Question

find an equation of the tangent line to the curve for the given value of t.
x=t^2-2t, y=t^2+2t when t=1 find an equation of the tangent line to the curve for the given value of t.
x=t^2-2t, y=t^2+2t when t=1 @Mathematics

Answer 1

ok gimme a sec

Answer 2

slope of tangent like = 0

Answer 3

dy/dt = 2t + 2
dx/dt = 2t - 2
dy/dx = (2t + 2)/(2t - 2)
plug in 1:
(2(1) + 2)/(2(1) - 2) = 4/0 => PROBLEM as dx/dt = 0 at t = 1

Answer 4

at t=1, x= 1-2= -1 and y= 1+2 =3. (-1,3)
the slope is 4/0 which indicates a vertical line tangent to the point (-1,3). so the equation of the tangent line is
x= -1