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

1. Michele_Laino

using the De Moivre's formula, and the tringle of Tartaglia or the triangle of Pascal, we get this: $\sin \left( {4\theta } \right) = 4{\left( {\cos \theta } \right)^3}\sin \theta - 4\left( {\cos \theta } \right){\left( {\sin \theta } \right)^3}$

2. Michele_Laino

now we can go to the cartesian coordinates using these formulas: $\begin{gathered} x = r\cos \theta \hfill \\ y = r\sin \theta \hfill \\ \end{gathered}$

3. Michele_Laino

so we get this: ${r^5} = 4{x^3}y - 4x{y^3}$

4. Michele_Laino

nevertheless our computation becomes more difficult

5. Michele_Laino

6. Michele_Laino

using cartesian coordinates, we can rewrite your equation as below: $\Large {\left( {{x^2} + {y^2}} \right)^3} = 16{x^2}{y^2}$

7. Michele_Laino

now we can compute the first derivative of both sides, so we get: $\large 3{\left( {{x^2} + {y^2}} \right)^2}\left( {2x + 2yy'} \right) = 32xy\left( {x + y} \right)$

8. Michele_Laino

now we can use: $\Large r = 1,\theta = \frac{\pi }{8}$ so we have:

9. Michele_Laino

$\Large \begin{gathered} x = r\cos \theta = 1 \cdot \cos \left( {\frac{\pi }{8}} \right) \hfill \\ y = r\sin \theta = 1 \cdot \sin \left( {\frac{\pi }{8}} \right) \hfill \\ \hfill \\ \end{gathered}$

10. Michele_Laino

we have to substitute those values into the last formula above, and solve it with respect to y'

11. Michele_Laino

$\Large 3\left( {x + yy'} \right) = 16xy\left( {x + y} \right)$

12. Michele_Laino

the exact values are: $\Large \begin{gathered} x = r\cos \theta = 1 \cdot \cos \left( {\frac{\pi }{8}} \right) = \frac{{\sqrt {2 + \sqrt 2 } }}{2} \hfill \\ \hfill \\ y = r\sin \theta = 1 \cdot \sin \left( {\frac{\pi }{8}} \right) = \frac{{\sqrt {2 - \sqrt 2 } }}{2} \hfill \\ \end{gathered}$

13. 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: $\Large \begin{gathered} {x_0} = \frac{{\sqrt {2 + \sqrt 2 } }}{2} \hfill \\ {y_0} = \frac{{\sqrt {2 - \sqrt 2 } }}{2} \hfill \\ \end{gathered}$

14. Michele_Laino

and solve that equation, with respect to y'

15. anonymous

having trouble with the fractions and keep getting the wrong answer, can u elaborate

16. Michele_Laino

17. Michele_Laino

I got this: $\Large y' = \frac{{{x_0}}}{{{y_0}}} \cdot \frac{{16y_0^2 - 3}}{{3 - 16x_0^2}}$

18. Michele_Laino

after the substitution, I get this: $\Large y' = \frac{{{x_0}}}{{{y_0}}} \cdot \frac{{16y_0^2 - 3}}{{3 - 16x_0^2}} = \frac{{57 - 40\sqrt 2 }}{{7\left( {\sqrt 2 - 1} \right)}}$

19. anonymous

thats wasn't correct

20. Michele_Laino

please wait, I'm checking my computation

21. 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: $\Large \begin{gathered} y' = \frac{{12x_0^2{y_0} - 4y_0^3 - 5{x_0}}}{{5{y_0} + 12{x_0}y_0^2 - 4x_0^3}} = \hfill \\ \hfill \\ = \left( {\frac{{{y_0}}}{{{x_0}}}} \right) \cdot \frac{{12x_0^2 - 4y_0^2 - 5\left( {{x_0}/{y_0}} \right)}}{{5\left( {{y_0}/{x_0}} \right) + 12y_0^2 - 4x_0^2}} \hfill \\ \end{gathered}$

22. Michele_Laino

$\Large \begin{gathered} y' = \frac{{12x_0^2{y_0} - 4y_0^3 - 5x_0^3}}{{5{y_0} + 12{x_0}y_0^2 - 4x_0^3}} = \hfill \\ \hfill \\ = \left( {\frac{{{y_0}}}{{{x_0}}}} \right) \cdot \frac{{12x_0^2 - 4y_0^2 - 5\left( {{x_0}/{y_0}} \right)}}{{5\left( {{y_0}/{x_0}} \right) + 12y_0^2 - 4x_0^2}} \hfill \\ \end{gathered}$

23. anonymous

great thanks

24. Michele_Laino

25. Michele_Laino

$\Large m = - \frac{1}{{\sqrt 2 - 1}}$

26. 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