How to solve the following:

-10t + 60/piSine(pi/12 * t) + 2400

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- Jun 18th 2006, 12:16 AMNanpaTrigonometry
How to solve the following:

-10t + 60/piSine(pi/12 * t) + 2400 - Jun 18th 2006, 12:35 AMnath_quam
isn't it -5t

- Jun 18th 2006, 02:27 AMNanpa
Nope. -10t

- Jun 18th 2006, 03:28 AMCaptainBlackQuote:

Originally Posted by**Nanpa**

equation to have a solution.

The best you can do with an expression is simplify it.

Also the expression is ambiguous as it stands, what you have written

could be intended to mean:

$\displaystyle

-10t + \frac{60}{\pi} \sin(\pi.t / 12 ) + 2400

$

or:

$\displaystyle

-10t + \frac{60}{\pi \sin(\pi.t / 12 )} + 2400

$

or:

$\displaystyle

-10t + \frac{60}{\pi} sin(\pi / (12t )) + 2400

$,

or:

$\displaystyle

-10t + \frac{60}{\pi \sin(\pi./ (12t) )} + 2400

$

Now you probably intend the first of these, but you should use brackets

to make the meaning unambiguous when typing maths in ASCII.

RonL - Jun 18th 2006, 04:43 AMNanpa
Sorry, I forgot to add equals zero...

- Jun 18th 2006, 04:47 AMnath_quam
he means the top one captain

- Jun 18th 2006, 05:03 AMCaptainBlack
t=240

- Jun 18th 2006, 05:14 AMNanpa
Thanks, but how?

- Jun 18th 2006, 05:14 AMnath_quam
How you work that out???

- Jun 18th 2006, 06:05 AMCaptainBlackQuote:

Originally Posted by**nath_quam**

-10t + \frac{60}{\pi} \sin(\pi.t / 12 ) + 2400=0

$

First look for an approximate solution. Since the sinusoidal term is always

between $\displaystyle \pm 60/pi \approx \pm20$, it is clear that a solution

is close to a solution of $\displaystyle -10t+2400=0$, which occurs when

$\displaystyle t=240$, now the $\displaystyle \sin$ term is identically zero

at that point, so this is an exact root.

That there are no other roots is obvious as the maximum rate of increase

of the middle term is less than $\displaystyle 10$ (in fact it is $\displaystyle 5$)

so the entire LHS is a decreasing function and so the root is unique.

RonL