Question

Calculus 3: Convert \(3x^2-8xy+3y^2+4z^2=7\) using cylindrical coordinates, I just need something checked, not solved.

Answer 1

I'll just work it out here since it's harder to see on Skype I guess? lol

Answer 2

Since you prefer being tagged @perl

Answer 3

3 ( x^2 + y^2 ) , thats a nice expression

Answer 4

\[3x^2 -8xy +3y^2 +4z^2 = 7\]\[3(rcos(\theta))^2 -8(r\cos(\theta))(r\sin(\theta)) +3(rsin(\theta))^2 +4z^2 = 7\]\[3r^2\cos^2(\theta) -8r^2\cos(\theta)\sin(\theta)+3r^2sin^2(\theta) + 4z^2 = 7\]\[3r^2(cos^2(\theta) +sin^2(\theta)) \color{red}{-8r^2cos(\theta)sin(\theta)}+4z^2=7\]\[3r^2 \color{red}{-8r^2\cos(\theta)sin(\theta)} +4z^2 = 7\]

Answer 5

3(x^2 + y^2) , that becomes 3*r^2

Answer 6

you're saying... and i'm repeating this to understand it better...\[\sin(2\theta) = 2sin(\theta)cos(\theta)\]\[-4\cdot sin(2\theta) = -4 \cdot 2sin(\theta)cos(\theta))\]

Answer 7

Yeah, I got that part @perl haha, I was confused about the red portion.

Answer 8

wow youre really good at Latex

Answer 9

you can even do colors

Answer 10

I have no idea how she highlights with colors, lol.

Answer 11

`\color{red}{@Concentrationalizing }`

Answer 12

cylindrical coordinates are ( r , theta, z )

Answer 13

\[\color{red}{Concentrationalizing}\]

Answer 14

so if you have an equation in terms of r, theta, z , you are done

Answer 15

Yep those are cylindrical coordinates.

Answer 16

FAIL!

Answer 17

i posted this question up because he posted up a neat trick in skype, and I just wanted to see if I could replicate it here, haha. I had forgotten that \(\sin(2\theta) = 2sin(\theta)cos(\theta)\)

Answer 18

is that the final answer, can we simplify it more

Answer 19

Possibly simplify it even more.

What is fail? @Concentrationalizing >:((

Answer 20

3r^2 - 4r^2 sin(2theta) = 7 - 4z^2

(3 - 4sin(2theta) r^2 = 7 - 4z^2

Answer 21

you have to identify what red is, which is a `color` :P

Answer 22

\[3r^2 \color{red}{-8r^2\cos(\theta)sin(\theta)} +4z^2 = 7\]\[3r^2 \color{red}{-4sin(2\theta)}+4z^2 =7\]\[4z^2 = 7 -3r^2+4\sin(2\theta)\]

Answer 23

Then we solve for z.....right? O_o

Answer 24

Wait - @perl why are you subtracting \(4z^2\)?

Answer 25

you can do it that way

Answer 26

thats cleaner

Answer 27

So that means we were solving for r^2 this whole time?!

Answer 28

Or we're trying to parallel the equation of a circle.. hm.

Answer 29

nope, i dont think we have to solve for r, or z , or theta, in particular

Answer 30

as long as you have an equation in terms of r, z, theta only, youre good to go

Answer 31

as long as you dont have any x or y

Answer 32

my attach file button is broken but

Answer 33

http://facultyfp.salisbury.edu/despickler/personal/archive/m310F01/exam3a.pdf

Answer 34

question 1

Answer 35

he didn't solve for r or z

Answer 36

I see

Answer 37

\[3r^2 \color{red}{-8r^2\cos(\theta)sin(\theta)} +4z^2 = 7\]\[r^2 (3-4\sin(2\theta)) +4z^2 = 7\]\[r^2 = \frac{7-4z^2}{3-4sin(2\theta)}\]

Answer 38

\begin{array}{r|lll}
6&1&-7&6\\
\\
&\hline
1
\end{array}