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

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- June 16th 2006, 10:47 PMnath_quamRates of Decay
In the question attached what is the answer to part b) i can do part a) and c) thanks

- June 16th 2006, 11:00 PMmalaygoelQuote:

Originally Posted by**nath_quam**

where x=

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.

or

hence I is stationary at - June 17th 2006, 01:10 AMCaptainBlack
What Malaygoel forgot to mention though it is implicit in what he wrote

is that a differentiable function is decreasing at if and only

if and is stationary if and only if

RonL - June 17th 2006, 02:57 AMnath_quamStationary Points
Thanks guys, for this question there must be other stationary points such as 18?? and so on

- June 17th 2006, 03:38 AMCaptainBlackQuote:

Originally Posted by**malaygoel**

so the stationary points are solutions of:

,

or: , .

RonL - June 17th 2006, 04:02 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

- June 17th 2006, 04:05 AMCaptainBlackQuote:

Originally Posted by**nath_quam**

Stationary means the rate of change is zero.

RonL - June 17th 2006, 04:08 AMnath_quam
Thanks your a legend

- June 17th 2006, 07:22 AMSoroban
Small correction, Cap'n . . .

only when x is a multiple of ,

so the stationary points are solutions of: .**2**

or: . .

- June 17th 2006, 07:28 AMmalaygoelQuote:

Originally Posted by**Soroban**

Does it make any sense that I is stationary at . - June 17th 2006, 08:01 AMSoroban
Hello, malaygoel!

Quote:

Does it make any sense that I is stationary at .

An analogous example:

A particle can be*at rest*at .

A nano-jiffy later it is moving due to some acceleration. - June 17th 2006, 08:40 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 - June 17th 2006, 07:36 PMmalaygoelQuote:

Originally Posted by**Soroban**

Let describes the motion of the particle. Consider the motion from to .Can we find at . 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'?