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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.
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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Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3using 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
 one year ago
Best ResponseYou've already chosen the best response.3now 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
 one year ago
Best ResponseYou've already chosen the best response.3so we get this: \[{r^5} = 4{x^3}y  4x{y^3}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3nevertheless our computation becomes more difficult

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3please wait a moment

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3using cartesian coordinates, we can rewrite your equation as below: \[\Large {\left( {{x^2} + {y^2}} \right)^3} = 16{x^2}{y^2}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3now 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
 one year ago
Best ResponseYou've already chosen the best response.3now we can use: \[\Large r = 1,\theta = \frac{\pi }{8}\] so we have:

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3\[\Large \begin{gathered} x = r\cos \theta = 1 \cdot \cos \left( {\frac{\pi }{8}} \right) \hfill \\ y = r\sin \theta = 1 \cdot \sin \left( {\frac{\pi }{8}} \right) \hfill \\ \hfill \\ \end{gathered} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3we have to substitute those values into the last formula above, and solve it with respect to y'

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3\[\Large 3\left( {x + yy'} \right) = 16xy\left( {x + y} \right)\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3the 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
 one year ago
Best ResponseYou've already chosen the best response.3sorry 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
 one year ago
Best ResponseYou've already chosen the best response.3and solve that equation, with respect to y'

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0having trouble with the fractions and keep getting the wrong answer, can u elaborate

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3ok! please wait...

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3I got this: \[\Large y' = \frac{{{x_0}}}{{{y_0}}} \cdot \frac{{16y_0^2  3}}{{3  16x_0^2}}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3after 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
 one year ago
Best ResponseYou've already chosen the best response.0thats wasn't correct

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3please wait, I'm checking my computation

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3I 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
 one year ago
Best ResponseYou've already chosen the best response.3\[\Large \begin{gathered} y' = \frac{{12x_0^2{y_0}  4y_0^3  5x_0^3}}{{5{y_0} + 12{x_0}y_0^2  4x_0^3}} = \hfill \\ \hfill \\ = \left( {\frac{{{y_0}}}{{{x_0}}}} \right) \cdot \frac{{12x_0^2  4y_0^2  5\left( {{x_0}/{y_0}} \right)}}{{5\left( {{y_0}/{x_0}} \right) + 12y_0^2  4x_0^2}} \hfill \\ \end{gathered} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3please wait: I got this:

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3\[\Large m =  \frac{1}{{\sqrt 2  1}}\]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1there'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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