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*.

^{2}= (dx

_{1}/ dt)

^{ 2}+ (x

_{1}

^{ })

^{ 2}(dx

^{2}/ dt)

^{ 2}+ (x

_{1}

^{ }sin x

^{ 2})

^{2}(dx

^{3}/ dt)

^{ }

^{2}

^{3}/ dt)

^{ 2}= 2t/ Ö2(t

^{2 }- 1) = t

^{2 }/ (t

^{2 }- 1)

^{2}=

^{2 }· 1/ t

^{2 }(t

^{2}

^{ }- 1) + (t · 1/ t

^{ }**)**

^{ 2 }· t

^{2 }/ (t

^{2 }- 1)

^{2 }/ (t

^{2}

^{ }- 1)

_{1 }

^{2}

^{ }Ö2 t/ (t

^{2 }- 1)

^{1/2}

^{ }dt = Ö2(t

^{2}

^{ }- 1) ]

_{1}

^{2 }

**= Ö6**

^{ }2) Obtain the arc length s of the curve:

f(x) = x^{3 }/ 2 - x^{2 }/ 3

Between x = 0 and x = 2

* Soln*.

f(x) = x^{3
}/ 2
- x^{2 }/ 3

d(f(x) / dx = 3 x^{2 }/ 2 -
2x/ 3

(d(f(x) / dx) ^{2} = (3 x^{2 }/ 2 -
2x/ 3 ) (3 x^{2 }/ 2 -
2x/ 3 )

= 9 x^{4 }/ 4 - 2 x^{3 } + 4 x^{2 }/ 9

Then:

L = ò** **^{2 }_{0}_{ } Ö**(1 + **9 x^{4 }/ 4
- 2 x^{3 } + 4 x^{2 }/ 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: **Ö**cosh ^{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:

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