Results 1 to 9 of 9

Math Help - Parametric Differentiation

  1. #1
    Junior Member
    Joined
    Oct 2009
    Posts
    31

    Parametric Differentiation

    The diagram (attached) shows a sketch of the curve with parametric equations 'x=t^3', 'y=3t^3 along with the normal at the point with parameter t=2.
    Parametric Differentiation-q7-11b.jpg

    Find the equation of this normal and the coordinates of this point where it cuts the curve again.

    I found dy/dx to be 6t/(3t^2-12), and am stumped on where to go from here. Clearly the normal at t=2 has something to do with this but I can't presently see how. Any ideas guys?

    Thanks for any help
    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 alibond07 View Post
    The diagram (attached) shows a sketch of the curve with parametric equations 'x=t^3', 'y=3t^3 along with the normal at the point with parameter t=2.
    Click image for larger version. 

Name:	Q7 11B.jpg 
Views:	11 
Size:	23.2 KB 
ID:	20830

    Find the equation of this normal and the coordinates of this point where it cuts the curve again.

    I found dy/dx to be 6t/(3t^2-12), and am stumped on where to go from here. Clearly the normal at t=2 has something to do with this but I can't presently see how. Any ideas guys?

    Thanks for any help
    The equation you gave us does not match the graph

    x=t^3,y=3t^3 \implies y=3(x) which is a line!
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Oct 2009
    Posts
    31
    Sorry the pair of equations is ' x=t^3-12t' , 'y=3t^2'
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Oct 2009
    Posts
    31
    Quote Originally Posted by TheEmptySet View Post
    The equation you gave us does not match the graph

    x=t^3,y=3t^3 \implies y=3(x) which is a line!
    Sorry the pair of equations is ' x=t^3-12t' , 'y=3t^2'
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by alibond07 View Post
    Sorry the pair of equations is ' x=t^3-12t' , 'y=3t^2'
    Okay (just FYI your normal is at the wrong point and normal mean perpendicular to the surface).

    Your Idea is sound first you need to take the derivative

    \displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=  \frac{2t}{t^2-4}

    So the slope of the normals line is perpendicular to the tangent this gives
    \displaystyle m_{\perp}=\frac{t^2-4}{2t}\bigg|_{t=2}=0

    So the equation of the normal line is horizontal and since y(2)=3(2)^2=12 So the equation of the normal line y_{N}=12

    Now we just need to find another value of t such that y(t)=12 \implies 3t^2=12

    Can you finish from here?
    Last edited by TheEmptySet; February 16th 2011 at 01:24 PM. Reason: typo
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Oct 2009
    Posts
    31
    Quote Originally Posted by TheEmptySet View Post
    Okay (just FYI your normal is at the wrong point and normal mean perpendicular to the surface).

    Your Idea is sound first you need to take the derivative

    \displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=  \frac{2t^2}{t^2-4}


    So the slope of the normals line is perpendicular to the tangent this gives
    \displaystyle m_{\perp}=\frac{t^2-4}{2t^2}\bigg|_{t=2}=0

    So the equation of the normal line is horizontal and since y(2)=3(2)^2=12 So the equation of the normal line y_{N}=12

    Now we just need to find another value of t such that y(t)=12 \implies 3t^2=12

    Can you finish from here?
    Okay thanks, how come out results for \displaystyle \frac{dy}{dx} are different?

    I don't think I'm getting you sadly.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by alibond07 View Post
    Okay thanks, how come out results for \displaystyle \frac{dy}{dx} are different?

    I don't think I'm getting you sadly.
    They are not, you need to simplify your answer. e.g factor out a 3 and reduce. Please be more specific about what you don't understand.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Oct 2009
    Posts
    31
    Quote Originally Posted by TheEmptySet View Post
    They are not, you need to simplify your answer. e.g factor out a 3 and reduce. Please be more specific about what you don't understand.
    I'm confused about how you got from (6t)/(3t^2-12) to (2t^2/ t^2 -4).
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by alibond07 View Post
    I'm confused about how you got from (6t)/(3t^2-12) to (2t^2/ t^2 -4).
    \displaystyle \frac{6t}{3t^2-12}=\frac{3\cdot 2t}{3(t^2-4)}=\frac{\not{3}\cdot 2t}{\not{3}(t^2-4)}=\frac{ 2t}{t^2-4}

    There is a typo this is correct.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Parametric differentiation
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 17th 2011, 04:47 AM
  2. Parametric Differentiation
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: April 29th 2009, 03:39 PM
  3. Parametric Differentiation help....
    Posted in the Calculus Forum
    Replies: 9
    Last Post: April 15th 2009, 01:34 PM
  4. Differentiation ( Parametric)
    Posted in the Calculus Forum
    Replies: 2
    Last Post: September 2nd 2008, 06:33 AM
  5. help with parametric differentiation
    Posted in the Calculus Forum
    Replies: 6
    Last Post: November 7th 2006, 05:06 AM

Search Tags


/mathhelpforum @mathhelpforum