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Parametric Equations


Date: 07/21/98 at 19:15:19
From: Ivan Li
Subject: Paramatric equations

I hope you can help me with this question.

A) A curve is given parametrically by the equations:

   x = (3 - 2k)^2   and  y = (2 + k)^2

  (i) Find dy/dx in terms of k and hence find the point on the curve
      at which the tangent is parallel to the  y-axis.

 (ii) Find the x-intercept of the curve and the equation of the normal
      at this point.  

(iii) Find point of intersection between the curve and another curve
      defined parametrically by the equations:

      x = 4/p + 4  and  y = (p + 1)/p

B) Find the cartesian equation of the curve which is defined
   parametrically by:

   x = cosec(t + pi/2)  and  y = tan (t + pi/4)


Date: 07/22/98 at 11:04:40
From: Doctor Anthony
Subject: Re: paramatric equations

Problem A(i):

            dy/dk     2(2+k)        2+k
   dy/dx =  ----- = ----------- = --------
            dx/dk   2(3-2k)(-2)   -2(3-2k)

            2+k
   dy/dx = -----
            4k-6 

The graph is parallel to the y-axis when 4k-6 = 0. Thus k = 3/2.
Then x = (3-3)^2 = 0 and y = (7/2)^2 = 49/4

The curve is parallel to y axis at the point (0, 49/4).

Problem A(ii):

At y = 0, k = -2 and then x = (3-2k)^2 = (3+4)^2 = 49.
So the x-intercept of curve is the point (49,0).

The slope of curve at this point is 0, and so the normal is the line  
x = 49.

Problem A(iii):

At the point of intersection:

   (3-2k)^2 = 4/p + 4  = 4(1 + p)/p    and

   (2+k)^2 =  (p + 1)/p  

Dividing these equations we get:

   (3-2k)^2
   -------- = 4
    (2+k)^2 

Taking square roots:

   3-2k
   ---- = +/- 2
   2+k

Taking positive value first, we find:

   3-2k = 2(2+k)
   3-2k = 4+2k   

Thus:

   4k=-1   
   k = -1/4

Putting this value into expressions for x and y in terms of k we get 
one point of intersection.

Taking negative value:

   3-2k = -2(2+k)
   3-2k = -4-2k   

Since this equation has no solution, there is just one point of 
intersection, given by k = -1/4

Problem B:

First note:

   x = cosec(t+pi/2) = sec(t)    
   y = [tan(t) + tan(pi/4)]/(1-tan(t)) = [1+tan(t)]/[1-tan(t)]

Multiply top and bottom by 1+tan(t):

  y = [1+tan(t)]^2/[1-tan^2(t)] 
  y = [1+2tan(t)+tan^2(t)]/[1-tan^2(t)]

We have x^2 = sec^2(t) = 1+tan^2(t), so:

   tan^2(t) = x^2 - 1
   tan(t) = sqrt(x^2-1)

Thus:

        1 + 2 sqrt(x^2-1) + x^2 - 1     x^2 + 2 sqrt(x^2-1)
   y =  ---------------------------  =  -------------------
              1 - (x^2 - 1)                  2 - x^2

- Doctor Anthony, The Math Forum
Check out our web site! http://mathforum.org/dr.math/   
    
Associated Topics:
High School Calculus
High School Equations, Graphs, Translations

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