# Rates of Decay

• Jun 16th 2006, 09:47 PM
nath_quam
Rates 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 PM
malaygoel
Quote:

Originally Posted by nath_quam
In the question attached what is the answer to part ii) i can do part 1 and 3 thanks

you could see the $\displaystyle \frac{dI}{dt}= -5(1-cosx)$
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 AM
CaptainBlack
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 AM
nath_quam
Stationary Points
Thanks guys, for this question there must be other stationary points such as 18?? and so on
• Jun 17th 2006, 02:38 AM
CaptainBlack
Quote:

Originally Posted by malaygoel
you could see the $\displaystyle \frac{dI}{dt}= -5(1-cosx)$
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 $\displaystyle t=6$

$\displaystyle \cos(x)=1$ only when x is a multiple of $\displaystyle 2\pi$,
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 AM
nath_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 AM
CaptainBlack
Quote:

Originally Posted by nath_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

I is stationary when dI/dt=0.

Stationary means the rate of change is zero.

RonL
• Jun 17th 2006, 03:08 AM
nath_quam
• Jun 17th 2006, 06:22 AM
Soroban
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 AM
malaygoel
Quote:

Originally Posted by Soroban
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: . $\displaystyle t \,= \,24n,\;\;n=0,\,1,\,2, ...$.

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

Quote:

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

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 AM
CaptainBlack
Quote:

Originally Posted by Soroban
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: . $\displaystyle t \,= \,24n,\;\;n=0,\,1,\,2, ...$.

Oops :o

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 PM
malaygoel
Quote:

Originally Posted by Soroban
Hello, malaygoel!

Yes, it does . . .

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.

Imagine the motion of the same particle.
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'?