Time-varying waveform My situation is as follows: i am trying to generate a waveform the hard way, by constructing the samples one by one and then saving the result to a .wav file using Python.

When the frequency is constant, everything is fine: i use $y(t) = \sin(2 \pi \cdot f \cdot t)$.
However, if i change the frequency to be a function of time, things go wrong. If the function is a linear one of the form $f(t) = a + bt$, it still works. But if i choose, for example, $f(t) = 40 + 10 \sin(t)$, the max frequency increases over time, reaching a maximum higher than the expect 50Hz.

I have read something about instantaneous frequency, namely, this: Why does a wave continuously decreasing in frequency start increasing its frequency past the half of its length?. But doing the integral evaluation makes the sound even weirder. And the method i currently have works for linear function of time, so i figured there must be something else wrong.

I also tried to generate chunks of sound in the frequency i need, at each time, and then glue them together. I calculated the period of the oscillation, so that a chunk has as many samples as necessary to make a whole period on that frequency, so that there are no "jumps" between different frequencies. But generates a cracking sound in the sample. 

Here is an example of the kind of a JavaScript implementation of the kind of sound i need: http://jsfiddle.net/m7US6/4/.

Any help would be appreciated. Thanks.
Consider the function $\sin(2\pi f t)$. When $2\pi f t$ goes from $0$ to $2\pi$ you get oscillation of the sine wave, $2\pi$ to $4\pi$, another, and so on. So every time the argument changes by $2\pi$ you get one oscillation.

Now lets plot $2\pi f(t) t$ where $f(t) = 10 + 10 \sin(2 \pi t)$. (I've modified it a bit to show the effect more):



As $t$ increases, this value changes faster and faster, meaning $\sin(2\pi f(t) t)$ will have higher and higher frequency.

Another way of thinking about it - the frequency is the derivative of $f(t) t$. For the example this is $20 \pi (\sin(2 \pi t)+2 \pi \mathbf{t} \cos(2 \pi t)+1)$. Note the $t$ multiplier in there - $t$ increases thus the frequency increases.

So what you really want is 
$\tfrac{d}{dt} f(t) t = 40 + 10 \sin(t)$ which gives $f(t) t = 40t - 10\cos(t) + c$ and for your example try $ \sin(2\pi(40t - 10 \cos(t)))$