The equation of the evolute is
r ⃗ ( t ) = x ( t ) ⋅ i ⃗ + y ( t ) ⋅ j ⃗ → { X ( t ) = x ( t ) − y ′ ( t ) ⋅ ( x ′ ( t ) 2 + y ′ ( t ) 2 ) x ′ ( t ) y ′ ′ ( t ) − x ′ ′ ( t ) y ′ ( t ) Y ( t ) = y ( t ) + x ′ ( t ) ⋅ ( x ′ ( t ) 2 + y ′ ( t ) 2 ) x ′ ( t ) y ′ ′ ( t ) − x ′ ′ ( t ) y ′ ( t ) \vec{r}(t)=x(t)\cdot\vec{i}+y(t)\cdot\vec{j}\to\\[0.3cm]
\left\{\begin{array}{l}
X(t)=x(t)-\displaystyle\frac{y'(t)\cdot\left(x'(t)^2+y'(t)^2\right)}{x'(t)y''(t)-x''(t)y'(t)}\\[0.5cm]
Y(t)=y(t)+\displaystyle\frac{x'(t)\cdot\left(x'(t)^2+y'(t)^2\right)}{x'(t)y''(t)-x''(t)y'(t)}
\end{array}\right. r ( t ) = x ( t ) ⋅ i + y ( t ) ⋅ j → ⎩ ⎨ ⎧ X ( t ) = x ( t ) − x ′ ( t ) y ′′ ( t ) − x ′′ ( t ) y ′ ( t ) y ′ ( t ) ⋅ ( x ′ ( t ) 2 + y ′ ( t ) 2 ) Y ( t ) = y ( t ) + x ′ ( t ) y ′′ ( t ) − x ′′ ( t ) y ′ ( t ) x ′ ( t ) ⋅ ( x ′ ( t ) 2 + y ′ ( t ) 2 )
(more infotmation : https://en.wikipedia.org/wiki/Evolute )
In our case,
{ x ( θ ) = a cos 3 ( θ ) y ( θ ) = a sin 3 ( θ ) → { x ′ ( θ ) = − 3 a cos 2 ( θ ) ⋅ sin ( θ ) x ′ ( θ ) = − 3 a 2 cos ( θ ) sin ( 2 θ ) x ′ ′ ( θ ) = − 3 a 2 ( − sin ( θ ) sin ( 2 θ ) + 2 cos ( θ ) cos ( 2 θ ) ) y ′ ( θ ) = 3 a sin 2 ( θ ) cos ( θ ) y ′ ( θ ) = 3 a 2 sin ( θ ) sin ( 2 θ ) y ′ ′ ( θ ) = 3 a 2 ( cos ( θ ) sin ( 2 θ ) + 2 sin ( θ ) cos ( 2 θ ) ) \left\{\begin{array}{c}
x(\theta)=a\cos^3(\theta)\\[0.3cm]
y(\theta)=a\sin^3(\theta)
\end{array}\right.\to\\[0.3cm]
\left\{\begin{array}{c}
x'(\theta)=-3a\cos^2(\theta)\cdot\sin(\theta)\\[0.3cm]
\boxed{x'(\theta)=\displaystyle\frac{-3a}{2}\cos(\theta)\sin(2\theta)}\\[0.5cm]
x''(\theta)=\displaystyle\frac{-3a}{2}(-\sin(\theta)\sin(2\theta)+2\cos(\theta)\cos(2\theta))\\[0.5cm]
y'(\theta)=3a\sin^2(\theta)\cos(\theta)\\[0.3cm]
\boxed{y'(\theta)=\displaystyle\frac{3a}{2}\sin(\theta)\sin(2\theta)}\\[0.5cm]
y''(\theta)=\displaystyle\frac{3a}{2}(\cos(\theta)\sin(2\theta)+2\sin(\theta)\cos(2\theta))
\end{array}\right. { x ( θ ) = a cos 3 ( θ ) y ( θ ) = a sin 3 ( θ ) → ⎩ ⎨ ⎧ x ′ ( θ ) = − 3 a cos 2 ( θ ) ⋅ sin ( θ ) x ′ ( θ ) = 2 − 3 a cos ( θ ) sin ( 2 θ ) x ′′ ( θ ) = 2 − 3 a ( − sin ( θ ) sin ( 2 θ ) + 2 cos ( θ ) cos ( 2 θ )) y ′ ( θ ) = 3 a sin 2 ( θ ) cos ( θ ) y ′ ( θ ) = 2 3 a sin ( θ ) sin ( 2 θ ) y ′′ ( θ ) = 2 3 a ( cos ( θ ) sin ( 2 θ ) + 2 sin ( θ ) cos ( 2 θ ))
Now we transform the expressions that enter the equations for the evolute :
x ′ ( θ ) 2 + y ′ ( θ ) 2 = ( − 3 a 2 cos ( θ ) sin ( 2 θ ) ) 2 + ( 3 a 2 sin ( θ ) sin ( 2 θ ) ) 2 = 9 a 2 4 ⋅ sin 2 ( 2 θ ) ⋅ ( cos 2 ( θ ) + sin 2 ( θ ) ) ≡ 9 a 2 4 ⋅ sin 2 ( 2 θ ) x ′ ( θ ) 2 + y ′ ( θ ) 2 = 9 a 2 4 ⋅ sin 2 ( 2 θ ) x ′ ( θ ) y ′ ′ ( θ ) − x ′ ′ ( θ ) y ′ ( θ ) = = − 3 a 2 cos ( θ ) sin ( 2 θ ) ⋅ 3 a 2 ( cos ( θ ) sin ( 2 θ ) + 2 sin ( θ ) cos ( 2 θ ) ) − − − 3 a 2 ( − sin ( θ ) sin ( 2 θ ) + 2 cos ( θ ) cos ( 2 θ ) ) ⋅ 3 a 2 sin ( θ ) sin ( 2 θ ) = = − 9 a 2 4 sin ( 2 θ ) ⋅ ( cos 2 ( θ ) sin ( 2 θ ) + 2 sin ( θ ) cos ( θ ) cos ( 2 θ ) + + sin 2 ( θ ) sin ( 2 θ ) − 2 sin ( θ ) cos ( θ ) cos ( 2 θ ) ) = = − 9 a 2 4 sin ( 2 θ ) ⋅ ( sin ( 2 θ ) ⋅ ( sin 2 ( θ ) + cos 2 ( θ ) ) ) = = − 9 a 2 4 sin 2 ( 2 θ ) x ′ ( θ ) y ′ ′ ( θ ) − x ′ ′ ( θ ) y ′ ( θ ) = − 9 a 2 4 sin 2 ( 2 θ ) x'(\theta)^2+y'(\theta)^2=\left(\frac{-3a}{2}\cos(\theta)\sin(2\theta)\right)^2+\left(\frac{3a}{2}\sin(\theta)\sin(2\theta)\right)^2\\[0.3cm]
=\frac{9a^2}{4}\cdot\sin^2(2\theta)\cdot\left(\cos^2(\theta)+\sin^2(\theta)\right)\equiv\frac{9a^2}{4}\cdot\sin^2(2\theta)\\[0.3cm]
\boxed{x'(\theta)^2+y'(\theta)^2=\frac{9a^2}{4}\cdot\sin^2(2\theta)}\\[0.3cm]
x'(\theta)y''(\theta)-x''(\theta)y'(\theta)=\\[0.3cm]
=\frac{-3a}{2}\cos(\theta)\sin(2\theta)\cdot\frac{3a}{2}(\cos(\theta)\sin(2\theta)+2\sin(\theta)\cos(2\theta))-\\[0.3cm]
-\frac{-3a}{2}(-\sin(\theta)\sin(2\theta)+2\cos(\theta)\cos(2\theta))\cdot\frac{3a}{2}\sin(\theta)\sin(2\theta)=\\[0.3cm]
=\frac{-9a^2}{4}\sin(2\theta)\cdot\left(\cos^2(\theta)\sin(2\theta)+2\sin(\theta)\cos(\theta)\cos(2\theta)+\right.\\[0.3cm]
\left.+\sin^2(\theta)\sin(2\theta)-2\sin(\theta)\cos(\theta)\cos(2\theta)\right)=\\[0.3cm]
=\frac{-9a^2}{4}\sin(2\theta)\cdot\left(\sin(2\theta)\cdot\left(\sin^2(\theta)+\cos^2(\theta)\right)\right)=\\[0.3cm]
=\frac{-9a^2}{4}\sin^2(2\theta)\\[0.3cm]
\boxed{x'(\theta)y''(\theta)-x''(\theta)y'(\theta)=\frac{-9a^2}{4}\sin^2(2\theta)} x ′ ( θ ) 2 + y ′ ( θ ) 2 = ( 2 − 3 a cos ( θ ) sin ( 2 θ ) ) 2 + ( 2 3 a sin ( θ ) sin ( 2 θ ) ) 2 = 4 9 a 2 ⋅ sin 2 ( 2 θ ) ⋅ ( cos 2 ( θ ) + sin 2 ( θ ) ) ≡ 4 9 a 2 ⋅ sin 2 ( 2 θ ) x ′ ( θ ) 2 + y ′ ( θ ) 2 = 4 9 a 2 ⋅ sin 2 ( 2 θ ) x ′ ( θ ) y ′′ ( θ ) − x ′′ ( θ ) y ′ ( θ ) = = 2 − 3 a cos ( θ ) sin ( 2 θ ) ⋅ 2 3 a ( cos ( θ ) sin ( 2 θ ) + 2 sin ( θ ) cos ( 2 θ )) − − 2 − 3 a ( − sin ( θ ) sin ( 2 θ ) + 2 cos ( θ ) cos ( 2 θ )) ⋅ 2 3 a sin ( θ ) sin ( 2 θ ) = = 4 − 9 a 2 sin ( 2 θ ) ⋅ ( cos 2 ( θ ) sin ( 2 θ ) + 2 sin ( θ ) cos ( θ ) cos ( 2 θ ) + + sin 2 ( θ ) sin ( 2 θ ) − 2 sin ( θ ) cos ( θ ) cos ( 2 θ ) ) = = 4 − 9 a 2 sin ( 2 θ ) ⋅ ( sin ( 2 θ ) ⋅ ( sin 2 ( θ ) + cos 2 ( θ ) ) ) = = 4 − 9 a 2 sin 2 ( 2 θ ) x ′ ( θ ) y ′′ ( θ ) − x ′′ ( θ ) y ′ ( θ ) = 4 − 9 a 2 sin 2 ( 2 θ )
Substitute the found expressions into the equations for the evolute, which were indicated at the very beginning :
X ( θ ) = x ( θ ) − y ′ ( θ ) ⋅ ( x ′ ( θ ) 2 + y ′ ( θ ) 2 ) x ′ ( θ ) y ′ ′ ( θ ) − x ′ ′ ( θ ) y ′ ( θ ) = = a cos 3 ( θ ) − 3 a 2 sin ( θ ) sin ( 2 θ ) ⋅ 9 a 2 4 sin 2 ( 2 θ ) − 9 a 2 4 sin 2 ( 2 θ ) = = a cos 3 ( θ ) + 3 a 2 sin ( θ ) sin ( 2 θ ) = a cos 3 ( θ ) + 3 a sin 2 ( θ ) cos ( θ ) = = a cos ( θ ) ( cos 2 ( θ ) + 3 sin 2 ( θ ) ) ≡ a cos ( θ ) ( 1 + 2 sin 2 ( θ ) ) X ( θ ) = a cos ( θ ) ( 1 + 2 sin 2 ( θ ) ) Y ( θ ) = y ( θ ) + x ′ ( θ ) ⋅ ( x ′ ( θ ) 2 + y ′ ( θ ) 2 ) x ′ ( θ ) y ′ ′ ( θ ) − x ′ ′ ( θ ) y ′ ( θ ) = = a sin 3 ( θ ) + − 3 a 2 cos ( θ ) sin ( 2 θ ) ⋅ 9 a 2 4 sin 2 ( 2 θ ) − 9 a 2 4 sin 2 ( 2 θ ) = = a sin 3 ( θ ) + 3 a 2 cos ( θ ) sin ( 2 θ ) = a sin 3 ( θ ) + 3 a cos 2 ( θ ) sin ( θ ) = = a sin ( θ ) ( sin 2 ( θ ) + 3 cos 2 ( θ ) ) ≡ a sin ( θ ) ( 1 + 2 cos 2 ( θ ) ) Y ( θ ) = a sin ( θ ) ( 1 + 2 cos 2 ( θ ) ) X(\theta)=x(\theta)-\frac{y'(\theta)\cdot\left(x'(\theta)^2+y'(\theta)^2\right)}{x'(\theta)y''(\theta)-x''(\theta)y'(\theta)}=\\[0.3cm]
=a\cos^3(\theta)-\frac{\displaystyle\frac{3a}{2}\sin(\theta)\sin(2\theta)\cdot\displaystyle\frac{9a^2}{4}\sin^2(2\theta)}{\displaystyle\frac{-9a^2}{4}\sin^2(2\theta)}=\\[0.3cm]
=a\cos^3(\theta)+\frac{3a}{2}\sin(\theta)\sin(2\theta)=a\cos^3(\theta)+3a\sin^2(\theta)\cos(\theta)=\\[0.3cm]
=a\cos(\theta)\left(\cos^2(\theta)+3\sin^2(\theta)\right)\equiv a\cos(\theta)\left(1+2\sin^2(\theta)\right)\\[0.3cm]
\boxed{X(\theta)=a\cos(\theta)\left(1+2\sin^2(\theta)\right)}\\[0.3cm]
Y(\theta)=y(\theta)+\frac{x'(\theta)\cdot\left(x'(\theta)^2+y'(\theta)^2\right)}{x'(\theta)y''(\theta)-x''(\theta)y'(\theta)}=\\[0.3cm]
=a\sin^3(\theta)+\frac{\displaystyle\frac{-3a}{2}\cos(\theta)\sin(2\theta)\cdot\displaystyle\frac{9a^2}{4}\sin^2(2\theta)}{\displaystyle\frac{-9a^2}{4}\sin^2(2\theta)}=\\[0.3cm]
=a\sin^3(\theta)+\frac{3a}{2}\cos(\theta)\sin(2\theta)=a\sin^3(\theta)+3a\cos^2(\theta)\sin(\theta)=\\[0.3cm]
=a\sin(\theta)\left(\sin^2(\theta)+3\cos^2(\theta)\right)\equiv a\sin(\theta)\left(1+2\cos^2(\theta)\right)\\[0.3cm]
\boxed{Y(\theta)=a\sin(\theta)\left(1+2\cos^2(\theta)\right)}\\[0.3cm] X ( θ ) = x ( θ ) − x ′ ( θ ) y ′′ ( θ ) − x ′′ ( θ ) y ′ ( θ ) y ′ ( θ ) ⋅ ( x ′ ( θ ) 2 + y ′ ( θ ) 2 ) = = a cos 3 ( θ ) − 4 − 9 a 2 sin 2 ( 2 θ ) 2 3 a sin ( θ ) sin ( 2 θ ) ⋅ 4 9 a 2 sin 2 ( 2 θ ) = = a cos 3 ( θ ) + 2 3 a sin ( θ ) sin ( 2 θ ) = a cos 3 ( θ ) + 3 a sin 2 ( θ ) cos ( θ ) = = a cos ( θ ) ( cos 2 ( θ ) + 3 sin 2 ( θ ) ) ≡ a cos ( θ ) ( 1 + 2 sin 2 ( θ ) ) X ( θ ) = a cos ( θ ) ( 1 + 2 sin 2 ( θ ) ) Y ( θ ) = y ( θ ) + x ′ ( θ ) y ′′ ( θ ) − x ′′ ( θ ) y ′ ( θ ) x ′ ( θ ) ⋅ ( x ′ ( θ ) 2 + y ′ ( θ ) 2 ) = = a sin 3 ( θ ) + 4 − 9 a 2 sin 2 ( 2 θ ) 2 − 3 a cos ( θ ) sin ( 2 θ ) ⋅ 4 9 a 2 sin 2 ( 2 θ ) = = a sin 3 ( θ ) + 2 3 a cos ( θ ) sin ( 2 θ ) = a sin 3 ( θ ) + 3 a cos 2 ( θ ) sin ( θ ) = = a sin ( θ ) ( sin 2 ( θ ) + 3 cos 2 ( θ ) ) ≡ a sin ( θ ) ( 1 + 2 cos 2 ( θ ) ) Y ( θ ) = a sin ( θ ) ( 1 + 2 cos 2 ( θ ) )
ANSWER
{ x ( θ ) = a cos 3 ( θ ) y ( θ ) = a sin 3 ( θ ) → { X ( θ ) = a cos ( θ ) ( 1 + 2 sin 2 ( θ ) ) Y ( θ ) = a sin ( θ ) ( 1 + 2 cos 2 ( θ ) ) \left\{\begin{array}{c}
x(\theta)=a\cos^3(\theta)\\[0.3cm]
y(\theta)=a\sin^3(\theta)
\end{array}\right.\to
\left\{\begin{array}{c}
X(\theta)=a\cos(\theta)\left(1+2\sin^2(\theta)\right)\\[0.3cm]
Y(\theta)=a\sin(\theta)\left(1+2\cos^2(\theta)\right)
\end{array}\right. { x ( θ ) = a cos 3 ( θ ) y ( θ ) = a sin 3 ( θ ) → { X ( θ ) = a cos ( θ ) ( 1 + 2 sin 2 ( θ ) ) Y ( θ ) = a sin ( θ ) ( 1 + 2 cos 2 ( θ ) )
#339914 on Dec 2023
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