In the question attached what is the answer to part b) i can do part a) and c) thanks

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- Jun 16th 2006, 09:47 PMnath_quamRates of Decay
In the question attached what is the answer to part b) i can do part a) and c) thanks

- Jun 16th 2006, 10:00 PMmalaygoelQuote:

Originally Posted by**nath_quam**

where x= $\displaystyle \frac {\pi*t}{12}$

now, 1- cosx is always greater than or equal to zero.

if 1- cosx is greater than zero., I is decreasing

I is stationary if cosx =1 i.e. $\displaystyle x=\frac{\pi}{2}$

or $\displaystyle \frac {\pi t}{12}=\frac{\pi}{2}$

hence I is stationary at $\displaystyle t=6$ - Jun 17th 2006, 12:10 AMCaptainBlack
What Malaygoel forgot to mention though it is implicit in what he wrote

is that a differentiable function $\displaystyle f(x)$ is decreasing at $\displaystyle x$ if and only

if $\displaystyle f'(x)<0$ and is stationary if and only if $\displaystyle f'(x)=0$

RonL - Jun 17th 2006, 01:57 AMnath_quamStationary Points
Thanks guys, for this question there must be other stationary points such as 18?? and so on

- Jun 17th 2006, 02:38 AMCaptainBlackQuote:

Originally Posted by**malaygoel**

so the stationary points are solutions of:

$\displaystyle

\frac{\pi t}{12}=n \pi

$,

or: $\displaystyle t=12n$, $\displaystyle n=0,\ 1,\ ...$.

RonL - Jun 17th 2006, 03:02 AMnath_quam
the question asks for when I is stationary, ain't we saying that at t = 12n the rate is stationary because for example at t = 12 -- I is decreasing still

- Jun 17th 2006, 03:05 AMCaptainBlackQuote:

Originally Posted by**nath_quam**

Stationary means the rate of change is zero.

RonL - Jun 17th 2006, 03:08 AMnath_quam
Thanks your a legend

- Jun 17th 2006, 06:22 AMSoroban
Small correction, Cap'n . . .

$\displaystyle \cos(x)=1$ only when x is a multiple of $\displaystyle 2\pi$,

so the stationary points are solutions of: . $\displaystyle \frac{\pi}{12}t \,=\,$**2**$\displaystyle \pi\cdot n$

or: . $\displaystyle t \,= \,24n,\;\;n=0,\,1,\,2, ...$.

- Jun 17th 2006, 06:28 AMmalaygoelQuote:

Originally Posted by**Soroban**

Does it make any sense that I is stationary at $\displaystyle t=0(n=0)$. - Jun 17th 2006, 07:01 AMSoroban
Hello, malaygoel!

Quote:

Does it make any sense that I is stationary at $\displaystyle t=0\,(n=0)$.

An analogous example:

A particle can be*at rest*at $\displaystyle t = 0\:(v = 0)$.

A nano-jiffy later it is moving $\displaystyle (v > 0)$ due to some acceleration. - Jun 17th 2006, 07:40 AMCaptainBlackQuote:

Originally Posted by**Soroban**

Thanks for spotting that.

Also perhaps a reminder to everyone to check what the helpers write,

after all we all make mistakes at some time :eek:

RonL - Jun 17th 2006, 06:36 PMmalaygoelQuote:

Originally Posted by**Soroban**

Let $\displaystyle x=f(t)$ describes the motion of the particle. Consider the motion from $\displaystyle t=0$ to $\displaystyle t=10$.Can we find $\displaystyle \frac{dx}{dt} $at$\displaystyle t=0.$. In other words, if a function is defined on the closed interval [a,b], can it be differentiable on one of its end-points e.g.'a'?