find min and max of f(t)=2COSt + SIN2t on [0, pi/2]
- Thread starter member566
- Start date
- Joined
- Feb 4, 2004
- Messages
- 16,582
If you haven't yet memorized the trig derivatives they gave you, find that table and/or section in your book now, and get to work.
As for finding the derivative, this is a simple application of the above formulas and the Chain Rule. The critical numbers are found by setting f' equal to zero, and solving, just like you solved trig equations back in trigonometry.
Once you have reviewed your trig derivatives, please attempt the exercise. If you get stuck, please reply showing all of your work and reasoning. Thank you!
Eliz.
As for finding the derivative, this is a simple application of the above formulas and the Chain Rule. The critical numbers are found by setting f' equal to zero, and solving, just like you solved trig equations back in trigonometry.
Once you have reviewed your trig derivatives, please attempt the exercise. If you get stuck, please reply showing all of your work and reasoning. Thank you!
Eliz.
find min and max
f(t) = 2Cost + Sin2t [0, pi/2]
f is continuous on all real numbers
f'(t) = -2Sint + 2Cos2t =0
2 2
-Sint + Cos2t = 2 f(pi/6) = 2Cospi/6 + Sin2 pi/6
2(2^1/2)/2 + (3^1/2)/2
-Sint + Sin^2t = 0 2^1/2 + 3^1/2
=3
2Sin^2t + sint -1 = 0
(2Sint - 1) (Sint + 1) = 0 f(0) = 2Cos(0) +Sin 2(0)
2(1) + Sin(0) pi
2Sint - 1 = 0 Sint + 1 = 0 2 + 0
=2
2/2Sint = 1/2 Sint = -1
Sint = 1/2 Sint = -1 f(pi/2) = 2Cos(pi/2) + Sin 2 pi/2
2(1/2) + sin(0)
= pi/6 =3pi/2 = 1
Am I missing anything,
Thank you
f(t) = 2Cost + Sin2t [0, pi/2]
f is continuous on all real numbers
f'(t) = -2Sint + 2Cos2t =0
2 2
-Sint + Cos2t = 2 f(pi/6) = 2Cospi/6 + Sin2 pi/6
2(2^1/2)/2 + (3^1/2)/2
-Sint + Sin^2t = 0 2^1/2 + 3^1/2
=3
2Sin^2t + sint -1 = 0
(2Sint - 1) (Sint + 1) = 0 f(0) = 2Cos(0) +Sin 2(0)
2(1) + Sin(0) pi
2Sint - 1 = 0 Sint + 1 = 0 2 + 0
=2
2/2Sint = 1/2 Sint = -1
Sint = 1/2 Sint = -1 f(pi/2) = 2Cos(pi/2) + Sin 2 pi/2
2(1/2) + sin(0)
= pi/6 =3pi/2 = 1
Am I missing anything,
Thank you