Results 1 to 4 of 4

Math Help - How many tangent lines pass through...

  1. #1
    Member
    Joined
    Oct 2006
    From
    Toronto
    Posts
    86

    How many tangent lines pass through...

    Hi

    Posting this because I don't know how to go about doing it (I'm sure the actual math is easy, just the process). Anyways, it asks how many tangent lines pass through the curve y = x^2 + 2x through the point (-1/2,-3). The differentiated equation is y'=2x + 2 but how exactly would you find the number of curves passing through that?


    My second question's on Lemniscate's curve (the figure 8). The equation is y^2 = x^2 - x^4. I don't understand what it means when asking "find dy/dx or y' at (1/2, sqrt3/4). Would I just plug in these two coordinates for y and x after implicitly differentiating? I get y' = 1-2x^2/1-x^2 after implicitly differentiating this.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by SportfreundeKeaneKent View Post
    Hi

    Posting this because I don't know how to go about doing it (I'm sure the actual math is easy, just the process). Anyways, it asks how many tangent lines pass through the curve y = x^2 + 2x through the point (-1/2,-3). The differentiated equation is y'=2x + 2 but how exactly would you find the number of curves passing through that?
    Any point on the curve y=x^2+2x is an ordered pair of the form (x,x^2+2x) We also have another point (-\frac{1}{2},-3) So now we can calculate the slope between these two points

    m=\frac{y_2-y_1}{x_2-x_1}=\frac{x^2+2x+3}{x+\frac{1}{2}}

    Bute we know that the slope of a tangent line is equal to the derivative

    m = f'(x) \iff \frac{x^2+2x+3}{x+\frac{1}{2}}=2x+2 \iff x^2+2x+3=2x^2+2x+x+1 \implies

    0=x^2+x-2 \iff 0=(x+2)(x-1) so the x=-2 and 1

    So these are the x=coordinates of the tangent lines passing thought the point. So there are two of them.

    I hope this helps.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    o_O
    o_O is offline
    Primero Espada
    o_O's Avatar
    Joined
    Mar 2008
    From
    Canada
    Posts
    1,407
    Quote Originally Posted by SportfreundeKeaneKent View Post
    My second question's on Lemniscate's curve (the figure 8). The equation is y^2 = x^2 - x^4. I don't understand what it means when asking "find dy/dx or y' at (1/2, sqrt3/4). Would I just plug in these two coordinates for y and x after implicitly differentiating? I get y' = 1-2x^2/1-x^2 after implicitly differentiating this.

    Let's do an example. Find y' at (1,2) of the equation x^{2} + y^{2} = 5.

    Implicit differentiation gives us:
    2x + 2yy' = 0
    2yy' = -2x
    y' = -\frac{x}{y} = -\frac{1}{2}

    Our expression for y' is in both terms of x and y but that's perfectly fine. We have a coordinate (1,2) to evaluate at so you simply plug it in to your expression for y' to find the slope of a line tangent to that particular point.

    Your expression for y' in your particular question doesn't seem right. Show us what you've done and we'll see if we can pinpoint any mistakes you've made.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,803
    Thanks
    114
    Quote Originally Posted by SportfreundeKeaneKent View Post
    ...

    My second question's on Lemniscate's curve (the figure 8). The equation is y^2 = x^2 - x^4. I don't understand what it means when asking "find dy/dx or y' at (1/2, sqrt3/4). Would I just plug in these two coordinates for y and x after implicitly differentiating? I get y' = 1-2x^2/1-x^2 (unfortunately wrong) after implicitly differentiating this.
    to #2:

    Only for the records: Your equation of the derivative is wrong:

    From

    2yy'=2x-4x^3~\implies~y'=\frac{x-2x^3}{\pm \sqrt{x^2-x^4}} ... \implies ... y'=\frac{x(1-2x^2)}{|x| \cdot (\pm \sqrt{1-x^2})}=\pm \frac{1-2x^2}{\sqrt{1-x^2}}

    All these transformations are not necessary because you know the coordinates of the tangent point. Plug in the coordinates into the first equation and calculate y':

    2 \cdot \left(\frac{\sqrt{3}}4 \right)y'=2\cdot \frac12-4\left(\frac12 \right)^3~\implies~y'=\frac{\sqrt{3}}{3}
    Attached Thumbnails Attached Thumbnails How many tangent lines pass through...-lemniskate_tangente.gif  
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 8
    Last Post: May 13th 2011, 06:26 PM
  2. Replies: 6
    Last Post: February 12th 2011, 11:37 AM
  3. Looking at tangent lines
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: April 16th 2010, 09:31 AM
  4. Replies: 3
    Last Post: January 26th 2010, 02:41 PM
  5. Replies: 2
    Last Post: January 25th 2009, 08:41 PM

Search Tags


/mathhelpforum @mathhelpforum