Submitted by jtedds on Fri, 04/21/2017 - 21:16
Forums:
For parts (i) and (ii) I've managed to muddle through,
For part (iii)
I assume you start by finding an expression for $E(x)$
So I used something similar to (i) and (ii) to obtain ${dE\over dx} = -2\left({dw\over dx} \right)^2(5cosh(w) - 4sinh(w) - 3)$
Integrating to get $E(x) =\left ( {dw \over dx} \right)^2 + wsinh(w) + cosh(w)$
Setting $5cosh(w) - 4sinh(w) - 3 = 0$ the solution is $w = ln(3)$
So the graph of $E(x)$ is an increasing function for $0 \le x < ln(3)$ and a decreasing function after, with $ln(3)$ as the maximum
I then know ${5\over 3} \ge \left({dw \over dx}\right)^2 + wsinh(w) + cosh(w)$ but I'm struggling to make anything else of it
This is a great question.
This is a great question.
Surely you'd integrate to get $E(x) = \left(\frac{\mathrm{d}w}{\mathrm{d}x}\right)^2 + 2(w\sinh w + \cosh w)$ (i.e: you're missing that factor of $2$?)
Anyway, I'm don't think you're tackling this the right way (but you're close!): Note that $$\frac{\mathrm{d}E}{\mathrm{d}x} = -2\left(\frac{dw}{dx}\right)^2 (5\cosh w - 4\sinh w - 3) = -2\left(\frac{\mathrm{d}w}{\mathrm{d}x}\right)^2 \frac{e^{-x}}{2}(e^x -3)^2$$
What can you say about the positivity/negativity about each factor in the above expression? Overall what does that mean for the positivity/negativity of $\frac{\mathrm{d}E}{\mathrm{d}x}$?
Does $E(x)$ have a initial value that you know of?
If so, what does that mean for $E(x)$, considering the positivity/negativity of the derivative and the functions initial value?
Once you have that, note that $w\sinh w $ is always positive. So what can you say about $\cosh w$ from your data about $E(x), \left(\frac{\mathrm{d}w}{\mathrm{d}x}\right)^2$ and $w\sinh w$? (it's precisely the same way you did (i) and (ii)).
So since $\frac{\mathrm{d}E}{
So since $\frac{\mathrm{d}E}{\mathrm{d}x}$ is always negative, then the initial value of $E(x)$ is the greatest - which is $\frac{5}{2}$
Since each term in $E(x)$ is positive then $\frac{5}{2}$ is the greatest value any of them can take
So $\frac{5}{2} \ge 2\cosh(w)$ from which our result follows
Thanks for the help with the derivative and integrating back - I'd done some of the work for the factorisation of $5\cosh w - 4\sinh w - 3$ but not used it in the right way and the missing factor of 2 was forgetting I'd factorised it out already in my derivative before integrating
Looks good! Although
Looks good! Although personally I'd write "since each term in [...] is positive then [...] is the greatest value any of them can take" as: $\cosh w = \frac{1}{2}\left(E(x) - \left(\frac{\mathrm{d}w}{\mathrm{d}x}\right)^2 - w\sinh w \right) \leq \frac{1}{2}E(x) \leq \frac{1}{2}\left(\frac{5}{2}\right)$ since both $w\sinh w$ and $\left(\frac{\mathrm{d}w}{\mathrm{d}x}\right)^2$ are positive.
But yours is equally fine - if a little informal.