1. freckles

hey

2. anonymous

can u help?

3. freckles

$\text{ we have to use this } \int\limits_{a(t)}^{b(t)} \sqrt{(x')^2+(y')^2} dt \text{ or try to use } \\ \int\limits_{a(x)}^{b(x)} \sqrt{ 1+(y')^2} dx$

4. anonymous

oh okay great

5. freckles

do you want to try to write y in terms of x?

6. anonymous

7. freckles

i'm asking that because if we try to write the x limits in terms of t limits we will end up with 4 numbers 2 of them being imaginary

8. freckles

this question is a tad weird for me it is 0<x<6 right?

9. anonymous

yea it says Find the length of the curve X=1+9t^2, and y+3+6t^3 (0 <t<6)

10. anonymous

that should clarify

11. freckles

oh that makes the problem tons easier you said 0<x<6

12. anonymous

yeah sorry

13. freckles

then just use that one parametric formula I gave you

14. freckles

$\int\limits\limits_{0}^{6}\sqrt{((1+9t^2)')^2+((3+6t^3)')^2} dt$

15. anonymous

i did but my answer wasn't right can u help me check the final answer again?

16. freckles

ok we can check you work what did you get for (1+9t^2)'? and for (3+6t^3)'?

17. anonymous

i got 461 for the final answer, but know thats wrong what did u get?

18. freckles

ok but what did you get for the derivatives of those two expressions I asked you above

19. anonymous

i did but just wanted to check my final answer first

20. freckles

is there anyway you can tell me what you got for the derivative of (1+9t^2) and (3+6t^3)

21. freckles

did you want me to check your problem?

22. freckles

In order for me to do that I need to see where you went wrong if anywhere

23. anonymous

yes i wanted to review the retriceer first then go over questions if possible

24. anonymous

18t and 18t^2 is the detractive

25. anonymous

i just like to see once i have the answer if i can work back and see my error it helps my learning

26. freckles

great! now inputting those in we get: $\int\limits_{0}^{6}\sqrt{(18t)^2+(18t^2)^2} dt \\ \sqrt{18^2} \int\limits_0^6\sqrt{t^2+t^4} dt \\ 18 \int\limits_0^6 \sqrt{t^2+t^4} dt$ did you get to this part?

27. anonymous

yes

28. anonymous

then do i solve from there?? because i got 29200?

29. freckles

notice both terms in the sqrt( ) have a t^2 in common $\text{ since } t>0 \text{ then we can write }\\ 18 \int\limits_0^6 \sqrt{t^2} \sqrt{1+t^2} dt \\ \\ 18 \int\limits_0^6 t \sqrt{1+t^2} dt \\ \text{ now use substituion }$

30. freckles

let u=1+t^2

31. anonymous

i have to go soon because my battery is about to die, could you share the answer?

32. anonymous

is mine right?

33. anonymous

please let me know mat 1%

34. freckles

I haven't done the problem. I was hoping we could do it together. If u=1+t^2 then du/dt=?

35. freckles

do you know power rule?

36. freckles

constant rule says (1)'=0 power rule says (t^2)=?

37. freckles

use wolframalpha if you just want answers