Question

what is the equation (not the answer) for the total arc length of an ellipse when x=4sin(theta) and y=3cos(theta)

Answer 1

line integral
integrate from 0 to 2 pi ;Sqrt[(x')^2 +(y')^2]

Answer 2

Its multiple choice and there are two answers with that, one where the whole integral is multplied by for and the other is multiplied by 16

Answer 3

4* not for

Answer 4

oh, that just symmetry

Answer 5

what are the limits?

Answer 6

they are both 0 to 2pi

Answer 7

what was the integrand

Answer 8

Sqrt[cos^2(x)+9/16 Sin^2(x)] ?

Answer 9

\[16\int\limits_{0}^{2\pi}\sqrt{16\sin^{2}\theta+9\cos^{2}\theta} d \theta\]
or
\[4\int\limits_{0}^{2\pi}\sqrt{16\sin^{2}\theta+9\cos^{2}\theta} d \theta\]

Answer 10

I don't know where 4 and 16 are coming from because we are not using symetry

Answer 11

thats what im confused about too.

Answer 12

there are 4 answers. all of them have the same thing under the radical, its the limits and the number its multiplied by that are varied

Answer 13

put all of them up

Answer 14

others are o to theta...

Answer 15

a) \[4\int\limits_{0}^{\pi/2}\]
b) \[16\int\limits_{0}^{2 \pi}\]
c)\[16\int\limits_{0}^{ \pi}\]
d)\[4\int\limits_{0}^{2 \pi}\]

Answer 16

a) because we are doing quater and using symmtary

Answer 17

thats what i thought too, but that was wrong

Answer 18

are you absolutely sure the integrand are same, lol
because other than that I think we are right

Answer 19

yes i am positive haha.

Answer 20

I have had 3 different people look at it and we all come to the same conclusion so im thinking maybe its a mistake on the problem

Answer 21

I used to see it as a sqrt (1- sin^2t) 0 to theta

Answer 22

The 1 - sin^2 is just a trig rewrite...

Answer 23

k sin^2, sorry...

Answer 24

estudier that is only if y is function of x

Answer 25

?

Answer 26

x=4sin(theta) and y=3cos(theta), right?

Answer 27

correct

Answer 28

\[\int \sqrt{1+(dy/dx)^2}\] ?

Answer 29

No, I mean u can use sin^2 + cos^2 = 1 to rewrite....

Answer 30

Here , confirmed
http://www.wolframalpha.com/input/?i=x^2%2F16+%2By^2%2F9%3D1++arclenght
http://www.wolframalpha.com/input/?i=int%20of%20sqrt%5B16%20sin^2%20theta%2B9%20cos^2%20theta%5D%20from%200%20to%202pi&t=crmtb01

Answer 31

there should be no number in front of integral

Answer 32

Anyway, we are talking about the same thing just restricted to certain angles..

Answer 33

there are two links up the , btw

Answer 34

They have a 12 and a 16 in the answers.

Answer 35

I meant the numerical answer 22 is same thing for letting wolfram figure out arc length based on equation and letting it figure out integral

Answer 36

Agree but E is the elliptic integral,,,

Answer 37

Which is what u are looking for here, right?

Answer 38

looking for arclength of ellipse,

Answer 39

That is the elliptic integral....

Answer 40

Based on the Wolfram links, I would go with the 16...

Answer 41

but we get same result for
\[\int\limits_{0}^{2\pi}\sqrt{16\sin^{2}\theta+9\cos^{2}\theta} d \theta\]=22.1

letting wolfram figure out arclength based on ellipse equation =22.1

Answer 42

It's the diff between E(7/16) and E(-7/9) which must be to do with the limits...

Answer 43

See, if u rewrite the cos^2 + sin^2 to sin^2 only, u get the 7.

Answer 44

but numerical answer is same which must mean the
\[\int\limits_{0}^{2\pi}\sqrt{16\sin^{2}\theta+9\cos^{2}\theta} d \theta\] is correct

Answer 45

Yes and the answer is 16* elliptic integral so u need the 16.

Answer 46

oh, he is doing elliptical integral?

Answer 47

The arc length is the elliptic integral...

Answer 48

The 12 E(-7/9) or 16E(7/16) gives same answer....

Answer 49

He is just trying to set up a line integral on ellipse

Answer 50

sqrt(16 sin^2 +9 cos^2)
sqrt(7 sin^2 +9)

-> 3* int ]sqrt(1+ 7/9 sin^2)]

is the elliptic integral (or arc length integral if u prefer)

Answer 51

Wait a minute, have I got sin and cos backwards?

Answer 52

If so, maybe it will come out to 4....

Answer 53

It does come out to 4 if sin and cos are reversed....

Answer 54

Is that what the problem is..?

Answer 55

Back to the beginning, let me look....

Answer 56

Isn't the usual parameters for ellipse

(a cos t, b sin t)?

Answer 57

yes, but it works too

Answer 58

Yes but the answers in the options are based on the standard paramaterization..

Answer 59

Yes, That's what I was trying to say

Answer 60

If u reverse them, then everything will work and the answer is 4*

Answer 61

So the question is wrong...

Answer 62

Not wrong, inconsistent with the proposed answers...

Answer 63

It should be

x=4cos(theta) and y=3sin(theta) to match the answers.

Answer 64

imranmeah, u agree?

Answer 65

well ive been discussing it with my friend and we have also come to the conclusion that the problem is wrong

Answer 66

For parametric curve we want the integral of

sqrt(x'^2 + y'^2) and if this is

sqrt(16 sin^2 + 9 cos^2) then

x' = 4 sin^2 -> x = 4 cos

and y = 3 sin

Answer 67

4 sin, sorry, not 4 sin^2

Answer 68

For paramteric curve we want the integral of

sqrt(x'^2 + y'^2) and if this is

sqrt(16 sin^2 + 9 cos^2) then

x' = 4 sin -> x = 4 cos

and y = 3 sin