## 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

i.e. about 2 meters long! |

Link to the commands in this file

Carlos Rodriguez <carlos@math.albany.edu> Last modified: Thu Oct 11 09:33:52 EDT 2001