# Math Help - Just starting calculus...computing the average...guidance/help please

1. ## Just starting calculus...computing the average...guidance/help please

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?

2. 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.

3. Originally Posted by seichan
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.
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?

4. If you want to solve for g, you might be interested in isolating the g, such that you have a function of the form
g=(something)

I'm pretty sure this is what your question is asking for

5. Except the whole thing is already equal to v .... hm

6. I am assuming you do not know the value of V or of g? If so, then you are done. If you know the value of V, than solving such that g=(something) will give you an answer. If you know g, then you should just substitute the value for g in...

7. Sorry if I am being unclear....we are solving for V....in terms of G. We do not know either, G on earth is 9.81 m/s, but he said "We could be on another planet for all we know, so leave it in terms of g."

8. Alright then, you should be done you already solved for V in terms of g, and, if he wants it left in terms of g, you're done!

9. Originally Posted by Dreamshot
Sorry if I am being unclear....we are solving for V....in terms of G. We do not know either, G on earth is 9.81 m/s, but he said "We could be on another planet for all we know, so leave it in terms of g."
Please do not confuse "g" and "G". they have completely different meanings!

10. 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 $\frac{f(b)- f(a)}{b- a}$, 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
$\frac{1}{b-a} \int_a^b f(x)dx$

The average you want is

$\frac{2g\int_{\pi/4}^{\pi/3} (1- cos(\theta))d\theta}{\frac{\pi}{3}- \frac{\pi}{4}}$

11. Originally Posted by HallsofIvy
The average of f(x) on the interval [a, b] is
$\frac{1}{b-a} \int_a^b f(x)dx$
provided $f$ is continuous on $[a,b]$.
Just an important addition.

12. Originally Posted by HallsofIvy
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 $\frac{f(b)- f(a)}{b- a}$, 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
$\frac{1}{b-a} \int_a^b f(x)dx$

The average you want is

$\frac{2g\int_{\pi/4}^{\pi/3} (1- cos(\theta))d\theta}{\frac{\pi}{3}- \frac{\pi}{4}}$
This helps a lot, but we haven't done derivatives yet or integration for that matter. Could you show me a process, if possible, of solving the problem without using those two techniques?

13. 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

14. Originally Posted by Calculus26
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
On this step:
2g[1-cos(pi/3)- (1-cos(pi/4)]/pi/12

Wouldn't it be 2g[1-cos(pi/3)-2g(1-cos(pi/4)]/pi/12 ?

Since its an average rate of change, do you include the entire equation?

15. I factored out the 2g so the entire equation for V was included