Curvature for parametrically defined curves

2 Curvature for parametrically defined curves

An expression for the curvature is also available if the curve is described parametrically:

$x = g (t) y = h (t) t_{0} \leq t \leq t_{1}$

We remember the basic formulae connecting derivatives

$\frac{d y}{d x} = \frac{ẏ}{ẋ} \frac{d^{2} y}{d x^{2}} = \frac{ẋ ÿ - ẏ ẍ}{ẋ^{3}}$

where, as usual $ẋ \equiv \frac{d x}{d t}, ẍ \equiv \frac{d^{2} x}{d t^{2}} etc.$

Then

$κ = |\frac{f^{″} (x)}{{1 + {[f^{'} (x)]}^{2}}^{3 ∕ 2}}| = |\frac{ẋ ÿ - ẏ ẍ}{ẋ^{3} {[1 + {(\frac{ẏ}{ẋ})}^{2}]}^{3 ∕ 2}}|$

$= |\frac{ẋ ÿ - ẏ ẍ}{{[ẋ^{2} + ẏ^{2}]}^{3 ∕ 2}}|$

Key Point 7

The formula for curvature in parametric form is $κ = |\frac{ẋ ÿ - ẏ ẍ}{{[ẋ^{2} + ẏ^{2}]}^{3 ∕ 2}}|$

Task!

An ellipse is described parametrically by the equations

$x = 2 cos t y = sin t 0 \leq t \leq 2 π$

Obtain an expression for the curvature $κ$ and find where the curvature is a maximum or a minimum.

First find $ẋ, ẏ, ẍ, ÿ$ :

$ẋ = - 2 sin t ẏ = cos t ẍ = - 2 cos t ÿ = - sin t$

Now find $κ$ :

$κ = |\frac{ẋ ÿ - ẏ ẍ}{{[ẋ^{2} + ẏ^{2}]}^{3 ∕ 2}}| = |\frac{2 {sin}^{2} t + 2 {cos}^{2} t}{{[4 {sin}^{2} t + {cos}^{2} t]}^{3 ∕ 2}}| = \frac{2}{{[1 + 3 {sin}^{2} t]}^{3 ∕ 2}}$

Find maximum and minimum values of $κ$ by inspection of the expression for $κ$ :

Denominator is max when $t = π ∕ 2$ . This gives minimum value of $κ = 1 ∕ 4$ ,

Denominator is min when $t = 0$ . This gives maximum value of $κ = 2.$

2 Curvature for parametrically defined curves

Key Point 7

Task!

Answer

Answer

Answer