your dy/dx should be
-2sint + 2cos2t
= 0
recall that one of the equivalents for cos 2t is
1 - 2sin^2 t
so
-2sint + 2(1 - 2sin^2 t) = 0
-sint + 1 - 2sin^2 t = 0
2sin^2 t + sint - 1= 0
(2sint - 1)(sint + 1) = 0
sint = 1/2 or sint = -1
but we want quadant I ,so sint = 1/2
t = π/6
how about the end points ?
for t = 0
y = 2cos0 + sin0 = 2 + 0 = 2
for t = π/2
y = 2cos π/2 + sin π = 0 + 0 = 0
for t = π/6
y = 2cos π/6 + sin π/3
= 2(√3/2) + 1/2
= √3 + 1/2 or (2√3 + 1)/2 which is > 2
so the max is (2√3 + 1)/2 and the min is 0
find the absolute maximum and minimum of the function y=2cos(t)+sin(2t) on the interval of [0, pi/2]
I have taken the derivative but I have no clue how to solve it for 0
1 answer