Posted by stephanie on Thursday, June 21, 2012 at 3:12pm.
π is PI not PIE!
a)
y'(x) = -3sinx + 4cosx
y'(π) = -3sin π + 4cos π = -4
y(π) = 3(-1) + 0 + 1 = -2
thus, the line through (π,-2) having slope -4 is
(y+2) = -4(x-pi)
b) y' = 1 + 3sinx
y'(5π/3) = 1 + 3(-√3/2) = 1 - 3√3/2
y(5π/3) = 5π/3 - 3(1/2) = 5π/3 - 3/2
(y-(5π/3 - 3/2) = (1 - 3√3/2)(x-5π/3)
thank you so much steve!
for b) (y-(5π/3 - 3/2) = (1 - 3√3/2)(x-5π/3)
is the square root for the whole 3/2 or is it just root 3 OVER 2.
I think by now you know the value of cos(pi/3)
(y-(5π/3 - 3/2) = (1 - (3√3)/2)(x-5π/3)
like that ? :(
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