## anonymous 5 years ago For position function s(t):=20-5t^2, if h is a small number, find the average rate of change of s(t) on the interval [2,2+h]. (Simplify fully)

1. amistre64

ug ..... i spose

2. anonymous

?

3. amistre64

s(2+h) -s(2) ----------- h

4. amistre64

its alot of work just to prove that derivatives work ....

5. anonymous

lol true.

6. amistre64

20-5(2+h)^2 -(20-5(2)^2) ------------------------ h 20-5(4 +4h +h^2) -20 +20 ------------------------ h 20-20-20h-5h^2 ---------------- h -20h-5h^2 ---------- = -20 -5h h

7. anonymous

I think I have to use the formula (f(a)-f(x))/(a-x)

8. anonymous

Can you explain to me how to use that formula?

9. amistre64

now for the derivative we get: -10t ; at t=2 we get -20 when h=0, we get -20

10. amistre64

its the same formula

11. anonymous

oh no derivatives yet

12. anonymous

can you explain to me how to use it? like what's a and what's x?

13. anonymous

on the interval

14. amistre64

yes, you are doing the proof for derivatives so that in the next few weeks your teacher will say, and this is how we do it the easy way...

15. anonymous

haha ok.

16. amistre64

f(a) and f(x) simply mean to use the formula with an a and an x instead of a t like this: f(a) - f(x) (20-5a^2) - (20-5x^2) ------------------- a-x

17. amistre64

20-5a^2 -20+5x^2 ----------------- a - x -5a^2+5x^2 ------------- a - x -5(a^2-x^2) ------------- a - x -5(a+x)(a-x) ------------- a - x -5(a+x)

18. anonymous

ok, what do you plug into f(a)?

19. anonymous

in [2,2+h]

20. amistre64

.............. a = 2 and x = h

21. anonymous

?

22. anonymous

is a=2 from the 2+h part or from just the 2 part.

23. amistre64

it doesnt matter..... a =2 OR a=h, OR x=2 OR x=h.... -5(a+x) -5(2+h) is the same things as -5(h+2)

24. anonymous

I'm confused.

25. anonymous

Like, I know the forumla, I just don't understand how to use it.

26. amistre64

then dont use it lol

27. amistre64

does it matter if f(a) = 2 and f(x) = (2+h)? no

28. amistre64

becasuse when h=0 it just disappears to -20

29. amistre64

$\frac{y-y_0}{x-x_0}$ do you understand this formula?

30. anonymous

yes.

31. amistre64

that is all it is; but they rename that parts ..... but it doesnt matter what they name them, they still do the same things

32. anonymous

so the bottom part is a-x... don't you put 2-h?

33. anonymous

why did you only put h?

34. amistre64

$\frac{f(a)-f(x)}{a-x} \iff \frac{y-y_0}{x-x_0} \iff \frac{f(x+h)-f(x)}{h}$

35. anonymous

oh ok.

36. amistre64

because $$(x+h)-(x)=h$$

37. anonymous

why?

38. amistre64

why what? why does x-x+h = 0+h = h?

39. anonymous

oh ok. just kidding.

40. amistre64

$(x+h)-(x)$ $x+h-x$ $x-x+h$ $(x-x)+h$ $0+h$ $h$

41. anonymous

so the formula would be $(f(2+h)-f(2))\div(h)$

42. amistre64

yes

43. anonymous

ok.

44. anonymous

so its ((20-5(2+h)^2)-(20-5(2)^2))/h

45. amistre64

yes

46. anonymous

so (20-5(2+h)^2)/h

47. anonymous

and finally -20-h?

48. amistre64

yes

49. anonymous

:) Thank you