anonymous
  • anonymous
Find the slope of the tangent line to the polar curve r=sin(4θ) at θ=π/8.
Mathematics
chestercat
  • chestercat
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Michele_Laino
  • Michele_Laino
using the De Moivre's formula, and the tringle of Tartaglia or the triangle of Pascal, we get this: \[\sin \left( {4\theta } \right) = 4{\left( {\cos \theta } \right)^3}\sin \theta - 4\left( {\cos \theta } \right){\left( {\sin \theta } \right)^3}\]
Michele_Laino
  • Michele_Laino
now we can go to the cartesian coordinates using these formulas: \[\begin{gathered} x = r\cos \theta \hfill \\ y = r\sin \theta \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
so we get this: \[{r^5} = 4{x^3}y - 4x{y^3}\]

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Michele_Laino
  • Michele_Laino
nevertheless our computation becomes more difficult
Michele_Laino
  • Michele_Laino
please wait a moment
Michele_Laino
  • Michele_Laino
using cartesian coordinates, we can rewrite your equation as below: \[\Large {\left( {{x^2} + {y^2}} \right)^3} = 16{x^2}{y^2}\]
Michele_Laino
  • Michele_Laino
now we can compute the first derivative of both sides, so we get: \[\large 3{\left( {{x^2} + {y^2}} \right)^2}\left( {2x + 2yy'} \right) = 32xy\left( {x + y} \right)\]
Michele_Laino
  • Michele_Laino
now we can use: \[\Large r = 1,\theta = \frac{\pi }{8}\] so we have:
Michele_Laino
  • Michele_Laino
\[\Large \begin{gathered} x = r\cos \theta = 1 \cdot \cos \left( {\frac{\pi }{8}} \right) \hfill \\ y = r\sin \theta = 1 \cdot \sin \left( {\frac{\pi }{8}} \right) \hfill \\ \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
we have to substitute those values into the last formula above, and solve it with respect to y'
Michele_Laino
  • Michele_Laino
\[\Large 3\left( {x + yy'} \right) = 16xy\left( {x + y} \right)\]
Michele_Laino
  • Michele_Laino
the exact values are: \[\Large \begin{gathered} x = r\cos \theta = 1 \cdot \cos \left( {\frac{\pi }{8}} \right) = \frac{{\sqrt {2 + \sqrt 2 } }}{2} \hfill \\ \hfill \\ y = r\sin \theta = 1 \cdot \sin \left( {\frac{\pi }{8}} \right) = \frac{{\sqrt {2 - \sqrt 2 } }}{2} \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
sorry I have made an error, the right first derivative, is: \[\Large 3\left( {{x_0} + {y_0}y'} \right) = 16{x_0}{y_0}\left( {{y_0} + {x_0}y'} \right)\] where: \[\Large \begin{gathered} {x_0} = \frac{{\sqrt {2 + \sqrt 2 } }}{2} \hfill \\ {y_0} = \frac{{\sqrt {2 - \sqrt 2 } }}{2} \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
and solve that equation, with respect to y'
anonymous
  • anonymous
having trouble with the fractions and keep getting the wrong answer, can u elaborate
Michele_Laino
  • Michele_Laino
ok! please wait...
Michele_Laino
  • Michele_Laino
I got this: \[\Large y' = \frac{{{x_0}}}{{{y_0}}} \cdot \frac{{16y_0^2 - 3}}{{3 - 16x_0^2}}\]
Michele_Laino
  • Michele_Laino
after the substitution, I get this: \[\Large y' = \frac{{{x_0}}}{{{y_0}}} \cdot \frac{{16y_0^2 - 3}}{{3 - 16x_0^2}} = \frac{{57 - 40\sqrt 2 }}{{7\left( {\sqrt 2 - 1} \right)}}\]
anonymous
  • anonymous
thats wasn't correct
Michele_Laino
  • Michele_Laino
please wait, I'm checking my computation
Michele_Laino
  • Michele_Laino
I have made an error of sign, here is the right first derivative: \[\Large y' = \frac{{12x_0^2{y_0} - 4y_0^3 - 5{x_0}}}{{5{y_0} + 12{x_0}y_0^2 - 4x_0^3}}\] which can be simplified as below: \[\Large \begin{gathered} y' = \frac{{12x_0^2{y_0} - 4y_0^3 - 5{x_0}}}{{5{y_0} + 12{x_0}y_0^2 - 4x_0^3}} = \hfill \\ \hfill \\ = \left( {\frac{{{y_0}}}{{{x_0}}}} \right) \cdot \frac{{12x_0^2 - 4y_0^2 - 5\left( {{x_0}/{y_0}} \right)}}{{5\left( {{y_0}/{x_0}} \right) + 12y_0^2 - 4x_0^2}} \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
\[\Large \begin{gathered} y' = \frac{{12x_0^2{y_0} - 4y_0^3 - 5x_0^3}}{{5{y_0} + 12{x_0}y_0^2 - 4x_0^3}} = \hfill \\ \hfill \\ = \left( {\frac{{{y_0}}}{{{x_0}}}} \right) \cdot \frac{{12x_0^2 - 4y_0^2 - 5\left( {{x_0}/{y_0}} \right)}}{{5\left( {{y_0}/{x_0}} \right) + 12y_0^2 - 4x_0^2}} \hfill \\ \end{gathered} \]
anonymous
  • anonymous
great thanks
Michele_Laino
  • Michele_Laino
please wait: I got this:
Michele_Laino
  • Michele_Laino
\[\Large m = - \frac{1}{{\sqrt 2 - 1}}\]
IrishBoy123
  • IrishBoy123
there's also a simple hack for this one at \(\theta = \pi / 8\) we have \(dr/d \theta = 4 \ cos 4 \theta = 4 cos (\pi / 2) = 0\) which means you are right at the extreme of the first petal, which lies on the line \(\theta = \pi/8\) or \(y = tan(\pi/8) x\) the end of the petal also lies on the circle r = 1, so the tangent at that point is the tangent to the circle, ie the normal to \(y = tan(\pi/8) x\) so the slope at that point is \(\large -\frac{1}{tan(\pi/8)} = -2.414\) same value as @Michele_Laino 's expression https://www.desmos.com/calculator/tntrwjwrtd

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