Question

Find length of perimeter of curve

x=2 cos^3 t, y= 2 sin^3 t

Answer 1

x=2 cos^3(t), y= 2 sin^3(t)

x'= -6 cos^2(t) sin(t), y' = 6 sin^2(t) cos(t)

\(\int_{a}^{b}\sqrt{(-6 cos^2(t) sin(t))^2+(6 sin^2(t) cos(t))^2}\)dt
maybe?

Since we only have 2 components; we dont need the f(x) part right?

Answer 2

its a star :)

Answer 3

Yup, bit of symmetry there....

Answer 4

so i guess we can get away with 4* int[0,2]

Answer 5

pi/2

Answer 6

well, that too

Answer 7

Works out exactly if anyone is looking to practice.....

Answer 8

I have a question, in "Calc 111" they teach line/surface integrals with vector formulations as well?

Answer 9

If its right; 24cos(t)sin(t), [0,pi/2] ??

Answer 10

that wouldnt work out well at all .... 0 - 0 = 0 lol

Answer 11

might be 12 in the end

Answer 12

Yes, 12.......(2 sint cost -> sin 2t...)

Answer 13

\[4\int_{a}^{b}\sqrt{(-6 cos^2(t) sin(t))^2+(6 sin^2(t) cos(t))^2}dt\]
\[4\int_{a}^{b}\sqrt{(36 cos^2(t)cos^2(t) sin^2(t) +36 sin^2(t)sin^2(t) cos^2(t)}dt\]
\[4\int_{a}^{b}\sqrt{36 cos^2(t)sin^2(t)\ (cos^2(t)+sin^2(t))}dt\]
\[4\int_{a}^{b}(6 sin(t))cos(t)dt\]
\[4*3 sin^2(t)dt;\ [0,pi/2]\]

Answer 14

12(1) - 12(0) = 12

Answer 15

the first time i simplified the sqrt and forgot to actually integrate lol

Answer 16

That will do as well..

Answer 17

YHum... http://openstudy.com/groups/mathematics/updates/4e6dfec20b8b4a2b95e77240

Answer 18

ESTUDIER COME BACK :(