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Two problems about infinite processes where smaller and smaller steps are taken and you have to discover what happens in the limit.

# Loch Ness

##### Age 16 to 18Challenge Level These following graphs are not monsters. They are humpy because the functions are periodic and involve sines, cosines and absolute values. This problem calls for you to describe and explain the features of the graphs.

1. Plot the graph of the function $y=f(x)$ where $f(x) = \sin x +|\sin x|$.
Find the first derivative of this function in the range $0 \le x \le 2 \pi$.  Is the first derivative defined everywhere?  Can you find an expression for the first derivative for all real $x$?

It might help to start by sketching $y=\sin x$.  Can you use this graph to sketch $y=|\sin x|$?

It might help to remember that if $y=x$ then \eqalign{ y &= x \ {\rm for } \ x\geq 0 \cr y&= -x \ {\rm for } \ x < 0 .}
You may find it useful to split up the domain into regions where $\sin x \ge 0$ and where $\sin x <0$.

See the Getting Started section for some hints on how to show where the derivative is not defined.

2. Express the function $f(x) = \sin x + \cos x$ in the form $f(x)=A\sin (x+\alpha)$, find $A$ and $\alpha$ (where $-\pi /2 < \alpha < \pi /2$) and plot the graph of this function.

Similarly express $g(x) = \sin x - \cos x$ in the form $g(x) = B\sin (x +\beta)$, find $B$ and $\beta$ (where $-\pi /2 < \beta < \pi /2$) and plot its graph on the same axes.

We have $A \sin (x + \alpha) = A \sin x \cos \alpha +A \cos x \sin \alpha$, and so equating coefficients gives \eqalign{1&=A \cos \alpha \cr 1&=A \sin \alpha}
You can then use the identities $\sin^2 \alpha + \cos^2 \alpha = 1$ and $\tan \alpha = \dfrac{\sin \alpha}{\cos \alpha}$ to find $A$ and $\alpha$.  You can use a similar idea for the second curve.

3. Plot the graph of the function $y=f(x)$ where $f(x)= \sin x + |\cos x|$. Find the first derivative of this function and say where it is defined and where it is not defined.

It might be helpful to consider where $\cos x$ is negative and where it is positive.  If we have $\cos x <0$ then $|\cos x| = - \cos x$ (so that we get a positive value).