Results 1 to 10 of 10

Math Help - equation for the line tangent to graph

  1. #1
    Super Member
    Joined
    Oct 2007
    From
    Santiago
    Posts
    517

    equation for the line tangent to graph

    hey i have im practising some problems and i dont know where to start on this question.

    find an equation for the line tangent to the graph of y = secx at (pie/3,2)

    any suggestions?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Aug 2007
    From
    USA
    Posts
    3,110
    Thanks
    2
    1) Verify that (pi/3,2) is ON the graph. This is a point on the line.

    2) Find the derivative of the function, y' = ???

    3) Evaluate the derivaive at x = pi/3. This is the slope of the line.

    4) Remember your algebra and use the Point-Slope form of a line.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Oct 2007
    From
    Santiago
    Posts
    517
    Quote Originally Posted by TKHunny View Post
    1) Verify that (pi/3,2) is ON the graph. This is a point on the line.

    2) Find the derivative of the function, y' = ???

    3) Evaluate the derivaive at x = pi/3. This is the slope of the line.

    4) Remember your algebra and use the Point-Slope form of a line.
    thank u ! ill give it a go and post it
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Oct 2007
    From
    Santiago
    Posts
    517
    <br />
y = secx<br />

    <br />
y' = tanx \cdot secx<br />

    input x = \frac{\pi}{3} into:

    <br />
y' = tan(\frac{\pi}{3}) \cdot sec(\frac{\pi}{3})<br />

    to get a gradient of

    <br />
m = 2 \cdot \sqrt{3}<br />

    now y - y1 = m(x - x1) formula to get the equation

    <br /> <br />
y - 2 = 2 \cdot \sqrt{3}(x - \frac{\pi}{3})<br /> <br />

    is that the correct path?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    13
    Yes.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Aug 2007
    From
    USA
    Posts
    3,110
    Thanks
    2
    Quote Originally Posted by jvignacio View Post
    <br />
y = secx<br />

    <br />
y' = tanx \cdot secx<br />

    input x = \frac{\pi}{3} into:

    <br />
y' = tan(\frac{\pi}{3}) \cdot sec(\frac{\pi}{3})<br />

    to get a gradient of

    <br />
m = 2 \cdot \sqrt{3}<br />

    now y - y1 = m(x - x1) formula to get the equation

    <br /> <br />
y - 2 = 2 \cdot \sqrt{3}(x - \frac{\pi}{3})<br /> <br />

    is that the correct path?
    Did you do my Step #1? You can be assured I would put one on the exam that was NOT actually on the curve.

    Other than that...AWESOME!!
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member
    Joined
    Oct 2007
    From
    Santiago
    Posts
    517
    Quote Originally Posted by TKHunny View Post
    Did you do my Step #1? You can be assured I would put one on the exam that was NOT actually on the curve.

    Other than that...AWESOME!!

    thakns mann!!
    how do i check if its on the graph?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member 11rdc11's Avatar
    Joined
    Jul 2007
    From
    New Orleans
    Posts
    894
    Plug in the x coordinate given into the function and see if it matches with the y coordinate given.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member
    Joined
    Oct 2007
    From
    Santiago
    Posts
    517
    Quote Originally Posted by 11rdc11 View Post
    Plug in the x coordinate given into the function and see if it matches with the y coordinate given.
    yes it does! both 2
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor
    Joined
    Aug 2007
    From
    USA
    Posts
    3,110
    Thanks
    2
    NOW we're done.

    When the sneaky one shows up, you'll be the only one to get it right.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Find the equation of the tangent line to the graph
    Posted in the Calculus Forum
    Replies: 18
    Last Post: January 16th 2011, 08:27 PM
  2. Equation of tangent line to the graph
    Posted in the Calculus Forum
    Replies: 5
    Last Post: May 31st 2009, 09:23 PM
  3. Replies: 3
    Last Post: May 31st 2009, 07:28 AM
  4. Equation For The Line Tangent To The Graph
    Posted in the Calculus Forum
    Replies: 7
    Last Post: September 7th 2008, 10:22 PM
  5. Replies: 9
    Last Post: September 2nd 2007, 10:58 PM

Search Tags


/mathhelpforum @mathhelpforum