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anonymous

  • one year ago

Find the slope of the tangent line to the polar curve r=sin(4θ) at θ=π/8.

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  1. Michele_Laino
    • one year ago
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    using the De Moivre's formula, and the tringle of Tartaglia or the triangle of Pascal, we get this: \[\sin \left( {4\theta } \right) = 4{\left( {\cos \theta } \right)^3}\sin \theta - 4\left( {\cos \theta } \right){\left( {\sin \theta } \right)^3}\]

  2. Michele_Laino
    • one year ago
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    now we can go to the cartesian coordinates using these formulas: \[\begin{gathered} x = r\cos \theta \hfill \\ y = r\sin \theta \hfill \\ \end{gathered} \]

  3. Michele_Laino
    • one year ago
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    so we get this: \[{r^5} = 4{x^3}y - 4x{y^3}\]

  4. Michele_Laino
    • one year ago
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    nevertheless our computation becomes more difficult

  5. Michele_Laino
    • one year ago
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    please wait a moment

  6. Michele_Laino
    • one year ago
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    using cartesian coordinates, we can rewrite your equation as below: \[\Large {\left( {{x^2} + {y^2}} \right)^3} = 16{x^2}{y^2}\]

  7. Michele_Laino
    • one year ago
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    now we can compute the first derivative of both sides, so we get: \[\large 3{\left( {{x^2} + {y^2}} \right)^2}\left( {2x + 2yy'} \right) = 32xy\left( {x + y} \right)\]

  8. Michele_Laino
    • one year ago
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    now we can use: \[\Large r = 1,\theta = \frac{\pi }{8}\] so we have:

  9. Michele_Laino
    • one year ago
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    \[\Large \begin{gathered} x = r\cos \theta = 1 \cdot \cos \left( {\frac{\pi }{8}} \right) \hfill \\ y = r\sin \theta = 1 \cdot \sin \left( {\frac{\pi }{8}} \right) \hfill \\ \hfill \\ \end{gathered} \]

  10. Michele_Laino
    • one year ago
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    we have to substitute those values into the last formula above, and solve it with respect to y'

  11. Michele_Laino
    • one year ago
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    \[\Large 3\left( {x + yy'} \right) = 16xy\left( {x + y} \right)\]

  12. Michele_Laino
    • one year ago
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    the exact values are: \[\Large \begin{gathered} x = r\cos \theta = 1 \cdot \cos \left( {\frac{\pi }{8}} \right) = \frac{{\sqrt {2 + \sqrt 2 } }}{2} \hfill \\ \hfill \\ y = r\sin \theta = 1 \cdot \sin \left( {\frac{\pi }{8}} \right) = \frac{{\sqrt {2 - \sqrt 2 } }}{2} \hfill \\ \end{gathered} \]

  13. Michele_Laino
    • one year ago
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    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: \[\Large \begin{gathered} {x_0} = \frac{{\sqrt {2 + \sqrt 2 } }}{2} \hfill \\ {y_0} = \frac{{\sqrt {2 - \sqrt 2 } }}{2} \hfill \\ \end{gathered} \]

  14. Michele_Laino
    • one year ago
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    and solve that equation, with respect to y'

  15. anonymous
    • one year ago
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    having trouble with the fractions and keep getting the wrong answer, can u elaborate

  16. Michele_Laino
    • one year ago
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    ok! please wait...

  17. Michele_Laino
    • one year ago
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    I got this: \[\Large y' = \frac{{{x_0}}}{{{y_0}}} \cdot \frac{{16y_0^2 - 3}}{{3 - 16x_0^2}}\]

  18. Michele_Laino
    • one year ago
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    after the substitution, I get this: \[\Large y' = \frac{{{x_0}}}{{{y_0}}} \cdot \frac{{16y_0^2 - 3}}{{3 - 16x_0^2}} = \frac{{57 - 40\sqrt 2 }}{{7\left( {\sqrt 2 - 1} \right)}}\]

  19. anonymous
    • one year ago
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    thats wasn't correct

  20. Michele_Laino
    • one year ago
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    please wait, I'm checking my computation

  21. Michele_Laino
    • one year ago
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    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: \[\Large \begin{gathered} y' = \frac{{12x_0^2{y_0} - 4y_0^3 - 5{x_0}}}{{5{y_0} + 12{x_0}y_0^2 - 4x_0^3}} = \hfill \\ \hfill \\ = \left( {\frac{{{y_0}}}{{{x_0}}}} \right) \cdot \frac{{12x_0^2 - 4y_0^2 - 5\left( {{x_0}/{y_0}} \right)}}{{5\left( {{y_0}/{x_0}} \right) + 12y_0^2 - 4x_0^2}} \hfill \\ \end{gathered} \]

  22. Michele_Laino
    • one year ago
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    \[\Large \begin{gathered} y' = \frac{{12x_0^2{y_0} - 4y_0^3 - 5x_0^3}}{{5{y_0} + 12{x_0}y_0^2 - 4x_0^3}} = \hfill \\ \hfill \\ = \left( {\frac{{{y_0}}}{{{x_0}}}} \right) \cdot \frac{{12x_0^2 - 4y_0^2 - 5\left( {{x_0}/{y_0}} \right)}}{{5\left( {{y_0}/{x_0}} \right) + 12y_0^2 - 4x_0^2}} \hfill \\ \end{gathered} \]

  23. anonymous
    • one year ago
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    great thanks

  24. Michele_Laino
    • one year ago
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    please wait: I got this:

  25. Michele_Laino
    • one year ago
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    \[\Large m = - \frac{1}{{\sqrt 2 - 1}}\]

  26. IrishBoy123
    • one year ago
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    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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