Can you help me on this

1. Mar 13, 2005

smith5029

hey guys i'm having trouble on these things.

ok here it is f(t):3cos^2(5t), the domain is [-pi/5,pi/5].

I have to find the concavity and where the curve increases and decreases. I took the second derivative and got the inflection point at .157.

2. Mar 14, 2005

HallsofIvy

Staff Emeritus
Well, lets take a look at it. f(t)= 3 cos2(5t) so
f'(t)= 6 cos(5t)(-sin(5t))(5)= -30 cos(5t)sin(5t).
f"(t)= (-30)(- 5 sin(5t)sin(5t)- 30 cos(5t)(5cos(5t)= -150(cos2(5t)- sin2(5t))= -150cos(10t). That will be 0 when 10t is an odd multiple of pi/2: that is, when pi= pi/10, -pi/10, 3pi/10, -3pi/10. None of those is equal to 0.157 (pi/20?). Since the second derivative can change sign only at those points, for t between -pi/10 and pi/10, f"(t) has the same sign as f"(0)= -150. f(t) is concave downward for -pi/10< t< pi/10, positive for pi/10< t< 3pi/10, etc.

Of course, the curve increases and decreases where f'(t) is positive and negative, respectively.