You might want to factor out a g on the right side... I'm assuming g is unknown? If not, you could just sub in for g... but, otherwise, factoring will do wonders.
A bonus question on one of my first calculus quizzes tomorrow for the semester:
V=Potential Energy
V=2g(1-costheta) ... compute the average Potential Energy over the interval [pi/4, pi/3]
Solve in terms of g
So I set it up and went through the motions...but am kind of stuck
v= 2g - 2gcos(pi/3) - 2g - 2gcos(pi/4) / pi/3 - pi/4
... then what?
ok continuing...
v= 2g - 2gcos(pi/3) - 2g - 2gcos(pi/4) / pi/3 - pi/4
v= -2g(1/2+sqrt(2)/2) / pi/3 - pi 4
...since this question involves cos, can you change pi/3 and pi/4 to 1/2 and sqrt(2)/2, respectively?
if so v = -2g(1/2+sqrt(2)/2) / 1/2 - sqrt(2)/2
...any more guidance?
Unfortunately, the original method is wrong so pretty much everything everyone has said is off the point. The problem is NOT just simplifying that equation, the equation itself is wrong.
You do NOT find the average of a function, f(x) on an interval, [a, b], by , any more than you find the average of a list of numbers by simply averaging the first and last values. you have to sum the numbers which, for functions, means "integrate".
The average of f(x) on the interval [a, b] is
The average you want is
Halls of Ivy is correct but for the sake of argument suppose the question
is what is the average rate of change of the potential energy then
[V(pi/3) -V(pi/4)]/(pi/3-pi/4)
Be careful with parenteheses 2g[1-cos(pi/3)- (1-cos(pi/4)]/pi/12
= 2g[1/2-1+sqrt(2)/2]/pi/12
= 24g[-1/2 +sqrt(2)/2]/pi
=g/pi [ -12 +12sqrt(2)]
g is of course the acceleration due to gravity = -32ft/sec^2 or
- 9.8m/sec^2
G is the universal gravitation constant G = 6.673 x10^(-11)Nm^2/kg^2