ATAR Notes: Forum
VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: JackSonSmith on April 19, 2015, 02:21:20 pm
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A curve has parametric equations x= t - cos(t) and y = sin(t)
Find the equation to the curve when t = pi/6
I keep on getting y = (root 3 /3) x - (root 3) pi / 18 + 1/2
However the answers say that it's y = (root 3 /3) x - (root 3) pi / 18 + 1
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A curve has parametric equations x= t - cos(t) and y = sin(t)
Find the equation to the curve when t = pi/6
I keep on getting y = (root 3 /3) x - (root 3) pi / 18 + 1/2
However the answers say that it's y = (root 3 /3) x - (root 3) pi / 18 + 1
r'(t) = (x'(t), y'(t)) = (1 + sin(t), cos(t))
r'(pi/6) = (3/2, sqrt(3)/2)
A tangent to the curve is r(pi/6) + t*r'(pi/6) = (pi/6 - sqrt(3/2), 1/2) + (3*t/2, sqrt(3)*t/2)
= (pi/6 - sqrt(3)/2 + 3*t/2, 1/2 + sqrt(3)*t/2)
Converting to cartesian we have
t = (2/3)*(x + sqrt(3)/2 - pi/6) = (y - 1/2)*(2/sqrt(3))
after algebra we get :
y = sqrt(3)*x/3 - sqrt(3)*pi/18 + 1