AmTran_Bus one year ago Find the arc length of ellipse x=4sintheta and y=3costheta

1. AmTran_Bus

So I know you can say$\int\limits_{a}^{b}\sqrt{1+(dy/dx)^2}$

2. AmTran_Bus

and also $\frac{ x }{ 4 }=\sin \theta , \frac{ y }{ 3 }= \cos \theta$

3. AmTran_Bus

so (x/4)^2 +(y/3)^2 =1

4. AmTran_Bus

Taking the derivative, you got 4 cos theta and -3 sin theta

5. AmTran_Bus

@amistre64

6. amistre64

id use: $\int \sqrt{(x')^2+(y')^2}~dt$

7. AmTran_Bus

That isnt the formula in my book? How did you get it and why does it work? Just from the trig identity?

8. amistre64

its the general setup for the arclength "s", which is just the pythag thrm s, the section, squared; is equal to sum of the square of the parts |dw:1438546179447:dw| s^2 = x^2 + y^2, or simply s = sqrt(x^2 + y^2) as s and x and y get small ...

9. amistre64

|dw:1438546232411:dw|

10. amistre64

ds = sqrt(dx^2 + dy^2) now when all of these are "with respect to x", then dx=1 ds = sqrt(1 + (y')^2) is the initial introducion

11. amistre64

if it is with respect to "t", dx = x'

12. AmTran_Bus

Ok, well thank you then. So all I have is$\int\limits_{0}^{2\pi}\sqrt{16\cos^2 \theta+9\sin^2 \theta }$

13. AmTran_Bus

*d theta

14. amistre64

yep

15. AmTran_Bus

Ok. And that is an answer choice.

16. AmTran_Bus

Thanks.

17. amistre64

lets hope memory serves us right then lol

18. AmTran_Bus

Yea, for real. I apparently do not get this real well, cause its still fuzzy why every other example they have said use the (dy/dx)^2 thing.

19. AmTran_Bus

That is, the very first one I posted.

20. amistre64

we could define y in terms of x if you are working with parameters of course .. but its just extra work

21. AmTran_Bus

Ok.

22. amistre64

the thing about arclength is, we dont want to be confined to a coordinant system ...

23. amistre64

anyways, thats my perspective :)