Question

Arc Lengths: Find the exact length of the polar curve r= 5sin(theta), 0<= theta <= pi/3.

Answer 1

LOLOLLOLOL

Answer 2

do you know the arclength formula?

Answer 3

LOLOLLOLOLLOLOLLOLOLLOLOLLOLOLLOLOLLOLOLLOLOLLOLOLLOLOLLOLOLLOLOLLOLOLLOLOLLOLOLLOLOLLOLOLLOLOLLOLOLLOLOLLOLOLLOLOLLOLOLLOLOLLOLOLLOLOLLOLOLLOLOLLOLOL

Answer 4

bro are you okay?

Answer 5

\[L=\int\limits_{0}^{pi/3} \sqrt{1+[dy/dx]^2} d\]

Answer 6

No, he's not. Im gonna report him cause he's annoying me.

Answer 7

yeah you got it in this case instead of dy/dx we will have dr/dtheta okay

Answer 8

and lol do it up so anyways

Answer 9

derivative of r with respect to theta is 5*cos(theta) alright now let us put this into our formula

KEEP in mind for these problems go inside out like do the derivative then square it then add 1 then square root it then take the integral alright ok so anyways now we got

1+ 25cos^2theta

Answer 10

now we have to take the square root of that and now we apply trignometric substitution here

Answer 11

to find the main integral

Answer 12

trignometric substitution !?!? Oh dear. I especially have issues with those.

Answer 13

yeah because there is no other way to actually integrate this :) so make the right substitution and work from there or do you want help with that?

Answer 14

Ummmmmmmmm, im gonna need help. You're help in two ways since I also need help with trig sub.

I know that 1+cos^2 = sin^2. and something something...

Answer 15

no 1-cos^2x = sin^2x

Answer 16

heck, smh

Answer 17

to solve this integral here take a look at this page for me it is really usefull it helped me all the way till 4th year uni

http://www.stewartcalculus.com/data/CALCULUS%20Concepts%20and%20Contexts/upfiles/3c3-TrigonometSubstitu_Stu.pdf

Answer 18

go down a little to see trignometric substitution table

Answer 19

i should use...the tan one?

Answer 20

exactly :) you got it where your x=5cosx alright :)

Answer 21

Ok, I tried and failed horribly, i am so sorry.

Answer 22

\[\sqrt{5^2-(5\cos(\theta))^2} = \sqrt{25-25\cos^2(\theta)} = \sqrt{5(1-\cos^2(\theta)} =\sqrt{5^2}\sqrt{1-\cos^2(\theta)} \]

Answer 23

\[= 5 \sqrt{\sin^2(\theta)} = 5 \sin(\theta)\]

Answer 24

no that is wrong my friend

Answer 25

good try though you were suppose to use tan remember from that table that table basically tells you if you have something of this format use this and apply this identity right?

Answer 26

yeaa. But because I have cos and sin, i got confused i guess

Answer 27

o wow which year are you in by any chance?

Answer 28

http://www.wolframalpha.com/input/?i=integral+of+sqrt%281+%2B+25cosx%5E2%29

according to wolfram you cannot integrate even with trig substituion you cannot apply it actually

ah mann

Answer 29

This is my second year of college, calc 2

Answer 30

How'd u make your wolfram link so short?

Answer 31

what do u meann

Answer 32

nvm. When i used wolfram, they said the answe was 4.4...
integral_0^(pi/3) sqrt(1+25 cos^2(t)) dt = sqrt(26) E(pi/3|25/26)~~4.45953

Answer 33

This is actually a test question. I dont see how he expected us to know how to do this

Answer 34

I realized the issue, lol. this is a polar curve., so \[L= \int\limits_{0}^{pi/3} \sqrt{(r)^2 + [dr/dt]^2} dt\]