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Finding the equation of a curve

  1. Mar 2, 2016 #1
    1. The problem statement, all variables and given/known data
    The normal to a curve at a point P cuts the y axis at T and N is the foot of the perpendiculor from P to the y axis. If for all P, T is always 1 unit below N, find the equation of the curve.


    2. Relevant equations


    3. The attempt at a solution

    Point T has co ordinates ## (0,y)##
    and point N has co ordinates ##(0,y+1)## Points T and N intersect at point P ,(a right angle ) having co ordinates##P (X,Y),## thus let the gradients be ##m## and ##-m## respectively, we then have ##(Y-y)/x=m## and ##(Y-(y+1))/x=-m## on solving we shall have ##(Y-y)=1/2## implying that ##(Y-y)/x=m##≡ ##1/2/x=m## , where ##m=1/2x## which is the gradient. we know that the gradient of a curve is always equal to the gradient of the tangent, to get the equation we need to ##∫dy/dx ## wrt x....

    the answer to this problem is ##y=k-1/2x^2##
     
    Last edited by a moderator: Mar 3, 2016
  2. jcsd
  3. Mar 3, 2016 #2

    Simon Bridge

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    I think your notation is confusing...
    i.e. if I'm following this:
    If N is the foot of the perpendicular from P (to the y-axis?) and N is (0,y+1) and P is (X,Y) then Y=y+1 ...

    Lets tidy it up a bit:

    You are looking for the curve y=f(x) ... which is the set of coordinates (x,y) which satisfy the conditions.
    Since P is on the curve, it has coordinates (x,y)
    The slope of the tangent to f at P is f'(x) = (need to work out the equation for the slope).
    You know two points on the tangent: express them in terms of the coordinates of P.
     
    Last edited: Mar 3, 2016
  4. Mar 3, 2016 #3
    ok sir are the points ## T(0,y), N(0,y+1) and P(x,y1) ?## i am unable to use the template to indicate y1 for pointP as subscript...
     
  5. Mar 3, 2016 #4

    Simon Bridge

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    Careful: if P is point (x,y), and N is the base of the perpendicular (to y) through P, then N has coordinate (0,?) what?
     
  6. Mar 3, 2016 #5
    N will have co ordinates ##(0,y)## in which case the gradient will be zero
     
  7. Mar 3, 2016 #6
    ie....## y-y/x-0=m ##
    i.e the gradient of PN is
    ##0/x=m##
    then it follows that for point T and P,which is perpendiculor to PN ##(0/x) b=-1##
    where b is gradient of PT
    ##b=-x/0##
    then ## y-(y-1)/x=-x/0##
    where
    ##1=-x^2##
    kindly advise
     
  8. Mar 3, 2016 #7
    now this is my second attempt , for line PT, the equation of the line is ##y+1-y/x-0=m##
    ##y+1-y=xm## then
    ##1=xm##
    ##m=1/x##
    for the line PN , its gradient will be(using ##m1×m2=-1)##
    ##(1/x)m=-1##, where ##m=-x##, which is the same as the gradient of the curve at the point P(x,y) ,then ##dy/dx=-x## it follows that ##y=k-0.5x^2##
     
    Last edited: Mar 3, 2016
  9. Mar 3, 2016 #8
    simon have a good day greetings from africa bro
     
  10. Mar 3, 2016 #9

    Simon Bridge

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    The gradient of the line through N and P is zero - yes,
    Any line perpendicular to the y axis has a gradient of zero - that is what "perpendicular to the y axis" means.
    You should draw a sketch.

    You know anothetr point though - if P is (x,y) and N is (0,y) then what is T?
    Hint, not (0,y+1)
     
  11. Mar 3, 2016 #10
    are you saying i am wrong? kindly look at my working...if you say that the gradient is zero... then it won't be possible to get an equation for ##y=f(x)##
     
  12. Mar 3, 2016 #11
    i still stand by my working, i dont see how you can use gradient 0 to get equation of the curve,kindly speak up.. it means that the gradient of the other line is either zero or math error. T can have the points##(0,y-1)##
     
  13. Mar 3, 2016 #12
    lines parallel to the x -axis have equations of the form ##y=k##, where gradient is zero. if this line is perpendiculor to another line, then the gradient of the other line will be undefined. if you check my workings i expressed the gradients of the two lines as ##m=1/x## and ##m=-x##, in which case if we substitute x=0 we shall get gradients that are undefined and zero respectively, that fully agrees with my earlier statement... in this question we were to simply use the logic of gradient to get ##y=f(x)##
     
  14. Mar 3, 2016 #13
    If your curve is parametrized by an arc ##\vec f(t) = (x(t),y(t))## with sufficient regularity,
    and if current point ##P## on the curve has coordinates ##P = \vec f(t) ##,
    then you have the relation ## \vec {f'(t)} . \vec {T(t)P(t)} = 0##.
    This implies that ##y'(t) = - x(t) x'(t) ##

    So I agree with your solution ;-)
     
  15. Mar 3, 2016 #14

    Simon Bridge

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    In post #7 you say:
    which doesn't make sense.
    I'm guessing you are saying the gradient of the line through P and T is [(y+1)-y]/x ... which suggests you put T=(0,y+1) ... is that correct?
    But from post #1 you wrote:
    ... but N = (0,y) ... one below (0,y) is surely (0,y-1)?

    However - it is possible that I have misunderstood the question. It is 3am here in NZ so I'll sleep on it.
    What does the question mean when it talks about the perpendicular through P?
    What is "the perpendicular" perpendicular to?
     
  16. Mar 3, 2016 #15
    I meant ##P(x,y+1)## and ##T(0,y)## and ##N(0,y+1)##
     
    Last edited: Mar 4, 2016
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