# given the function (theta sign) = 3 cos (theta sign - pie/6) +1 a) Calculate the average rate of change of the function from (theta) = pie/2 to 2(pie)/3 b) calculate the instantaneous rate of...

given the function (theta sign) = 3 cos (theta sign - pie/6) +1

a) Calculate the average rate of change of the function from (theta) = pie/2 to 2(pie)/3

b) calculate the instantaneous rate of change at (theta) = 2(pie)/3

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### 2 Answers

a) You need to remember the average rate of change formula such that:

`(dy)/(dx) = (f(2pi/3) - f(pi/2))/(2pi/3 - pi/2)`

You need to evaluate `f(2pi/3), ` hence you need to substitute `2pi/3` for theta in `f(theta) = 3cos(theta - pi/6)+ 1` such that:

`f(2pi/3) = 3cos(2pi/3 - pi/6)+ 1 =gt f(2pi/3) = 3cos(3pi/6) + 1` (you need to bring the terms in brackets to a common denominator)

`f(2pi/3) = 3cos(pi/2) + 1 `

You need to substitute 0 for cos `pi/2 ` such that:

`f(2pi/3) =3*0 + 1 =gt f(2pi/3) = 1 `

You need to evaluate `f(pi/32), ` hence you need to substitute `pi/2` for theta in `f(theta) = 3cos(theta - pi/6)+ 1` such that:

`f(pi/2) = 3cos(pi/2 - pi/6)+ 1`

`f(pi/2) = 3cos(2pi/6)+ 1`

`f(pi/2) = 3cos(pi/3)+ 1`

`f(pi/2) = 3(1/2)+ 1 =gt f(pi/2) = 3/2+ 1 = 5/2`

You need to evaluate the average rate of change such that:

`(dy)/(dx) = (1 - 5/2)/(2pi/3 - pi/2)`

`(dy)/(dx) = (-3/2)/(4pi/6 - 3pi/6)`

`(dy)/(dx) = (-3/2)/(pi/6) =gt (dy)/(dx) =-18/(2pi)`

`(dy)/(dx) =-9/pi`

**Hence, evaluating the average rate of change of function between `pi/2` and `2pi/3` yields `(dy)/(dx) =-9/pi` .**

**b) **You need to evaluate the instantaneous rate of change at `theta = 2pi/3` , hence you need to evaluate`f'(2pi/3), ` but first you need to find `f'(theta)` such that:

`f'(theta) = -3sin(theta - pi/6)*(theta - pi/6)'`

`f'(theta) = -3sin(theta - pi/6)`

You need to substitute `2pi/3` for theta in `f'(theta)` such that:

`f'(2pi/3) = -3sin(2pi/3 - pi/6)`

You need to bring the terms inside brackets to a common denominator such that:

`f'(2pi/3) = -3sin(4pi/6 - pi/6)`

`f'(2pi/3) = -3sin(3pi/6) =gt f'(2pi/3) = -3sin(pi/2)`

`f'(2pi/3) = -3 (sin pi/2 = 1)`

**Hence, evaluating the instantaneous rate of change at `theta = 2pi/3` yields `f'(2pi/3) = -3` .**

- You are asked to find the average rate of change of the function from pie/2 to 2pie/3 which now I will call a and b respectively to make it easier for you to comprehend. First it is very important that you remember average rate of change of a function f with respect to theta over the given interval [a, b] is {f(b)-f(a)}/(b-a). Thus it can be interpreted as the slope of the secant line connecting points at a and b. Now that you know what average rate of change is, calculate f(a) and f(b). f(a)=3.9995 f(b)=3.99887. Now divided f(b)-f(a) by (b-a). You should get (using your graphing calculator) about -0.001196.

- Then you are asked to calculate the instantaneous rate of change which means that it is the limit of the average rate or simply put, differentiate the function and plug in 2pie/3. If you differentiate function f(theta) you should get f'(thetha)= -3sin(theta-pie/6). Now all you have to do is plug is 2pie/3.