Results 1 to 4 of 4

Thread: 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 $\displaystyle x\neq 0$).

    So you need to solve:

    $\displaystyle 4y^3-2y=0$

    Then plug the points back into the function to find the $\displaystyle x$ 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 $\displaystyle x\neq 0$).

    So you need to solve:

    $\displaystyle 4y^3-2y=0$

    Then plug the points back into the function to find the $\displaystyle x$ 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 $\displaystyle y=\pm\frac{\sqrt{2}}{2}$ and $\displaystyle y=0$. Subbing $\displaystyle y=0$ into the first equation gives $\displaystyle x=\pm\sqrt{5}$.

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

    So the points are:

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

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

Search Tags


/mathhelpforum @mathhelpforum