Question
find equation of tangent line to the curve at the point corresponding to the given value of the parameter. x=t^(7)+1, y=t^(8)+t; t=-1
Answers
oobleck
dy/dx = (dy/dt)/(dx/dt) = (8t^7+1)/(7t^6)
at t = -1, dy/dx = (-8+1)/(7) = -1 at the point (0,1)
so the tangent line is
y-1 = -1(x-0)
or
y = -x+1
at t = -1, dy/dx = (-8+1)/(7) = -1 at the point (0,1)
so the tangent line is
y-1 = -1(x-0)
or
y = -x+1