Question

Find the arc length of the polar equation attached

Answer 1

\[r=8(1+\cos \theta) ....Interval -\pi \le0\]

Answer 2

Interval should be -pi less than equal 0 less than equal pi

Answer 3

\[\int\limits_{-\pi}^{\pi}\sqrt{64+16\cos \theta+64 \cos \theta ^{2}+64\sin \theta ^{2}}\]

Answer 4

integral sqrt (r^2 + (dr/dtheta)^2 ) d theta

Answer 5

then\[\int\limits_{-\pi}^{\pi} \sqrt{64(1+1/4\cos \theta+1}\]

Answer 6

i think there is a mistake

Answer 7

\[8\int\limits_{-\pi}^{\pi}\sqrt{2+1/4\cos}\]

Answer 8

There si I just don't know where

Answer 9

you have
integral sqrt( 64 ( 1 + cos t)^2 + 64 sin^2(t) , correct?

Answer 10

\[16\cos(\theta)\]

Answer 11

not good :)

Answer 12

should be
sqrt( 64 ( 1 + 2cos(t) + cos^2(t) ) + 64 sin^2(t) )

Answer 13

yep, thats the term that is at fault here

Answer 14

So 16 cos theta should be what and why?

Answer 15

did you expand ( 1 + cos(t)) ^2 correctly?

( 1 + cos(t)) * ( 1 + cos(t)) = 1 + 2cos(t) + cos^2(t)

Answer 16

Actually I multiplied distributed 8 over (1+cos theta) then squared

Answer 17

that works too

Answer 18

( 8 + 8cost)^2 = 8^2 + 2*8*8cos(t) + 8^2 cos^2(t)

Answer 19

in general
(a + b)^2 = a^2 + 2ab + b^2

Answer 20

then it would be (8+8cos theta) ^ 2....thus 64 +16 cos theta + 64 cos theta

Answer 21

8*8 not 8+8

Answer 22

Brain lock

Answer 23

then mult by 2

Answer 24

It might help to foil it out (slowly)
( 8 + 8cost ) * ( 8 + 8 cos t )

Answer 25

I know your are correct I just need to look at it closely.