helping me with those... but i still need help with part B on my second question.
    Agreed. This is very messy, but post what you've done and I'll see if I can help work it out.
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  6. #6
    hpe
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    Quote Originally Posted by Jameson
    Agreed. This is very messy, but post what you've done and I'll see if I can help work it out.
    Ok, here goes. Define f(x) = \frac{x-\sin 2x}{\sin x} for x \ne 0,  f(0) = -1. The goal is to show that f'(0) = 0, using the definition.

    So set up the difference quotient:
    \frac{\frac{x-\sin 2x}{\sin x} + 1}{x}=\frac{x - \sin 2x + \sin x}{x \sin x}= \frac{x - 2 \sin x \cos x + \sin x}{x \sin x}.

    Recall that \lim_{x \to 0} \frac{\sin x}{x} = 1; thus it suffices to determine the limit of \frac{x - 2 \sin x \cos x + \sin x}{x^2} = \frac{x - \sin x}{x^2} +  \frac{2\sin x}{x} \cdot \frac{1 - \cos x}{x} .

    Since \lim_{x \to 0} \frac{1 - \cos x}{x} = \lim_{x \to 0} \frac{(1 - \cos x)(1+ \cos x)}{x(1 + \cos x)} which becomes  \lim_{x \to 0} \frac{-\sin^2 x}{x(1 + \cos x)} = 0, we only need to discuss \lim_{x \to 0}\frac{x -  \sin x}{x^2}.

    Now look at the proof of \lim_{x \to 0} \frac{\sin x}{x} = 1. In the course of that proof, the estimate  \sin x \cos x \le x \le \tan x is shown.

    See the picture below: the big right triangle has an angle x and cathetes 1 and  \tan x. The red right triangle has area \frac{1}{2}\sin x \cos x, the red plus blue sector has area \frac{x}{2}, and red + blue + yellow has an area equal to \frac{1}{2} \tan x.

    It follows that  x\cos x \le \sin x \le x\sec x.

    This implies  \frac{x - x \sec x}{x^2} \le \frac{x - \sin x}{x^2} \le \frac{x - x\cos x}{x^2} or  \frac{\cos x - 1}{x \cos x} = \frac{1 -  \sec x}{x} \le \frac{x - \sin x}{x^2} \le \frac{1 - \cos x}{x} .

    As shown above, \lim_{x \to 0}\frac{1-\cos x}{x} = 0. Therefore also \lim_{x \to 0}\frac{x -  \sin x}{x^2} = 0.

    QED
    Attached Thumbnails Attached Thumbnails BC calc problems-triangle.jpg  
    Last edited by hpe; November 19th 2005 at 09:09 AM.
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