Asked by Anonymous
Find the equation of the tangent line to
f(x)=sec x - 2 cos x at (pi/3, 1).
f(x)=sec x - 2 cos x at (pi/3, 1).
Answers
Answered by
Damon
f' = sec x tan x + 2 sin x
at x = pi/3 which is 60 deg
f' = m = (2 * sqrt 3) + 2 (sqrt 3/2)
so m = 3 sqrt 3
so y = 3 sqrt 3 * x + b
at (pi/3 , 1)
1 = 3 (pi/3) sqrt 3 + b
b = 1 - pi sqrt 3
so
y = (3 sqrt 3) x + (1-pi sqrt 3)
or approx
y = 5.20 x - 4.44
at x = pi/3 which is 60 deg
f' = m = (2 * sqrt 3) + 2 (sqrt 3/2)
so m = 3 sqrt 3
so y = 3 sqrt 3 * x + b
at (pi/3 , 1)
1 = 3 (pi/3) sqrt 3 + b
b = 1 - pi sqrt 3
so
y = (3 sqrt 3) x + (1-pi sqrt 3)
or approx
y = 5.20 x - 4.44
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