The Length of the DNA Helix

Problem:

The DNA molecule has the shape of a double helix. The radius of each helix is about 10 angstroms (1 angstrom = 10^(-8) cm). Each helix rises about 34 angstroms during each complete turn and there are about 2.9 x 10^(8) complete turns. Estimate the length of each helix.

Solution:

 and as usual the i,j,k basis vectors:

> alias(v=vector): i:=v([1,0,0]): j:=v([0,1,0]): k:=v([0,0,1]):
 The radius of the helix is a, and the hight of each turn is h, in angstroms.

> a := 10: h := 17/Pi:
 The vector function for this helix is then, r

> r := a*cos(t)*i + a*sin(t)*j + h*t*k;

```                                                        t k
r := 10 cos(t) i + 10 sin(t) j + 17 ---
Pi```
 The velocity vector is:

> V := diff(r,t);

```                                                         k
V := -10 sin(t) i + 10 cos(t) j + 17 ----
Pi```
 from where we obtain the speed at time t by computing the length of V. The length of one complete revolution is then the integral of the speed,

> L := Int(len(evalm(V)),t=0..2*Pi);

```                              2 Pi
/            2       1/2
|      (100 Pi  + 289)
L :=  |      ------------------ dt
|              Pi
/
0```
 which evaluates to:

> L := evalf(L);

`                               L := 71.44117693`
 This is given in angstroms. The total length of 290 million revolutions is (in centimeters):

> TotL := 2.9*L;

`                              TotL := 207.1794131`