Find an equation of the tangent line to the curve at the given point . y = sec(x) − 2 cos(x), P = (𝜋 /3 , 1)

Question

Don't know how to due my calc homework due at 11:59.

Answer 1

y = secx - 2cosx

y' = secxtanx + 2sinx

y'(pi/3) = sec(pi/3)tan(pi/3) + 2sin(pi/3)

= 2(sqr3/2) + 2sqr3/2 = sqr3 + sqr3 = 2sqr3

= about 2(1.732 = 3.464 = slope of the tangent line at x=pi/3

the tangent line equation is y= 2sqr3(x) + b. plug in the point (pi/3, 1) to solve for b

1 = 2sqr3(pi/3) + b

b = 1 -(2/3)pisqr3 = about -2.627

y = 2sqr3(x) + 1 - (2/3)pi(sqr3)

= about 3.464x - 2.627