Solutions To Differential Geometry Problems - Part 4

 1)A curve is given in spherical coordinates x _i  by:

x ₁ = t,    x ₂ = arcsin 1/t,    x ₃  =  (t ₂ – 1) ^1/2

 Compute the length of the arc between t = 1 and t = 2  

Soln.

(ds/ dt) ² =    (dx₁/ dt) ²  +  (x₁ ) ²  (dx²/ dt) ²    +   (x₁  sin x  ²) ² (dx³/ dt) ² 

And:

(dx₁/ dt) ²  =   1

(dx₁/ dt) ²    =   [ -1/  t²   / Ö {1 – (  1/  t²  )} = 1/ t²  (t²  -   1)

  (dx³/ dt) ²    = 2t/ Ö2(t²  -   1) =  t² / (t²  -   1)

Whence:

(ds/ dt)² =     

1 +  t²   · 1/ t²  (t²  -   1) + (t ·  1/  t ) ²   ·  t² / (t²  -   1)

=   2 t² / (t²  -   1)

Then the length of the curve is:

L = ò ₁ ²   Ö2 t/ (t²  -   1)^1/2  dt = Ö2(t²  -   1) ] ₁²      =  Ö6

 2) Obtain the arc length s of the curve:

f(x) =  x³ / 2   -   x² / 3   

Between x = 0 and x = 2

Soln.

f(x) =  x³ / 2   -   x² / 3  

d(f(x) / dx =   3 x² / 2 -  2x/ 3  

(d(f(x) / dx) ² =  (3 x² / 2 -  2x/ 3 ) (3 x² / 2 -  2x/ 3 ) 

=    9 x⁴ / 4  -   2 x³    +   4 x² / 9

Then: 

L  =  ò ² ₀   Ö(1 +  9 x⁴ / 4  -   2 x³    +   4 x² / 9)

=  3.775 units

Check vs. Mathcad calculation:

3) Determine the arc length of a catenary with parametric representation:  x(t) =  (t,  a cosh (t/a), 0)  

Soln.: 

Using the arc length computation equation we have:

x'(t) =  1, sinh (t/a), 0

And: 

x' · x' = 1  +  sinh ² (t/a) =  cosh ² (t/a)

But:  Öcosh ² (t/a)  =   cosh  (t/a)

So:  s (t)  =   ò ^t ₀  cosh  (u/a) du =  a sinh(t/a)

where s = 0 corresponds to t = 0

4) Find the full arc length of the Archimedian spiral shown by changing the integral to the correct limits.

Soln.

By inspection of the graph for r  =  q - sin  q, we need to have  the limits of integration from:    0  to  5.2 p/2:

Then: L  = 2.545  units, e.g. from Mathcad computation: