Question

1. Find the radius of curvature (ROC) at (a cos3 θ, a sin3

θ) on the curve x

2/3 +

y

2/3 = a

2/3

2. Find the radius of curvature (ROC) for the curve x = a(cos θ + θ sin θ), y =

a(sin θ − θ cos θ).

Answer 1

1.

The parametric equations are

$$x=a\cos^3 \theta=\dfrac{a}{4}(3\cos \theta+\cos 3\theta),$$

$$y=a\sin^3 \theta=\dfrac{a}{4}(3\sin \theta-\sin 3\theta),$$

$$x'=\dfrac{a}{4}(-3\sin \theta-3\sin 3\theta)$$

$$x''=\dfrac{a}{4}(-3\cos \theta-9\cos 3\theta)$$

$$y'=\dfrac{a}{4}(3\cos \theta-3\cos 3\theta)$$

$$y''=\dfrac{a}{4}(-3\sin \theta+9\sin 3\theta)$$

$$R=\dfrac{((x')^2+(y')^2)^{3/2}}{|x'y''-x''y'|}$$

$$(x')^2+(y')^2$$

$$=\dfrac{a^2}{16}(9\sin^2 \theta+9\sin^2 3\theta+18\sin\theta \sin 3\theta)$$

$$+\dfrac{a^2}{16}(9\cos^2 \theta+9\cos^2 3\theta-18\cos\theta \cos 3\theta)$$

$$=\dfrac{9a^2}{8}(1-\cos4\theta)=\dfrac{9a^2}{4}\sin^22\theta$$

$$((x')^2+(y')^2)^{3/2}=\dfrac{27a^3}{8}\sin^32\theta$$

$$x'y''-x''y'$$

$$=\dfrac{a^2}{16}(9\sin^2 \theta-27\sin^2 3\theta-18\sin\theta \sin 3\theta)$$

$$-\dfrac{a^2}{16}(-9\cos^2 \theta+27\cos^2 3\theta-18\cos\theta \cos 3\theta)$$

$$=-\dfrac{9a^2}{8}(1-\cos4\theta)=-\dfrac{9a^2}{4}\sin^22\theta$$

$$R=\dfrac{\dfrac{27a^3}{8}\sin^32\theta}{|-\dfrac{9a^2}{4}\sin^22\theta|}$$

$$=\dfrac{3a}{2}\sin2\theta=3a\sin\theta \cos\theta$$

$$R=3a\sin\theta \cos\theta$$

2.

$$x=a(\cos \theta+\theta \sin \theta)$$

$$y=a(\sin \theta-\theta \cos \theta)$$

$$x'=a(-\sin \theta+\sin \theta+\theta\cos \theta)=a(\theta \cos \theta)$$

$$x''=a(\cos \theta-\theta \sin \theta )$$

$$y'=a(\cos \theta-\cos \theta+\theta \sin \theta)=a(\theta \sin \theta)$$

$$y''=a(\sin \theta+\theta \cos \theta )$$

$$R=\dfrac{((x')^2+(y')^2)^{3/2}}{|x'y''-x''y'|}$$

$$(x')^2+(y')^2=a^2\theta^2(\cos62\theta+\sin^2\theta)=a^2\theta^2$$

$$((x')^2+(y')^2)^{3/2}=a^3\theta^3$$

$$x'y''-x''y'=a^2(\theta\sin\theta \cos\theta+\theta^2\cos^2\theta)$$

$$-a^2(\theta\sin\theta \cos\theta-\theta^2\sin^2\theta)=a^2\theta^2$$

$$R=\dfrac{a^3\theta^3}{|a^2\theta^2|}=a\theta$$