Parametric EquationsDate: 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/ |
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