Question

2. Find the radius of curvature at any point a on the curve x=3acosa – acos3a
y= 3asina – asin3a​

Answer 1

Answer:

The radius of curvature is |3asinФcosФ|

Step-by-step explanation:

Given:

x=a\sin^3\thetax=asin

3

θ

y=a\cos^3\thetay=acos

3

θ

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

∣(x

′

)

2

+(y

′

)

2

∣

2

3

∣x

′

y

′′

−y

′

x

′′

∣

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

′

=3asin

2

θcosθ

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

′′

=a(6sinθ⋅cos

2

θ−3sin

3

θ)

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

′

=−3acos

2

θsinθ

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

′′

=−3a(cos

3

θ−2sin

2

θcosθ)

Substitute into formula

K=\dfrac{|(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)]|}{|(3a\sin^2\theta\cos\theta)^{2}+(-3a\cos^2\theta\sin\theta)^{2}|^{\frac{3}{2}}}K=

∣(3asin

2

θcosθ)

2

+(−3acos

2

θsinθ)

2

∣

2

3

∣(3asin

2

θcosθ)[−3a(cos

3

θ−2sin

2

θcosθ)]−[−3acos

2

θsinθ][a(6sinθ⋅cos

2

θ−3sin

3

θ)]∣

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

3a∣sinθcosθ∣

1

Radius of curvature is inverse of curvature.

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

K

1

R=|3a\sin\theta\cos\theta|R=∣3asinθcosθ∣

Answer 2

Answer:

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