Sunday, July 2, 2017

Calculus of a Single Variable, Chapter 2, 2.3, Section 2.3, Problem 68

You need to evaluate the equation of the tangent line to the curve f(x) = sec x , t the point ((pi)/3, 2), using the following formula, such that:
f(x) - f((pi)/3) = f'((pi)/3)(x - (pi)/3)
Notice that f((pi)/3) = 2.
You need to evaluate f'(x) and then f'((pi)/3):
f'(x) = (sec x)' f'(x) = sec x*tan x => f'((pi)/3) = sec ((pi)/3) *tan ((pi)/3)
sec ((pi)/3) = 1/(cos((pi)/3)) = 2
tan ((pi)/3) = sqrt 3
f'((pi)/3) = 2sqrt 3
You need to replace the values into the equation of tangent line:
f(x) - 2 = 2sqrt 3*(x - (pi)/3)
Hence, evaluating the equation of the tangent line to te given curve , at the given point, yields f(x) = 2*(sqrt 3)*x - 2*(sqrt 3)*(pi)/3 + 2.

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