Asked by Lielle
find the tangent line equation at x=cost+sin2t, y=sint+cos2t, t=0
Answers
Answered by
Damon
y = sin t + cos^2 t - sin^2 t
dy/dt = cos t -2cost sint -2 sint cost
at t = 0 dy/dt = 1
x = cos t +2 sin t cos t
dx/dt = -sint -2sin^2t+2 cos^2t
at t = 0 dx/dt = 2
dy/dt / dx/dt = 1/2 = slope
x = 1 and y = 1 at t = 0
y = 1 x + b
b = 0
y = x
dy/dt = cos t -2cost sint -2 sint cost
at t = 0 dy/dt = 1
x = cos t +2 sin t cos t
dx/dt = -sint -2sin^2t+2 cos^2t
at t = 0 dx/dt = 2
dy/dt / dx/dt = 1/2 = slope
x = 1 and y = 1 at t = 0
y = 1 x + b
b = 0
y = x
Answered by
Damon
x = 1 and y = 1 at t = 0
y = .5 x + b
1 = .5 + b
b = .5
y = .5 x + .5
y = .5 x + b
1 = .5 + b
b = .5
y = .5 x + .5
Answered by
Lielle
the 2s aren't exponents, is the process the same?
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