Results 1 to 7 of 7

Math Help - Limits

  1. #1
    Member
    Joined
    Nov 2007
    Posts
    232

    Limits

    Find the equation of the tangent line to the curve f(x) = \sqrt {x - 3} at the point P(7,2).


    m = \lim_{x \to \-a} \frac{f(x) - f(a)}{x - a}

    m = \lim_{x \to \-7} \frac{\sqrt{x - 3} - 2}{x - 7} \times \frac{\sqrt{x - 3} + 2}{\sqrt{x - 3} + 2}

    m = \lim_{x \to \-7} \frac{x - 7}{(x - 7)(\sqrt {x - 3} + 2)}

    m = \lim_{x \to \-7} \frac{1}{\sqrt {x - 3} + 2}

    m = \frac {1}{4}



    y - \frac {1}{4}x + b

    2 = (\frac {1}{4})(7) + b

    \frac {-3}{4} = b

    Therefore, y = \frac {1}{4}x + \frac {-3}{4}



    Textbook Answer:
    x - 4y + 1 = 0

    I'm confused? Don't know what I did wrong
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641

    I think

    Quote Originally Posted by Macleef View Post
    Find the equation of the tangent line to the curve f(x) = \sqrt {x - 3} at the point P(7,2).


    m = \lim_{x \to \-a} \frac{f(x) - f(a)}{x - a}

    m = \lim_{x \to \-7} \frac{\sqrt{x - 3} - 2}{x - 7} \times \frac{\sqrt{x - 3} + 2}{\sqrt{x - 3} + 2}

    m = \lim_{x \to \-7} \frac{x - 7}{(x - 7)(\sqrt {x - 3} + 2)}

    m = \lim_{x \to \-7} \frac{1}{\sqrt {x - 3} + 2}

    m = \frac {1}{4}



    y - \frac {1}{4}x + b

    2 = (\frac {1}{4})(7) + b

    \frac {-3}{4} = b

    Therefore, y = \frac {1}{4}x + \frac {-3}{4}



    Textbook Answer:
    x - 4y + 1 = 0

    I'm confused? Don't know what I did wrong
    That you confused the equation for a tangent line wrong y-2=f'(7)(x-7)\Rightarrow{y-2=\frac{1}{4}\bigg(x-7\bigg)}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Nov 2007
    Posts
    232
    ?? still confused, why does the equation contain y - 2 and x - 7? Isn't it just the slope equation that you use to find the equation of the tangent line to the curve?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641

    Ok

    Quote Originally Posted by Macleef View Post
    ?? still confused, why does the equation contain y - 2 and x - 7? Isn't it just the slope equation that you use to find the equation of the tangent line to the curve?
    You I am sure have heard of the point slope form y-y_1=m(x-x_1) yeah? so adapting that we replace y_1=f(x_0) which makes sense since if you evaluate a function at a point it gives the y co-ordinate...next we replace x_1=x_0 which is the x co-ordinate lastly we need the slope at the point we want the tangent curve to go through which is x_0 and to get the slope of f(x) at x-0 we need to find f'(x_0)...giving us the adapattion of hte pointslope formula the tangent line equation y-x_0=f'(x_0)(x-x_0) and that is how/why we use the tangent line...any more questions just ask
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Nov 2007
    Posts
    232
    do i always use this formula whenever determining the equation of the tangent line? because i've been taught to only use "y = mx + b", but i guess the point slope and the slope formulas are basically the same thing...?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641

    Well

    Quote Originally Posted by Macleef View Post
    do i always use this formula whenever determining the equation of the tangent line? because i've been taught to only use "y = mx + b", but i guess the point slope and the slope formulas are basically the same thing...?
    Don't differ from your teacher's curriculum...but I would because if you think about it they give you a point and a derivative which to evaluate it and get a slope....therefore you have a slope and a point....haha point-slope was made for this situation
    Follow Math Help Forum on Facebook and Google+

  7. #7
    o_O
    o_O is offline
    Primero Espada
    o_O's Avatar
    Joined
    Mar 2008
    From
    Canada
    Posts
    1,407
    They are equivalent:

    y - y_{1} = m(x - x_{1})

    You use this when you have the slope but no points of the equation. If you are given a point (x_{1},y_{1}) you can substitute it above and arrive at your desired form of y = mx + b
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Using limits to find other limits
    Posted in the Calculus Forum
    Replies: 7
    Last Post: September 18th 2009, 05:34 PM
  2. Function limits and sequence limits
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: April 26th 2009, 01:45 PM
  3. HELP on LIMITS
    Posted in the Calculus Forum
    Replies: 4
    Last Post: September 23rd 2008, 11:17 PM
  4. Limits
    Posted in the Calculus Forum
    Replies: 5
    Last Post: September 21st 2008, 10:52 PM
  5. [SOLVED] [SOLVED] Limits. LIMITS!
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 25th 2008, 10:41 PM

Search Tags


/mathhelpforum @mathhelpforum