# Calculus - find average rate of change of the function over a given interval

1. Oct 6, 2009

### mastdesi

Calculus - plz help find average rate of change of the function over a given interval

1. The problem statement, all variables and given/known data

h(t) = sin t, [3pi/4,4pi/3]

plz help me solve this, try to give me an explanation on every step plz. i checked but i can't find anything on this in the book. i am basically having problem with this because of the sin.

2. Relevant equations

3. The attempt at a solution

i got to there also but i dont know how to solve the sin part. how does the square root come in.

2. Oct 6, 2009

### Staff: Mentor

Re: Calculus - plz help find average rate of change of the function over a given inte

Do you know sin($\pi/3$) = sin(60 deg.)?
Do you know sin($\pi/4$) = sin(45 deg.)?
The sines of these angles are numerically equal to the sines, respectively, of the two angles you showed. There are a few angles whose sine, cosine, and tangent you should memorize.

Last edited: Oct 6, 2009
3. Oct 6, 2009

### mastdesi

Re: Calculus - plz help find average rate of change of the function over a given inte

[sin(4π/3) - sin(3π/4)] / (4π/3 - 3π/4)
= (-√3/2 - √2/2) / (7π/12)
= -6(√3 - √2) / (7π)

this is what i got but its none of the 4 multiple choice answers. whats wrong in here?

4. Oct 6, 2009

### Office_Shredder

Staff Emeritus
Re: Calculus - plz help find average rate of change of the function over a given inte

Your last line should be -6(√3+√2)/(7π)

You didn't factor the - sign properly

5. Oct 6, 2009

### mastdesi

Re: Calculus - plz help find average rate of change of the function over a given inte

Thank you very much. but if you guyz can plz explain a lil clearly to me how do u go from this step : (-√3/2 - √2/2) / (7π/12))
to this:
= -6(√3 - √2) / (7π)

6. Oct 6, 2009

### Staff: Mentor

Re: Calculus - plz help find average rate of change of the function over a given inte

(-√3/2 - √2/2) / (7π/12)) = -1/2(√3 + √2) * 12/(7π) = -6(√3 + √2)/(7π)

BTW, my earlier post was slightly off: sin(π/3) = -sin(4π/3). Looks like you caught that.