Question

arccos (-3/2)?

Answer 1

i've forgotten everything about trig and i'm putting this in my calculator, it says ERROR. how do you get this?

Answer 2

Do you mean -sqrt(3)/2 for the inside?

Answer 3

@zepdrix no, i mean (-3/2)

Answer 4

no square roots here

Answer 5

Ok then you're not making any mistakes.
-3/2 is not in the domain of inverse cosine.

I assume this is part of a larger problem?
Integration?
Did you forget to aquire new bounds after making a substitution or something?

Answer 6

It's actually a part of the problem attached and I'm trying to do it as said here: http://tutorial.math.lamar.edu/Classes/CalcII/PolarArea.aspx =/

Answer 7

Oh ok :U sorry gimme few.. gotta wrap something up

Answer 8

take your time

Answer 9

Hmm where is this inverse cosine coming from?\[\large\rm r=3+2\cos \theta\]It completes it full loop in 2pi, yes?
So that gives us our bounds,\[\large\rm A=\frac12\int\limits_0^{2\pi}r^2d \theta\]

Answer 10

Plugging in our function,\[\large\rm A=\frac12\int\limits_0^{2\pi}(3+2\cos \theta)^2d \theta\]

Answer 11

And then we would expand the square, ya?
I'm really confused where the -3/2... and the inverse cosine are coming from. hmm

Answer 12

i got that from the web site, the first thing they do to find the range of the integral?

Answer 13

i only need 0 to 2Pi?

Answer 14

Umm um um um thinking

Answer 15

Thanks for showing me trig gets even harder, this school year is going to be so much fun with so many different subjects in one. :(

Answer 16

lol XD

Answer 17

hahaha calculus is the real demon, my friend! @Vuriffy

Answer 18

My class is Aice Math 2 which has trig, calculus 2, statistics 2, and probability 2. :( I forgot everything from last year, so woot.

Answer 19

@Vuriffy OUCHH! all the best for that deadly combo!!!

fingers crossed, if i pass this exam, it's my last math class ever

Answer 20

In the first example on Paul's Notes,
he wasn't using the whole cycle,
he only wanted the area inside one tiny part of it.

So it required a bit more work.

Answer 21

@dramaqueen1201 Yeah, I hope I will do well in the class, not so sure about the exam, anyways, sorry for side-tracking you too. Good luck! (: Have a goodnight as well.

Answer 22

ooh alright. so i can only use 0 to 2pi for the problem and i should be fine, yeah?

Answer 23

Ya I was able to come up with one of the answers :d
I think it's right...

Try it and see what you get :D

Answer 24

@Vuriffy it's okay. no problem. good night to you too!

Answer 25

@zepdrix i'll be right back!

Answer 26

I don't understand why you have arcos(-3/2) considering your problem

Answer 27

I suppose she was looking for a starting point for her function,\[\large\rm 0=3+2\cos \theta\qquad\to\qquad -\frac{3}{2}=\cos \theta\]But our function never has a radial length of 0,
that's why it doesn't work out for this problem.

Answer 28

11pi, right?

Answer 29

Ah I see, yeah this one is a bit tricky so you have to be careful with the domain, \[A = 1/2 \int\limits_{a}^{b}r^2 d \theta = 1/2 \int\limits_{0}^{2 \pi}(3+2\cos \theta)^2 d \theta \] looks good!

Answer 30

Oooo yay that's what I came up with also \c:/

Answer 31

@zepdrix thank you!!!