## anonymous one year ago A bee with velocity vector r'(t) starts out at the origin at t=0 and flies around for T seconds. Where is the bee located at time T if ∫(0-T) r'(u)du = 0? Also, what does the quantity ∫(0-T) ||r'(u)||du represent?

1. anonymous

2. amistre64

what does (0-T) represent?

3. anonymous

integral from 0 to T

4. anonymous

sorry it's the only way I know how to type it

5. amistre64

so those are the limits, ok

6. amistre64

do you recall the definition of: $\int_{a}^{b}f(x)~dx=??$

7. amistre64

the fundamental thrm of calculus is also what it is called

8. anonymous

It's F(b) - F(a)

9. amistre64

correct, so if F(b)-F(a) = 0, then either a=b, or F has the same value at multiple input levels similar to a periodic function.

10. amistre64

for example, cos(pi) - cos(0) = 0 since cos(pi)=cos(0)

11. amistre64

but the question is, what does taking the integral of a function tell us?

12. anonymous

it can tell you what the area under a curve is

13. amistre64

and my example is spose to read cos(2pi) = cos(0), but the site is lagging for me

14. anonymous

it's ok

15. amistre64

the area under a curve, yes ... but that reflects displacement

16. amistre64

ever wonder why in some cases we get a negative area? and in other cases we dont?

17. anonymous

Yes because the integral of velocity is displacement

18. anonymous

right?

19. anonymous

negative displacement ?

20. amistre64

when we want to find the area between 2 curves, area is a positive value always ... its an absolute value. And in those cases we have to concern ourselves with the places that one curve crosses over the other.

21. amistre64

displacement on the other hand reflects is a vector, it has magnitude and direction.

22. amistre64

F(b)-F(a) = 0 assumes that at a and b, we are in the same place. Either a = b, or we have simply ended were we started.

23. anonymous

Then the location of the bee is still the origin at time T?

24. amistre64

yes, at time T, the bee is simply back in the same place it was at when the time was equal to 0

25. anonymous

But why substitute u in for t for the integration? That's the most confusing part. Does that have any effect on the result?

26. amistre64

0 and T are values that the variable u can take on. the function itself is dependant on the value of u ...

27. anonymous

What is the difference between ∫(0-T) r'(u)du = 0 and ∫(0-T) r'(t)dt = 0 Why did they substitute ?

28. amistre64

hmm, im not absolutely sure, but i think it has something to do with what is called a 'dummy' variable - meant to avoid confusion between the overuse of the letter T

29. anonymous

oh ok .. well thank you !

30. amistre64

good luck