Is there any way to find the equation for this situation? $\mathbb N = \{1,2,3,4\dots\}$
Let us suppose we are starting at a point with coordinates $(0,0)$. Now draw a line from $(0,0)$ to $(1,0)$ and from $(1,0)$ to $(1,2)$. Now by the Pythagorean theorem, I will draw the hypotenuse i.e. a line from $(0,0)$ to $(1,2)$ of length $\sqrt{5}$. Now I will draw a line of length $3$ from the last point $(1,2)$ and perpendicular to the last hypotenuse we got. It will give another point, suppose $(x_1,y_1)$. Now again I will draw the hypotenuse and again draw a perpendicular line of length $4$ from this hypotenuse and so on. The values of length of the lines are coming from the set $\mathbb N$ of natural numbers.
Now the question is: How can I find out the equation of the curve satisfying the points $(0,0), (1,0), (1,2) , (x_1,y_1) , (x_2,y_2), \dots$ ?
It seems that the curve will be a spiral, but again I don't know how to find out the equation.
A similar figure can be - 

 A: It is probably good to also keep track of the displacement angle. Let's say we call it $α_n$ and it starts with $α_1 = 0$. Then the orthogonal line added is rotated from the up direction by $α_{n-1}$. So if $(x_0, y_0) = (0, 0)$ and $(x_1, y_1) = (1, 0)$, the coordinates change as
$$ x_n = x_{n-1} - n\cdot \sin α_{n-1} \\
y_n = y_{n-1} + n\cdot \cos α_{n-1}$$
Then you can calculate the new angle as
$$ α_n  = α_{n-1} + \arctan\left(\frac n {\sqrt{x_{n-1}^2+y_{n-1}^2}}\right)$$
Of course this is the most simplistic approach and you could probably avoid using $\arctan$, $\sin$ and $\cos$ with a little more calculation.
As you are always adding the length $n$, the radius will be
$$r_n := \sqrt{x_n^2+y_n^2} = \sqrt{\sum_{k=1}^n{k^2}} = \sqrt{\frac{n}{6}(n+1)(2n+1)}$$
(Compare this formula.)
So then you could write either of
$$ α_n - α_{n-1} = \arccos\left(\frac{r_{n-1}}{r_n}\right) = \arccos\sqrt{\frac{(n-1)(2n-1)}{(n+1)(2n+1)}}\\
= \arctan\left(\frac{n}{r_{n-1}}\right) = \arctan \sqrt{\frac{6n}{(n-1)(2n-1)}}\\
= \arcsin\left(\frac{n}{r_n}\right) = \arcsin\sqrt{\frac{6n}{(n+1)(2n+1)}}$$
This way the calculation can be improved by defining
$$ s_n := \sin α_n, \quad c_n := \cos α_n $$
Then you will get
$$s_{n+1} = \sin(α_n + (α_{n+1} - α_n)) = \sin α_n \cos (α_{n+1} - α_n) + \cos α_n \sin (α_{n+1} - α_n) = \frac{s_nr_n}{r_{n+1}}+\frac{c_n(n+1)}{r_{n+1}}\\
c_{n+1} = \cos(α_n + (α_{n+1} - α_n)) = \cos(α_n)\cos(α_{n+1}-α_n)-\sin{α_n}\sin(α_{n+1}-α_n) = \frac{c_nr_n}{r_{n+1}} - \frac{s_n(n+1)}{r_{n+1}}$$
So you can put
$$x_n = x_{n-1} - ns_{n-1}\\y_n = y_{n-1} + nc_{n-1}$$

Wikipedia gives a formula for the Spiral of Theodorus, like meantioned by pritam. Probably either of the there referenced solutions can be adapted to this problem. The complex function equation would here take the form
$$f(x+1) = \left(1+i\frac{x+1}{r_x}\right)f(x) = \left(1+i\sqrt{\frac{6(x+1)}{x(2x+1)}}\right)f(x)$$
A: The curve is given by $$f(x)=\prod_{k=1}^\infty\frac{1+i/\sqrt {k}}{1+i/\sqrt {x+k}}$$
See the last part titled, 'Continuous curve' in wikipedia link: http://en.wikipedia.org/wiki/Spiral_of_Theodorus
