calculus
posted by Bonaventutre .
Calculate the equation of the tangent and normal lines for the curve f(x) = tan 2x, at the point where the xcoordinate is: x = π/8.

dy/dx = 2sec^2 (2x)
when x = π/8
f(π/8) = tan π/4 = 1
so our point of contact is (π/8, 1)
slope of tangent = 2 sec^2 (π/4)
= 2(√2/1)^2 = 4
equation of tangent:
y  1 = 4(x  π/8)
y  1 = 4x  π/2
y = 4x + 1π/2
normal has slope 1/4
y = (1/4)x + b
but the point (π/8,1) lies on it, so
1 = (1/4)(π/8) + b
b = 1 + π/32
y = (1/4)x + 1+π/32
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