Another rate of change question (sorry)

[cole92]

As I asked previously, I need help with a rate of change question.

I have to find the R.O.C. of the function over the indicatedinterva;, then compare the rate of change with the instantaneous rate of change at the endpoints of the interval:

Function: f(x) = sin x
Interval: [0, pi/6]

My rate of change is pi/3, but then to find the inst. rates I need to take the derivative of pi/3 .... (this is where I am having issues) and plug in my interval points. What is the derivative of pi over 3?

 

[Deleted member 4993]

As I asked previously, I need help with a rate of change question.

I have to find the R.O.C. of the function over the indicated interval, then compare the rate of change with the instantaneous rate of change at the endpoints of the interval:

Function: f(x) = sin x
Interval: [0, pi/6]

My rate of change is pi/3, but then to find the inst. rates I need to take the derivative of pi/3 ... NO!

You are trying find ROC of sin (x) - NOT pi/3 (which is a constant and does not change)

So you need to get derivative of sin(x)



.... (this is where I am having issues) and plug in my interval points. What is the derivative of pi over 3?

By the way:

My rate of change is pi/3 ........... is incorrect. How did you find that value? Please show work.

 

[cole92]

As I asked previously, I need help with a rate of change question.

I have to find the R.O.C. of the function over the indicated interval, then compare the rate of change with the instantaneous rate of change at the endpoints of the interval:

Function: f(x) = sin x
Interval: [0, pi/6]

My rate of change is pi/3, but then to find the inst. rates I need to take the derivative of pi/3 ... NO!

You are trying find ROC of sin (x) - NOT pi/3 (which is a constant and does not change)

So you need to get derivative of sin(x)



.... (this is where I am having issues) and plug in my interval points. What is the derivative of pi over 3?

I may have confused you with my explination, but I thought I was doing it correctly based on how I was told to solve my previous question....

With the function f(x) = sin x I take the interval, [0, pi/6] and put it in an equation such that:
[(1/2) - 0] \ [(pi/6) - 0] = [(1/2) / (pi/6)] which = pi/3

From there I am supposed to plug in my two interval points into the derivative of that answer to find the instantaneous rates.


FOR EXAMPLE:

In my previous question the function was f(t) = t^2 - 3 ..... doing the same thing mentioned above you get an avg rate of 4.1 and inst. rates of 4 and 4.2


Correct?

 

[Deleted member 4993]

With the function f(x) = sin x I take the interval, [0, pi/6] and put it in an equation such that:
[(1/2) - 0] \ [(pi/6) - 0] = [(1/2) / (pi/6)] which = pi/3...No!

You are correct upto

Average ROC = [(1/2) / (pi/6)]

Then

= 1/2 * 6/pi = 3/pi

From there I am supposed to plug in my two interval points into the derivative of that answer to find the instantaneous rates.

That is not correct - you need to read it again. You have to take the derivative of the original function - it has "almost" nothing to do with the average ROC you have calculated.



FOR EXAMPLE:

In my previous question the function was f(t) = t^2 - 3 ..... doing the same thing mentioned above you get an avg rate of 4.1

You had said - in that post

The derivative would be 2t, so do I plug in 2 and 2.1 for t?
If so that would make the rates 4 and 4.2, correct?

and inst. rates of 4 and 4.2 <<< Did you take derivative of (4.1) here - or (t[sup:7plxjtr9]2[/sup:7plxjtr9] -3)


Correct?

 

[cole92]

Oh, I think I may understand now. So the Avg. R.O.C. is 3/pi, but I take the derivative of sin x, which I believe is (- cos x)?

So then I would plug in my two interval values, giving me
f(0) = -1
and
f(pi/6) = negative [sqrt(3)/2]

???

 

[Deleted member 4993]

Oh, I think I may understand now. So the Avg. R.O.C. is 3/pi, but I take the derivative of sin x, which I believe is (+ cos x)?

So then I would plug in my two interval values, giving me
f(0) = -1
and
f(pi/6) = negative [sqrt(3)/2]

???

 

[cole92]

Oh yea. I knew that, sorry.

It is when you take the drivative of cos x that you get (- sin x)

So then my answers would e the same, just positive, correct?