Question

The tangent of the curve f(t) =2+nt+mt^2 at point (1,1/2) is parallel to the normal g(t)= t^2+6t+13(-2,2), calculate the values m and n.​

Answer 1

Explanation:

The slope of g(t) as (-2,2) is given by

`g'(t)=2t+6`

`g'(-2)=2(-2)+6=2`

Since the normal is parallel to the tangent of f(t) at (1, 1/2),

The tangent line of f(t) has a slope of

`(d)/(dx)f(t)=n+2mt=2`

At point (1, 1/2), we know

`f'(1)=n+2m=2`

`f(1)=2+n(1)+m(1)^2=2+n+m=\frac12`

Solving,

`n=5, m=-\frac72`