Expert: Alon Mandes - 10/9/2008

Question
QUESTION: I have some homework problems that are being graded as a quiz in my Calculus 3 class.  There is one I cannot figure out.  It deals with DNA and the length of one helix in it.  The given information is that the radius is 10 angstroms, rises 34 angstroms for each complete turn, and there are 2.9 x 10^8 turns in the entire molecule.  I need to find the length of the helix.  I'm fairly certain this is the arc length (given that's the section it's in).  That I can do, but I have no function to integrate or the time values on which to do it.  I don't even know where to start.  I figured I'd need the function first and foremost, but I'm clueless as to how to get it.  Can you help?

ANSWER: The shape of the DNA is 3D spiral . Where the radius grows in
fashion of r(θ)=10+(34/2π)θ. So the curve will be s(θ)=(r(θ)e^(iθ). Or if you want [(10+18θ/π)cosθ,(10+18θ/π)sinθ]. Note that this is a
projection on the z-axis obtaining 2D curve.So, as we know the arc length will be : L=∫|s(θ)|dθ {where θ goes from 0 to 2π*(2.9*10^8).
Note that |s(θ)|= { [x'(θ)]^2+[y'(θ)]^2 } ^ (1/2).
You may proceed from here & calculate the integral. Good luck !

Alon.

---------- FOLLOW-UP ----------

QUESTION: Why does the radius of DNA molecule grow?  I'm pretty certain it's a fixed radius.  You lost me there and I can't follow most of the rest of it.

ANSWER: I quote what you wrote to me : " the radius is 10 angstroms, rises 34 angstroms for each complete turn " . I understood from this,
that the radius is 10° at first loop, then every loop it increase additional 34°. (" ° is angstrom ").
If that's not the case, then I think you meant, that the radius is
always 10°, but the height growth of the shape is 34°. If this is
what you meant, then the theory will be changed & the integral will be different. If it's not, then please explain it to me again.
Thank you,
Alon.


---------- FOLLOW-UP ----------

QUESTION: Yes, I meant the radius to be a fixed 10 and the vertical height (z) between each loop is a constant 34.  Thanks again!

Answer
Ok, in this case our space trajectory or our 3D curve will be :
[10cos(θ),10sin(θ),34θ/2π] which means :
x(θ)=10cos(θ)
y(θ)=10sin(θ)
z(θ)=(34/2π)θ
As I explained before :
L=∫|s(θ)|dθ
|s(θ)|= { [x'(θ)]^2+[y'(θ)]^2+[z'(θ)]^2 } ^ (1/2).
So,the arc length will be :
L=∫ {[-10sin(θ)]^2+[10cos(θ)]^2+ [34/2π]^2 }^(1/2) dθ
{where θ goes from 0 to 2π*(2.9*10^8).
L=∫ { 100+289/π^2 }^(1/2) dθ {where θ goes from 0 to 2π*(2.9*10^8).
L=2π*(2.9*10^8)*(100+289/π^2)^(1/2)=(207.17*10^8)° .

Alon.