Results 1 to 14 of 14

Math Help - [SOLVED] Curve Sketching

  1. #1
    Junior Member
    Joined
    Sep 2008
    Posts
    59

    [SOLVED] Curve Sketching

    a)Find the equation of the tangent at the point (t, t^3) on the curve y=x^3.Find also the coordinates of the point where this tangent meets the curve again.

    b) For the curve, y=(4/x^2)+x , show the following ,
    i) the entire curve lies above the line y=x

    Red colour fonts are parts which I need help, thanks in advance!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Oct 2005
    From
    Earth
    Posts
    1,599
    a) y' = 3x^2, so the slope at point x=t is 3t^2. Now we have a point and a slope, thus we're able to make a line.

    y-t^3=3t^2(x-t), plugging in the info into the point-slope equation for a line. Simplify that out and you'll get your tangent equation.

    To see where these two lines meet again, set them equal to each other and solve.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2008
    Posts
    59
    Quote Originally Posted by Jameson View Post
    a) y' = 3x^2, so the slope at point x=t is 3t^2. Now we have a point and a slope, thus we're able to make a line.

    y-t^3=3t^2(x-t), plugging in the info into the point-slope equation for a line. Simplify that out and you'll get your tangent equation.

    To see where these two lines meet again, set them equal to each other and solve.
    Set these two equations together?
    y=x^3 and y=3t^2(x-t)+t^3
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Oct 2005
    From
    Earth
    Posts
    1,599
    Yessir.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Jun 2008
    Posts
    792
    y=(4/x^2)+x

    b) Assume that the curve y=(4/x^2)+x and the line y=x intersects. Therefore:
    x = (4/x^2)+x

    Canceling x, you would get the ridiculous statement that:
    0 = \frac{4}{x^2}

    0 = 4

    There is a contradiction, therefore our original assumption is not true.

    EDIT: Continue with the next post below.
    Last edited by Chop Suey; September 16th 2008 at 03:05 AM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Sep 2008
    Posts
    59
    Quote Originally Posted by Jameson View Post
    Yessir.
    I'm sorry to trouble you again, but I cannot solve the equation for x.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Sep 2008
    Posts
    59
    Quote Originally Posted by Chop Suey View Post
    y=(4/x^2)+x

    b) Assume that the curve y=(4/x^2)+x and the line y=x intersects. Therefore:
    x = (4/x^2)+x

    Canceling x, you would get the ridiculous statement that:
    0 = \frac{4}{x^2}

    0 = 4

    There is a contradiction, therefore our original assumption is not true.

    EDIT: I didn't read your question correctly. I will edit again.

    Isn't that an inclined asymptote for the curve?
    I can't explain why it is above the line y=x.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Joined
    Oct 2005
    From
    Earth
    Posts
    1,599
    No problem

    So x^3=3t^2(x)-3t^3+t^3

    Simply x^3=3t^2(x)-2t^3

    Group and factor x(x^2-3t^2)=-2t^3

    Made a mistake... working on it
    Last edited by Jameson; September 16th 2008 at 02:55 AM. Reason: mistake
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member
    Joined
    Jun 2008
    Posts
    792
    Quote Originally Posted by ose90 View Post
    Isn't that an inclined asymptote for the curve?
    I can't explain why it is above the line y=x.
    f(x) = \frac{4}{x^2} + x

    g(x) = x

    Let's assume that f(x) < g(x). So that:
    \frac{4}{x^2} + x < x

    But if you cancel the x, you see that:
    \frac{4}{x^2}<0

    4<0

    Which is a contradiction.

    \therefore f(x) > g(x)

    The reason why I didn't put the equality sign is because I showed that they do not intersect.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Junior Member
    Joined
    Sep 2008
    Posts
    59
    Quote Originally Posted by Chop Suey View Post
    f(x) = \frac{4}{x^2} + x

    g(x) = x

    Let's assume that f(x) < g(x). So that:
    \frac{4}{x^2} + x < x

    But if you cancel the x, you see that:
    \frac{4}{x^2}<0

    4<0

    Which is a contradiction.

    \therefore f(x) > g(x)

    The reason why I didn't put the equality sign is because I showed that they do not intersect.
    Thanks a lot
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Junior Member
    Joined
    Sep 2008
    Posts
    59
    Quote Originally Posted by Jameson View Post
    No problem

    So x^3=3t^2(x)-3t^3+t^3

    Simply x^3=3t^2(x)-2t^3

    Group and factor x(x^2-3t^2)=-2t^3

    Made a mistake... working on it
    Thanks for helping me. I have tried to solve it for many times but still can't get the value of x. Maybe there is no solution.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor
    Joined
    Oct 2005
    From
    Earth
    Posts
    1,599
    You're welcome. I think there's a trick I'm not seeing. Cubics that have no x^2 term have ways of being solved. Either someone else will see it or I'll solve it soon.
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Super Member
    Joined
    Jun 2008
    Posts
    792
    Notice that x-t is a factor, and by polynomial division, you can reduce it to:
    (x-t)(x^2+tx-2t^2)=0

    And it can be factored even further to:
    (x-t)(x-t)(x+2t)=0
    Last edited by Chop Suey; September 16th 2008 at 03:45 AM.
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Junior Member
    Joined
    Sep 2008
    Posts
    59
    thanks for your help!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] curve sketching problem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: November 6th 2009, 09:39 PM
  2. [SOLVED] Curve Sketching - vertical asymptotes
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 28th 2009, 07:41 PM
  3. Curve Sketching
    Posted in the Calculus Forum
    Replies: 3
    Last Post: August 24th 2009, 12:56 PM
  4. Replies: 2
    Last Post: February 22nd 2009, 01:47 PM
  5. Curve Sketching
    Posted in the Calculus Forum
    Replies: 2
    Last Post: February 21st 2009, 01:07 PM

Search Tags


/mathhelpforum @mathhelpforum