Asked by ChrismB
Give the equation of the tangent line to the curve at given point x(t)=2 cos(t) y=2 sin(t) at t=pi/6?
Answers
Answered by
Reiny
x = 2cos t
y = 2sin t
dx/dt = -2sin t
dy/dt = 2cos t
slope = dy/dx
= (dy/dt) / )dx/dt)
= 2cost/-2sint
= -cost/sint
when t = pi/6
sint = 1/2
cost= √3/2
slope = (-√3/2)/(1/2) = -√3
point: (2cos pi/6, 2sin pi/6)
or (√3 , 1)
So now you have m = -√3, and the point (√3,1)
equation of tangent:
y - 1 = √3(x - √3)
y - 1 = √3x - 3
or
y = √3x - 2
y = 2sin t
dx/dt = -2sin t
dy/dt = 2cos t
slope = dy/dx
= (dy/dt) / )dx/dt)
= 2cost/-2sint
= -cost/sint
when t = pi/6
sint = 1/2
cost= √3/2
slope = (-√3/2)/(1/2) = -√3
point: (2cos pi/6, 2sin pi/6)
or (√3 , 1)
So now you have m = -√3, and the point (√3,1)
equation of tangent:
y - 1 = √3(x - √3)
y - 1 = √3x - 3
or
y = √3x - 2
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