Results 1 to 4 of 4

Math Help - Question regarding finding the tangent line to a point on a curve

  1. #1
    Newbie
    Joined
    Sep 2011
    Posts
    11

    Question regarding finding the tangent line to a point on a curve

    So, I have been working problems finding the derivative of polynomial equations. I am having some problems, however. The problem is as follows:

    Find the line tangent to the curve of Y = 4th root of X, on the point (1, 1).

    This is my working of the problem...
    y= 4th root of x
    y' = x ^ 1/4 = 1/4 x ^ -3/4 = m

    therefore, the line tangent to this should be Y - Y1 = m(x - x1) so:
    y - 1 = 1/4 x ^ -3/4(x - 1)
    y - 1 = 1/4 x ^ 1/4 - 1/4x ^ -3/4
    y = 1/4 x ^ 1/4 - 1/4x ^ -3/4 + 1
    y = 1/4x + 1 <<<<<<

    So by my calculations the line equation should be y = 1/4x + 1. However the book has the answer as y = 1/4x + 3/4. Where have I gone wrong? I don't see it.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,386
    Thanks
    1476
    Awards
    1

    Re: Question regarding finding the tangent line to a point on a curve

    Quote Originally Posted by iAmKrizzle View Post
    Find the line tangent to the curve of Y = 4th root of X, on the point (1, 1).
    I cannot read what you posted.
    If y=x^{\frac{1}{4}} then y'=\frac{1}{4}x^{\frac{-3}{4}}.

    So at (1,1) the slope is \frac{1}{4}.

    Tangent line is (y-1)=\frac{1}{4}(x-1)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    14,973
    Thanks
    1121

    Re: Question regarding finding the tangent line to a point on a curve

    Quote Originally Posted by iAmKrizzle View Post
    So, I have been working problems finding the derivative of polynomial equations. I am having some problems, however. The problem is as follows:

    Find the line tangent to the curve of Y = 4th root of X, on the point (1, 1).

    This is my working of the problem...
    y= 4th root of x
    Okay, y= x^1/4)

    y' = x ^ 1/4
    Oh dear, no! If y= x^(1/4) then y' is NOT the same thing!
    I know that's not what you meant to say, but is is what you actually said. Be careful what you write. Mathematics is very precise.
    What you meant to say was
    y'= 1/4 x ^ -3/4 = m

    therefore, the line tangent to this should be Y - Y1 = m(x - x1) so:
    y - 1 = 1/4 x ^ -3/4(x - 1)
    No, the slope or derivative at x= 1 is (1/4)1^(-3/4)= 1/4. It is a constant, not a function of x.

    y - 1 = 1/4 x ^ 1/4 - 1/4x ^ -3/4
    y = 1/4 x ^ 1/4 - 1/4x ^ -3/4 + 1
    y = 1/4x + 1 <<<<<<

    So by my calculations the line equation should be y = 1/4x + 1. However the book has the answer as y = 1/4x + 3/4. Where have I gone wrong? I don't see it.
    y- 1= (1/4)(x- 1)= (1/4)- 1/4. Adding 1 to both sides, y= (1/4)x+ 1- 1/4= (1/4)x+ 3/4.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Sep 2011
    Posts
    11

    Re: Question regarding finding the tangent line to a point on a curve

    HallsofIvy and Plato,

    Thanks your for your responses I see now where I went astray. With your help I've come to the correct answer also. Thank you again.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 6
    Last Post: May 16th 2011, 04:57 AM
  2. Replies: 6
    Last Post: February 9th 2010, 05:43 PM
  3. Replies: 3
    Last Post: December 12th 2009, 01:53 PM
  4. Replies: 5
    Last Post: November 4th 2009, 06:49 PM
  5. Replies: 1
    Last Post: October 17th 2009, 05:57 PM

/mathhelpforum @mathhelpforum