Results 1 to 4 of 4

Math Help - How to solve this problem

  1. #1
    Newbie
    Joined
    Oct 2009
    Posts
    19

    How to solve this problem

    If the derivative dx/dy exists at a point and dx/dy = 0, then the tangent line is vertical. Use this information and the given equation to answer the following questions-
    y4 + 5 = y2 + x2

    1) Find dx/dy

    Ans:
    y4 + 5 = y2 + x2
    4y3 =2 y + 2x dx/dy
    (4y3 -2 y)/ 2x = dx/dy

    2) Find the points on the graph where the tangent line is vertical. Arrange the values from smallest to largest.

    How do I do this one? Some one please help me with this.
    (P.S.-please show the step so I will understand)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member redsoxfan325's Avatar
    Joined
    Feb 2009
    From
    Swampscott, MA
    Posts
    943
    Quote Originally Posted by skboss View Post
    If the derivative dx/dy exists at a point and dx/dy = 0, then the tangent line is vertical. Use this information and the given equation to answer the following questions-
    y4 + 5 = y2 + x2

    1) Find dx/dy

    Ans:
    y4 + 5 = y2 + x2
    4y3 =2 y + 2x dx/dy
    (4y3 -2 y)/ 2x = dx/dy

    2) Find the points on the graph where the tangent line is vertical. Arrange the values from smallest to largest.

    How do I do this one? Some one please help me with this.
    (P.S.-please show the step so I will understand)
    If the derivative is zero, than the numerator of that fraction needs to be zero (and also x\neq 0).

    So you need to solve:

    4y^3-2y=0

    Then plug the points back into the function to find the x coordinates.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Oct 2009
    Posts
    19
    Quote Originally Posted by redsoxfan325 View Post
    If the derivative is zero, than the numerator of that fraction needs to be zero (and also x\neq 0).

    So you need to solve:

    4y^3-2y=0

    Then plug the points back into the function to find the x coordinates.

    Thanks, that gave me two of the points which are (+-2 and 0) But there are 4 other values. How can I find those?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member redsoxfan325's Avatar
    Joined
    Feb 2009
    From
    Swampscott, MA
    Posts
    943
    Quote Originally Posted by skboss View Post
    Thanks, that gave me two of the points which are (+-2 and 0) But there are 4 other values. How can I find those?
    Well the three points here are y=\pm\frac{\sqrt{2}}{2} and y=0. Subbing y=0 into the first equation gives x=\pm\sqrt{5}.

    Subbing in y=\pm\frac{\sqrt{2}}{2} gives x=\pm\frac{\sqrt{19}}{2}

    So the points are:

    \left\{\left(\pm\frac{\sqrt{19}}{2},\pm\frac{\sqrt  {2}}{2}\right),\left(\pm\sqrt{5},0\right)\right\}

    for a total of six points. Is that what you meant by four more?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. How do you solve this problem?
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 7th 2010, 11:01 AM
  2. solve for x problem
    Posted in the Algebra Forum
    Replies: 8
    Last Post: August 21st 2009, 08:00 PM
  3. Can anyone help me solve this problem?
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: November 28th 2008, 06:50 AM
  4. Replies: 10
    Last Post: July 30th 2006, 10:26 AM
  5. Solve this problem please
    Posted in the Math Topics Forum
    Replies: 3
    Last Post: December 14th 2005, 03:24 PM

Search Tags


/mathhelpforum @mathhelpforum