Results 1 to 6 of 6

Math Help - find tangent

  1. #1
    Member
    Joined
    Oct 2006
    Posts
    88

    find tangent

    i need help finding the horizontal tangents on this curve. i got the verticals down correctly, but i cant figure the horizontals out...

    C: r=2+2cos(theta)

    y=rsin(theta)
    y=(2+2cos(theta))sin(theta)
    =2sin(theta)+2cos(theta)sin(theta)

    dy/d(theta) = 2cos(theta) - 2 sin^2(theta) + 2cos^2(theta)
    = 2(cos(theta) - sin^2(theta) + cos^2(theta))
    =2(cos(theta) - 1)

    2cos(theta) - 2 = 0
    2cos(theta) = 2
    cos(theta) = 1
    (theta) = 0 , 2pi

    i was supposed to somehow get pi/3 , 5pi/3 , and pi. where did i go wrong?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    = 2(cos(theta) - sin^2(theta) + cos^2(theta))
    = 2[cos(theta) - (sin^2(theta) - cos^2(theta))]

    not= 2(cos(theta) - 1) since sin^2(theta) - cos^2(theta) not= 1
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Oct 2006
    Posts
    88
    so i would have to put that in parenthesis first before i try the trig identity? how would i solve it to find theta?

    2[cos(theta) - (sin^2(theta) - cos^2(theta))]
    -2[-cos(theta) + (sin^2(theta) + cos^2(theta))]
    -2[-cos(theta) + 1]
    2cos(theta) - 2 = 0
    cos(theta) = 1

    i still end up with the same thing ><
    Follow Math Help Forum on Facebook and Google+

  4. #4
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by jeph View Post
    so i would have to put that in parenthesis first before i try the trig identity? how would i solve it to find theta?

    2[cos(theta) - (sin^2(theta) - cos^2(theta))]
    -2[-cos(theta) + (sin^2(theta) + cos^2(theta))]
    -2[-cos(theta) + 1]
    2cos(theta) - 2 = 0
    cos(theta) = 1

    i still end up with the same thing ><
    no, you won't. i only did the contraction to show you that you would not get 1, but to actually do the problem, we would expand differently.

    2[cos(theta) - (sin^2(theta) - cos^2(theta))] = 0
    => 2cos(theta) - 2sin^2(theta) + 2cos^2(theta) = 0
    => 2cos(theta) - 2(1 - cos^2(theta)) + 2cos^2(theta) = 0
    => 4cos^2(theta) + 2cos(theta) - 2 = 0
    => 2cos^2(theta) + cos(theta) - 1 = 0
    => (2cos(theta) - 1)(cos(theta) + 1) = 0



    Note: if you simplified correctly using the method you did before, you'd end up with 2(cos(theta) + cos(2*theta))
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Oct 2006
    Posts
    88
    i dont get it...

    so if there was a + instead of a -

    2[cos(theta) - (sin^2(theta) - cos^2(theta))]
    2[cos(theta) + sin^2(theta) + cos^2(theta)]

    would that equal

    2[cos(theta) + 1] ?

    where did i go wrong in simplifying it?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by jeph View Post
    i dont get it...

    so if there was a + instead of a -

    2[cos(theta) + sin^2(theta) + cos^2(theta)]

    would that equal

    2[cos(theta) + 1] ?
    yes, that would be correct

    where did i go wrong in simplifying it?
    you expanded the brackets wrong in the first case, and factored out the -1 wrong in the second case.

    2[cos(theta) - (sin^2(theta) - cos^2(theta))]
    -2[-cos(theta) + (sin^2(theta) + cos^2(theta))]
    the red plus sign above should have remained a minus sign. if you expand everything you would realize this is the case
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 8
    Last Post: March 22nd 2011, 05:57 PM
  2. Find eguation of tangent
    Posted in the Calculus Forum
    Replies: 8
    Last Post: April 13th 2009, 07:38 PM
  3. use dy/dx to find the equ. of tangent
    Posted in the Calculus Forum
    Replies: 5
    Last Post: October 11th 2007, 09:01 AM
  4. How Do U Find The Tangent Of This??
    Posted in the Calculus Forum
    Replies: 5
    Last Post: June 23rd 2007, 07:10 AM
  5. Find the tangent.
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 29th 2006, 10:02 PM

Search Tags


/mathhelpforum @mathhelpforum