Find the total arc length of this astroid: x(@)=14cos^3(@), y(@)=14sin^3(@)
- Thread starter rmcginn
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D
Deleted member 4993
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What are your thoughts?
Please share your work with us ...even if you know it is wrong
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This problem looks interesting. I would start by trying to find a way to relate x and y...
\(\displaystyle \displaystyle \begin{align*} x &= 14\cos^3{\left( \theta \right) } \\ x^2 &= 196\cos^6{\left( \theta \right) } \\ \\ y &= 14\sin^3{ \left( \theta \right) } \\ y^2 &= 196\sin^6{\left( \theta \right) } \\ \\ x^2 + y^2 &= 196\cos^6{ \left( \theta \right) } + 196\sin^6{\left( \theta \right) } \\ &= 196\,\left\{ \left[ \cos^2{\left( \theta \right) } \right] ^3 + \left[ \sin^2{ \left( \theta \right) } \right] ^3 \right\} \\ &= 196\,\left[ \cos^2{\left( \theta \right) } + \sin^2{ \left( \theta \right) } \right] \left[ \cos^4{\left( \theta \right) } - \cos^2{ \left( \theta \right) } \sin^2{ \left( \theta \right) } + \sin^4{ \left( \theta \right) } \right] \\ &= 196\,\left( 1 \right) \left[ \cos^4{\left( \theta \right) } - \cos^2{ \left( \theta \right) } \sin^2{ \left( \theta \right) } + \sin^4{ \left( \theta \right) } \right] \\ &= 196\,\left\{ \cos^4{ \left( \theta \right) } - \cos^2{ \left( \theta \right) } \left[ 1 - \cos^2{ \left( \theta \right) } \right] + \left[ 1 - \cos^2{ \left( \theta \right) } \right] ^2 \right\} \\ &= 196\,\left[ \cos^4{ \left( \theta \right) } - \cos^2{ \left( \theta \right) } + \cos^4{ \left( \theta \right) } + 1 - 2\cos^2{ \left( \theta \right) } + \cos^4{ \left( \theta \right) } \right] \\ &= 196\,\left[ 3\cos^4{ \left( \theta \right) } - 3\cos^2{ \left( \theta \right) } + 1 \right] \\ &= 588\cos^4{ \left( \theta \right) } - 588\cos^2{ \left( \theta \right) } + 196 \\ &= 3\cdot 14^{\frac{2}{3}}\,\left[ 14\cos^3{ \left( \theta \right) } \right]^{\frac{4}{3}} - 42\cdot 14^{\frac{1}{3}}\,\left[ 14\cos^3{ \left( \theta \right) } \right] ^{\frac{2}{3}} + 196 \\ &= 3 \cdot 14^{\frac{2}{3}}\,x^{\frac{4}{3}} - 42 \cdot 14^{\frac{1}{3}}\,x^{\frac{2}{3}} + 196 \\ y^2 &= 3 \cdot 14^{\frac{2}{3}}\,x^{\frac{4}{3}} - 42\cdot 14^{\frac{1}{3}} \, x^{\frac{2}{3}} - x^2 + 196 \end{align*}\)
Hopefully you can now get \(\displaystyle \displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} \end{align*}\) (you may have to use Implicit Differentiation) so that you can apply the arclength formula \(\displaystyle \displaystyle \begin{align*} \mathcal{l} = \int_a^b{ \sqrt{1 + \left( \frac{\mathrm{d}y}{\mathrm{d}x} \right) ^2 } \,\mathrm{d}x } \end{align*}\).
D
Deleted member 4993
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cos2(Θ) = (x/14)2/3
sin2(Θ) = (y/14)2/3
(x/14)2/3 + (y/14)2/3 = 1
dy/dx = (dy/dΘ)/(dx/dΘ)
dy/dx = [-42 * sin(Θ) * cos2(Θ)]/[42 * cos(Θ) * sin2(Θ)] = - cot(Θ)
Now continue.....
That is definitely an easier way to relate x and y
Why not just calculate \(\displaystyle \displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} \end{align*}\) from the relation that you have found, instead of putting it in terms of \(\displaystyle \displaystyle \begin{align*} \theta \end{align*}\) though?
\(\displaystyle \displaystyle \begin{align*} \frac{\mathrm{d}}{\mathrm{d}x}\,\left[ \left( \frac{x}{14} \right) ^{\frac{2}{3}} + \left( \frac{y}{14} \right) ^{\frac{2}{3}} \right] &= \frac{ \mathrm{d} }{ \mathrm{d} x } \, \left( 1 \right) \\ 14^{ - \frac{2}{3} } \cdot \frac{2}{3} \, x^{ - \frac{1}{3} } + 14^{ - \frac{2}{3} } \cdot \frac{2}{3} \, y^{ -\frac{1}{3} } \, \frac{ \mathrm{d} y }{ \mathrm{d} x } &= 0 \\ x^{ -\frac{1}{3} } + y^{ -\frac{1}{3} } \, \frac{\mathrm{d} y }{ \mathrm{d} x } &= 0 \\ y^{ - \frac{1}{3} } \, \frac{ \mathrm{d} y }{ \mathrm{d} x } &= - x^{ - \frac{1}{3} } \\ \frac{ \mathrm{d} y }{ \mathrm{d} x} &= -x^{-\frac{1}{3}}\,y^{\frac{1}{3}} \end{align*}\)
and since \(\displaystyle \displaystyle \begin{align*} \left( \frac{x}{14} \right) ^{\frac{2}{3}} + \left( \frac{y}{14} \right) ^{\frac{2}{3}} &= 1 \end{align*}\) that means
\(\displaystyle \displaystyle \begin{align*} x^{\frac{2}{3}} + y^{\frac{2}{3}} &= 14^{\frac{2}{3}} \\ y^{\frac{2}{3}} &= 14^{\frac{2}{3}} - x^{\frac{2}{3}} \\ y^{\frac{1}{3}} &= \pm \left( 14^{\frac{2}{3}} - x^{\frac{2}{3}} \right) ^{\frac{1}{2}} \end{align*}\)
and thus
\(\displaystyle \displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} &= -x^{-\frac{1}{3}}\,\left[ \pm \left( 14^{\frac{2}{3}} - x^{\frac{2}{3}} \right)^{\frac{1}{2}} \right] \\ &= \mp x^{-\frac{1}{3}}\,\left( 14^{\frac{2}{3}} - x^{\frac{2}{3}} \right) ^{\frac{1}{2}} \\ &= \mp \left( x^{-\frac{2}{3}} \right) ^{\frac{1}{2}} \,\left( 14^{\frac{2}{3}} - x^{\frac{2}{3}} \right) ^{\frac{1}{2}} \\ &= \mp \left[ x^{-\frac{2}{3}} \,\left( 14^{\frac{2}{3}} - x^{\frac{2}{3}} \right) \right] ^{\frac{1}{2}} \\ &= \mp \left[ \left( \frac{14}{x} \right) ^{\frac{2}{3}} - 1 \right] ^{\frac{1}{2}} \end{align*}\)
D
Deleted member 4993
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That is definitely an easier way to relate x and y
Why not just calculate \(\displaystyle \displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} \end{align*}\) from the relation that you have found, instead of putting it in terms of \(\displaystyle \displaystyle \begin{align*} \theta \end{align*}\) though?
\(\displaystyle \displaystyle \begin{align*} \frac{\mathrm{d}}{\mathrm{d}x}\,\left[ \left( \frac{x}{14} \right) ^{\frac{2}{3}} + \left( \frac{y}{14} \right) ^{\frac{2}{3}} \right] &= \frac{ \mathrm{d} }{ \mathrm{d} x } \, \left( 1 \right) \\ 14^{ - \frac{2}{3} } \cdot \frac{2}{3} \, x^{ - \frac{1}{3} } + 14^{ - \frac{2}{3} } \cdot \frac{2}{3} \, y^{ -\frac{1}{3} } \, \frac{ \mathrm{d} y }{ \mathrm{d} x } &= 0 \\ x^{ -\frac{1}{3} } + y^{ -\frac{1}{3} } \, \frac{\mathrm{d} y }{ \mathrm{d} x } &= 0 \\ y^{ - \frac{1}{3} } \, \frac{ \mathrm{d} y }{ \mathrm{d} x } &= - x^{ - \frac{1}{3} } \\ \frac{ \mathrm{d} y }{ \mathrm{d} x} &= -x^{-\frac{1}{3}}\,y^{\frac{1}{3}} \end{align*}\)
and since \(\displaystyle \displaystyle \begin{align*} \left( \frac{x}{14} \right) ^{\frac{2}{3}} + \left( \frac{y}{14} \right) ^{\frac{2}{3}} &= 1 \end{align*}\) that means
\(\displaystyle \displaystyle \begin{align*} x^{\frac{2}{3}} + y^{\frac{2}{3}} &= 14^{\frac{2}{3}} \\ y^{\frac{2}{3}} &= 14^{\frac{2}{3}} - x^{\frac{2}{3}} \\ y^{\frac{1}{3}} &= \pm \left( 14^{\frac{2}{3}} - x^{\frac{2}{3}} \right) ^{\frac{1}{2}} \end{align*}\)
and thus
\(\displaystyle \displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} &= -x^{-\frac{1}{3}}\,\left[ \pm \left( 14^{\frac{2}{3}} - x^{\frac{2}{3}} \right)^{\frac{1}{2}} \right] \\ &= \mp x^{-\frac{1}{3}}\,\left( 14^{\frac{2}{3}} - x^{\frac{2}{3}} \right) ^{\frac{1}{2}} \\ &= \mp \left( x^{-\frac{2}{3}} \right) ^{\frac{1}{2}} \,\left( 14^{\frac{2}{3}} - x^{\frac{2}{3}} \right) ^{\frac{1}{2}} \\ &= \mp \left[ x^{-\frac{2}{3}} \,\left( 14^{\frac{2}{3}} - x^{\frac{2}{3}} \right) \right] ^{\frac{1}{2}} \\ &= \mp \left[ \left( \frac{14}{x} \right) ^{\frac{2}{3}} - 1 \right] ^{\frac{1}{2}} \end{align*}\)
You are getting dy/dx to be constant - that turns into a straight line!! ← wrong statement did not see the 'x' in the denominator
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Did you realise that the 14 is being divided by x???
D
Deleted member 4993
Guest
No I didn't - that's what happens when I try to deal with algebra before that cup of coffee.....
I think it may be easier to go like this: (abbreviations used to save chalk dust, er... ink?)
x = a cos^3(t) << never mind that stupid '14'.
dx = - 3a cos^2 sin dt
dx^2 = 9a^2 cos^4 sin^2 dt^2
..................
y = a sin^3(t)
dy = - 3a cos sin^2 dt
dy^2 = 9a^2 cos^2 sin^4 dt^2
..................
Sum = 9a^2 cos^2 sin^2(cos^2 + sin^2)
= 9a^2 cos^2 sin^2
arc length = 3a INTEGRAL[0,2PI] cos t sin t dt
etc. etc..
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