Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

nevertheless our computation becomes more difficult

Answer 5

please wait a moment

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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:
\[\Large \begin{gathered}
{x_0} = \frac{{\sqrt {2 + \sqrt 2 } }}{2} \hfill \\
{y_0} = \frac{{\sqrt {2 - \sqrt 2 } }}{2} \hfill \\
\end{gathered} \]

Answer 14

and solve that equation, with respect to y'

Answer 15

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

Answer 16

ok! please wait...

Answer 17

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

Answer 18

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

Answer 19

thats wasn't correct

Answer 20

please wait, I'm checking my computation

Answer 21

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:

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

Answer 22

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

Answer 23

great thanks

Answer 24

please wait:
I got this:

Answer 25

\[\Large m = - \frac{1}{{\sqrt 2 - 1}}\]

Answer 26

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