In the question attached what is the answer to part b) i can do part a) and c) thanks
you could see the $\displaystyle \frac{dI}{dt}= -5(1-cosx)$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 $\displaystyle t=6$
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
$\displaystyle \cos(x)=1$ only when x is a multiple of $\displaystyle 2\pi$,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
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, malaygoel!
Yes, it does . . .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.
Imagine the motion of the same particle.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'?