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Math Help - problem with limit & derivative

  1. #1
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    problem with limit & derivative

    i have problems with limit and derivative
    first problem :
    Code:
    l i m      Cos X - Cos(X+h)
    h-->0  ---------------------
                       h
    second one
    Code:
    find dy/dx :
    
    
    
    
    
    X^2+3xy+2y^2 = 9
    third one :
    Code:
    find points on curve y=tan(x/2) , x belong to ]0,360[ and the tangent is perpendicular to straight line 3x=8-3y
    i got 4 solutions in this problem .... 2 of them i think refused
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    Re: problem with limit & derivative

    Quote Originally Posted by mido22 View Post
    \lim_{h\to0}\frac{\cos x-\cos(x+h)}h
    Use the sum and difference formula for cosine.

    Alternatively, recognize that \lim_{h\to0}\frac{\cos x-\cos(x+h)}h=-\lim_{h\to0}\frac{\cos(x+h)-\cos x}h=-\frac d{dx}\left[\cos x\right].

    Quote Originally Posted by mido22 View Post
    find dy/dx :
    x^2+3xy+2y^2 = 9
    Differentiate implicitly and then solve for \frac{dy}{dx}. If you're really stuck, show us what you've tried.

    Quote Originally Posted by mido22 View Post
    find points on curve y=tan(x/2) , x belong to ]0,360[ and the tangent is perpendicular to straight line 3x=8-3y
    i got 4 solutions in this problem .... 2 of them i think refused
    I'm not even sure what this is saying. Are you looking for points on y=\tan\frac x2 whose tangent lines are perpendicular to the given line, or what?
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    Re: problem with limit & derivative

    for the third problem ..yes i 'm looking for points on y=tan x/2 whose tangent perpen. to 3x=8-3y
    here is my attempt for the 2nd one :
    y^2 = 9 - 3xy - x^2
    y = +- \sqrt{9-3xy-x^2}
    dy/dx = +-\frac{2x+3y+(3x*(dy/dx))}{2*\sqrt{9-3xy-x^2}}
    Last edited by mido22; June 11th 2012 at 11:28 AM.
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    Re: problem with limit & derivative

    Quote Originally Posted by mido22 View Post
    y^2 = 9 - 3xy - x^2<br />
y = +- sqrt(9-3xy-x^2) <br />
dy/dx = +- (2x+3y+(3x*(dy/dx)))/(2*sqrt(9-3xy-x^2))<br />
    That's a mess, but it looks like you're trying to solve for y first. That isn't really necessary:

    x^2+3xy+2y^2 = 9

    \Rightarrow\frac d{dx}\left[x^2+3xy+2y^2\right] = \frac d{dx}[9]

    \Rightarrow\frac d{dx}\left[x^2\right]+3\frac d{dx}\left[xy\right]+2\frac d{dx}\left[y^2\right] = 0

    And continue...
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    Re: problem with limit & derivative

    Quote Originally Posted by mido22 View Post
    for the third problem ..yes i 'm looking for points on y=tan x/2 whose tangent perpen. to 3x=8-3y
    What do we know about the slopes of perpendicular lines?
    Thanks from mido22
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    Re: problem with limit & derivative

    the perpen . line equation : 3x=8-3y so its slope is -1
    so the slope of tangent is -1 / -1 = 1
    \frac{dy}{dx} = 0.5 * \sec^2\frac{x}{2} = 1
    then i get x/2 =90 or 315 or 135 or 450
    then i cancelled the 2nd and fourth solutions becz x belong to ]0,360[

    is this right?
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    Re: problem with limit & derivative

    and i get the solution of the 2nd problem
    2x+3y+3x\frac{dy}{dx}+4y\frac{dy}{dx}
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    Re: problem with limit & derivative

    Hello, mido22!

    Find points on curve y\,=\,\tan\left(\tfrac{x}{2}\right),\; x \in [0^o,\,360^o]
    and the tangent is perpendicular to straight line 3x\,=\,8-3y

    The straight line is: . y \:=\:-x + \tfrac{8}{3}
    . . Its slope is -1.

    The slope of the tangent is: . y' \:=\:\tfrac{1}{2}\sec^2(\tfrac{x}{2})

    We want the slope of the tangent to be +1.

    . . \tfrac{1}{2}\sec^2(\tfrac{x}{2}) \:=\:1 \quad\Rightarrow\quad \sec^2(\tfrac{x}{2}) \:=\:2 \quad\Rightarrow\quad \sec(\tfrac{x}{2}) \:=\:\pm\sqrt{2}

    . . \tfrac{x}{2} \:=\:45^o,\:135^o,\:225^o,\:315^o\:\hdots

    . . x \:=\:90^o,\:270^o,\:450^o,\:630^o\:\hdots

    Therefore: . x \;=\;90^o,\:270^o

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