calculus
posted by Anonymous .
Find an equation of the tangent line to the curve at the given point.
y = 7 sec x
P = (π/3, 14)

y(x)=7sec(x)
y'(x)=7sec(x)tan(x)
at x=π/3,
slope of tangent, m =
y'(x)= 7*sec(π/3)tan(π/3)
=7*2*sqrt(3)
=14sqrt(3)
We now look for a line passing through P1(x0,y0)=(π/3,14) with a slope of m=14sqrt(3).
The equation to use is:
(yy0)=m(xx0)
substitute x0=π/3, y0=14, m=14sqrt(3) and simplify to get equation of tangent.