## jennychan12 2 years ago AP Calculus AB 2003 #2 http://www.collegeboard.com/prod_downloads/ap/students/calculus/calculus_ab_frq_03.pdf I don't understand how to do part c).

1. bahrom7893

total distance is integral of velocity but u gotta take into consideration which way it's going

2. jennychan12

I looked at the sample respoinses and they did like absolute value of v(t). But I don't understand why. http://apcentral.collegeboard.com/apc/public/repository/ap03_calculusab_q2_27980.pdf

3. bahrom7893

so you found where it changes direction right?

4. bahrom7893

the thing is...|dw:1357322365854:dw|

5. bahrom7893

if u take just the integral, the area on the bottom gets subtracted

6. bahrom7893

but that's for displacement. So if u want to get the distance you need to add that area instead of subtracting it.. So you have to integrate from start till when it changed direction, then from when it changed direction till when it changed direction again, etc.. and finally take the absolute value of all integrals and add them up

7. bahrom7893

|dw:1357322510296:dw| see if u start at pt A, displacement, which is the regular integral, is 0, but distance traveled, is 5

8. jennychan12

i thought to find the distance u have to take integral of the velocity.

9. bahrom7893

yes you do.. integral of velocity is indeed the area under the velocity curve, which is the displacement

10. bahrom7893

so in order to get the DISTANCE.. you want to add all the areas that end up below y- axis, instead of subtracting them

11. bahrom7893

if you were looking for DISPLACEMENT, you would just take the integral of that function from 0 to 3. But, since you're looking for DISTANCE, take the integral of that function from 0 to a, where a is the time at which particle changed direction.

12. bahrom7893

okay listen, where did the particle change direction?

13. jennychan12

14. bahrom7893

0?

15. jennychan12

yeah those were the solutions but only rad(2pi) works

16. bahrom7893

no it was just t = 2.507

17. jennychan12

18. bahrom7893

so integrate that function from 0 to 2.507, then from 2.507 to 3

19. bahrom7893

once you get the results, say Int from 0 to 2.507 = A, and Int from 2.507 to 3 = B; just add up the absolute values: |A| + |B|

20. bahrom7893

that's all he's doing by integrating the absolute value of V over 0 < t < 3

21. jennychan12

how'd u integrate the function tho? like algebraically?

22. bahrom7893

isn't that the calculator part?

23. bahrom7893

i dont think u can integrate exponents inside trig functions yet.. lol

24. jennychan12

or do u just plug it into the graphing calculator? ohhh -_- ok.

25. bahrom7893

i mean i dont think there is a way that anyone has found how to do that

26. bahrom7893

lol yea

27. jennychan12

ohhh yeah i got the answer when i plugged it into the calculator

28. bahrom7893

You mean to tell me.... I WASTED MY TIME???!!!!!!

29. jennychan12

nooooooo...... like after you told me all that...

30. bahrom7893

k good lol

31. jennychan12

sorry..if anyone's still viewing this for part d) why did they add 1 to rad(2pi) ?