Author Topic: Complex #s + velocity/time graphs (Read 767 times) Tweet Share

Chavi · « **on:** October 06, 2010, 11:26:20 pm »

Hey, just a couple of questions:

1. The complex number $z=cis(\theta - \frac{\pi}{2})$ can be written as:

according to the answer it's $sin(\theta) + icos(\theta)$ but why can't it also be $sin(\theta) -icos(\theta)$ given that $sin(\theta -\frac{\pi}{2}) = + and -icos(\theta)$ ?

2. The velocity of a particle is given by $v= \begin{cases}<br /> \sqrt{16-t^2}, 0\le t \le 4\\<br /> 4-t, t>4 <br />\end{cases}<br />$
The particle returns to it's starting point when t =?

I got $4 + \sqrt{8\pi}$ but the answer is $4 + \sqrt{4\pi}$ . What went wrong??

moekamo · « **Reply #1 on:** October 07, 2010, 07:39:32 am »

1. Try subbing in a number, when $\theta = 0, z=\text{cis} \left ( -\frac{\pi}{2} \right )=-i$

however when theta=0 in $\sin \theta +i \cos \theta$ it equals $i$ not $-i$

in the second one when theta=0, $\sin \theta - i \cos \theta = -i$ which it should, so the answer is:

$z=\text{cis} \left ( \theta -\frac{\pi}{2} \right )= \sin \theta - i\cos \theta$

2. I got the same answer as you, the area of the quarter circle part is $4\pi$ and when equating this to the area between the line and x axis and solving for t, i get your one... maybe the answers wrong or someone else knows?

Mao · « **Reply #2 on:** October 07, 2010, 10:54:34 pm »

Quote from: Chavi on October 06, 2010, 11:26:20 pm

given that $sin(\theta -\frac{\pi}{2}) = + and -icos(\theta)$ ?

Where do you get that? The correct expression is $\sin (\theta - \pi/2) =- \cos \theta$

Mao · « **Reply #3 on:** October 07, 2010, 10:57:22 pm »

2. Solve $\int_{0}^{4} \sqrt{16-t^2} \; dt + \int_{4}^{t_c} 4-t\; dt = 0$ for tc, the correct answer is $t_c = 4 + \sqrt{8\pi}$ , you are correct.

Chavi · « **Reply #4 on:** October 08, 2010, 12:47:30 am »

Quote from: Mao on October 07, 2010, 10:54:34 pm

Quote from: Chavi on October 06, 2010, 11:26:20 pm
given that $sin(\theta -\frac{\pi}{2}) = + and -icos(\theta)$ ?

Where do you get that? The correct expression is $\sin (\theta - \pi/2) =- \cos \theta$

How is that so?

Search the forums now!

We have moved!

Author Topic: Complex #s + velocity/time graphs (Read 767 times) Tweet Share

Chavi

Complex #s + velocity/time graphs

moekamo

Re: Complex #s + velocity/time graphs

Mao

Re: Complex #s + velocity/time graphs

Mao

Re: Complex #s + velocity/time graphs

Chavi

Re: Complex #s + velocity/time graphs

Recent Posts

User:
Password: