Results 1 to 5 of 5

Math Help - tangent line & normal line question

  1. #1
    Newbie
    Joined
    Mar 2009
    Posts
    3

    Red face tangent line & normal line question

    here's the problem
    Find the equations of the tangent line & normal line to the graph of the equation at the indicated point..
    ____
    y = √x+1; (3,2)

    i got the slope and its
    __1___
    2√x+1

    so i have to plug it in the tangent line & normal line formula which is
    <br />
y - y1 = m(x-x1)      - tangent line

    and
    <br />
y - y1 = 1/m(x-x1)    - normal line

    so its goin' to be:

    y-2 = 1/2√x+1 (x-3)


    and my answer is:
    ___ ___
    0 =1/2√x+1 + 1/6√x+1


    pls help..thanx
    Last edited by emjez15; March 14th 2009 at 09:40 PM. Reason: typo error
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,807
    Thanks
    116
    Quote Originally Posted by emjez15 View Post
    here's the problem
    Find the equations of the tangent line & normal line to the graph of the equation at the indicated point..
    ____
    y = √x+1; (3,2)

    i got the slope and its
    __1___
    2√x+1

    so i have to plug it in the tangent line & normal line formula which is
    <br />
y - y1 = m(x-x1)      - tangent line

    and
    <br />
y - y1 = {\bold{\color{red}\ -\ }}1/m(x-x1)    - normal line<<<<you forgot the negative sign

    so its goin' to be:

    y-2 = 1/2√x+1 (x-3)


    and my answer is:
    ___ ___
    0 =1/2√x+1 + 1/6√x+1


    pls help..thanx
    From f(x)=\sqrt{x+1} you got f'(x)=\dfrac1{2\sqrt{x+1}} . That's correct!

    Therefore you know that the slope of the tangent in T(3, 2) is:

    f'(3)= \dfrac1{2\sqrt{3+1}} = \dfrac1{2\cdot 2} = \dfrac14

    Now plug in the coordinates of the point T and the value of the slope to get the equation of the tangent:

    y - 2 = \dfrac14 \cdot (x-3) ~~~~~~~ \leftarrow\ \text{tangent line}

    The slope of the normal is m_n =-\ \dfrac1{\frac14} = -4

    Now plug in the coordinates of the point T and the value of the slope to get the equation of the normal:

    y - 2 = -4 \cdot (x-3) ~~~~~~~ \leftarrow\ \text{normal line}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Mar 2009
    Posts
    3
    yah, thanx dude,
    i got it ..

    but i've got 1 more problem

    how can i solve this

    y = 4/x2 ; (2,1)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,621
    Thanks
    426
    Quote Originally Posted by emjez15 View Post
    yah, thanx dude,
    i got it ..

    but i've got 1 more problem

    how can i solve this

    y = 4/x2 ; (2,1)
    you mean y = \frac{4}{x^2} = 4x^{-2} ?

    \frac{dy}{dx} = -8x^{-3} = -\frac{8}{x^3}

    now find the tangent and normal lines at the given point on the curve.

    same drill as before.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Mar 2009
    Posts
    3

    Talking

    ahhh,, i see, got it..thats why..^^
    thanks sir..^^
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Normal Line and Tangent Plane
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 9th 2011, 04:31 PM
  2. Normal plane and tangent line
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: March 31st 2011, 04:54 PM
  3. Derivatives: normal and tangent line
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 20th 2008, 02:58 PM
  4. tangent line/normal line
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 29th 2008, 08:21 AM
  5. Tangent plane and normal line..
    Posted in the Calculus Forum
    Replies: 4
    Last Post: May 14th 2008, 09:05 AM

Search Tags


/mathhelpforum @mathhelpforum