Question

.
Find the radius of curvature of the following : x = a sin^3theta; y = a cos^3theta​

Answer 1

Answer:

r=2f

f=r/2

Step-by-step explanation:

tnxxxx hope it helps

Answer 2

The radius of curvature is |3asinФcosФ|

Step-by-step explanation:

Given:

$x=a\sin^3\theta$

$y=a\cos^3\theta$

Curvature, $K=\dfrac{\left|x'y''-y'x''\right|}{\left|\left(x'\right)^{2}+\left(y'\right)^{2}\right|^{\frac{3}{2}}}$

$x'=3a\sin^2\theta\cos\theta$

$x''=a (6\sin \theta\cdot cos^2\theta - 3 \sin^3\theta)$

$y'=-3a\cos^2\theta\sin\theta$

$y''=-3 a (\cos^3\theta - 2 \sin^2\theta \cos \theta)$

Substitute into formula

$K=\dfrac{\left|(3a\sin^2\theta\cos\theta)[-3 a (\cos^3\theta - 2 \sin^2\theta \cos \theta)]-[-3a\cos^2\theta\sin\theta][a (6\sin \theta\cdot cos^2\theta - 3 \sin^3\theta)]\right|}{\left|\left(3a\sin^2\theta\cos\theta\right)^{2}+\left(-3a\cos^2\theta\sin\theta\right)^{2}\right|^{\frac{3}{2}}}$

$K=\dfrac{1}{3a|\sin\theta\cos\theta|}$

Radius of curvature is inverse of curvature.

$R=\dfrac{1}{K}$

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

#Learn more:

https://brainly.com/question/10642118