Results 1 to 4 of 4

Math Help - Equation of Line tangent to Curve.

  1. #1
    Senior Member topher0805's Avatar
    Joined
    Jan 2008
    From
    Vancouver
    Posts
    336

    Equation of Line tangent to Curve.

    Find an equation of the tangent line to the curve at the point (0, 0).

    y=sin(6 x) + sin^2(6 x)

    I found the derivative to be 6cos(6x)(1+2sin(6x)).

    When I plug in 0 for x I get y = 6 but I am not sure how that helps me. Galactus explained this to me the other night but I looked over that thread and can't apply what he showed me to this problem.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Feb 2008
    From
    Westwood, Los Angeles, CA
    Posts
    176
    Thanks
    1
    Remember that when you find a derivative, you're finding a formula that will give you the slope of the tangent line to any point on your function. So, when you plug in x = 0 into your derivative and get out y = 0 (good!), that value y = 0 is in SLOPE of the tangent line to your function at the point (0,0).

    So now you now your tangent line has a slope of 0. Think about what kinds of lines have a slope of 0 - this should give you an idea of what your line is. Remember, lastly, that your line has to pass through your original point, (0,0). So what is the equation of a line with a slope of 0 that passes through (0,0)?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member topher0805's Avatar
    Joined
    Jan 2008
    From
    Vancouver
    Posts
    336
    But when I plugged 0 for x into the derivative I got 6.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Feb 2008
    From
    Westwood, Los Angeles, CA
    Posts
    176
    Thanks
    1
    Topher,

    Sorry - should've proofread. Here's an updated response:

    Remember that when you find a derivative, you're finding a formula that will give you the slope of the tangent line to any point on your function. So, when you plug in x = 0 into your derivative and get out y = 6 (good!), that value y = 6 is in SLOPE of the tangent line to your function at the point (0,0).

    So now you now your tangent line has a slope of 6. You also know that it goes through (0,0). So you need the equation of a line given a point and the slope. This suggests using point-slope form to write your equation.

    Point-slope form, in case you've forgotten, is:

    (y - y_1) = m (x - x_1).

    Your slope is m and your point, (0,0), plays the role of (x_1, y_1). Plug in those three things and you should be in good shape.

    Sorry about the confusion! Too many zeros floating around, I guess.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Equation of tangent line to curve
    Posted in the Calculus Forum
    Replies: 2
    Last Post: September 10th 2011, 02:38 PM
  2. [SOLVED] equation of the line that is tangent to the curve
    Posted in the Pre-Calculus Forum
    Replies: 9
    Last Post: July 24th 2011, 10:48 PM
  3. Equation of tangent line to curve
    Posted in the Calculus Forum
    Replies: 4
    Last Post: October 26th 2009, 10:37 AM
  4. Equation of tangent line to curve
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 22nd 2009, 07:30 PM
  5. Equation of tangent line to following curve.
    Posted in the Calculus Forum
    Replies: 3
    Last Post: November 5th 2007, 10:45 AM

Search Tags


/mathhelpforum @mathhelpforum