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Parametric differentiation

  1. Apr 17, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the equation of normal to curve x=cos b , y=sin b at the point where b=pi/4 . Find the coordinates of the point on the curve where the normal meets the curve again .

    2. Relevant equations

    3. The attempt at a solution

    Gradient of normal :

    dx/db=-sin b

    dy/db=cos b

    dy/dx=-cot b

    gradient of normal = tan b , when b=pi/4 , gradient of normal =1

    The normal cuts the curve at point (1/sqrt(2) , 1/sqrt(2))

    equation of normal is y-1/sqrt(2)=1(x-1/sqrt(2))

    y=x --- 1

    The cartesian equation of the curve is cos^(-1) x=sin^(-1) y

    sin y=cos x --- 2

    combine 1 and 2 , sin x=cos x

    tan x=0

    x=0 , pi , 2pi

    Since the domain is not mentioned , i take it as 0<=x<=2pi

    Therefore , the normal would cut the curve at points (0,pi/2) , (pi,3pi/2), (2pi, 5pi/2)

    Am i correct ?
  2. jcsd
  3. Apr 17, 2010 #2
    I am confused by eq. 1 and 2. You converted (1) from cartesian to "wonky" cartesian. You have the equation of the normal line as y=x, which is correct, but that second equation seems to come out of nowhere.

    What does the original function actually represent? If you graph the curve, and y=x (the normal line) the answers become more clear.

    As a side note, limiting the domain might be a bad idea ;-)
  4. Apr 18, 2010 #3
    ok , so how do i get the equation of the curve given its parametric equation ? I tried to eliminate the parameters and got that second equation .
  5. Apr 18, 2010 #4
    Well, the curve given is a circle with radius 1.
    This can be verified: The equation of a circle is
    since you were given:
    x2+y2=r2 goes to:

    If you draw the circle and the line y=x, which according to the problem is:

    As for the intersections:
    tan(b)=1=x2+y2 I think you had zero here by mistake.
    If you check your unit circle, this occurs at Pi/4 +kPi, where k is an integer. (This works for all real domains, not just zero to 2Pi)

    So you were on the right path from the get go, but the extra transform threw things off.
  6. Apr 18, 2010 #5
    thanks a tons , m.w.Clever
  7. Apr 18, 2010 #6
    You're welcome, but no C in lever :) It can be tricky to visualize a lot of parametric problems, but in some cases (like this one) it really helps!
  8. Apr 18, 2010 #7


    Staff: Mentor

    You really should post calculus problems in the Calculus & Beyond section, not the Precalculus section
  9. Apr 18, 2010 #8
    sorry to trouble you with a few more questions ,

    (1) b= pi/4+kpi , is this the coordinate ? Or rather the coordinate is x=cos pi/4+kpi and
    y= sin pi/4+kpi ?

    because from y=x , x^2+y^2=1 , i could have solved for x and y but thats for b=pi/4 only right ?

    Mark , sorry for posting on the wrong subforum coz i thought this is just a high school question .
  10. Apr 18, 2010 #9


    Staff: Mentor

    Doesn't matter if it's high school or college or whatever. Calculus questions should go to the Calculus & Beyond section.

    The line y = x comes about because the problem asks for the normal on the parametric curve at b = pi/4. This point corresponds to the point (sqrt(2)/2, sqrt(2)/2) on the circle. Because the curve is a circle, the normal to the curve intersects the circle at a point directly across the circle, at (-sqrt(2)/2, -sqrt(2)/2). The x and y values in these points are the coordinates.
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