What is the equation of the line that is normal to the polar curve $f (θ) = 2 sin (3 θ + \frac{π}{3}) - 2 θ cos θ$ $f(theta)=2sin(3theta+pi/3) - 2thetacostheta$ at $θ = \frac{π}{4}$ $theta = pi/4$ ?

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What is the equation of the line that is normal to the polar curve $f (θ) = 2 sin (3 θ + \frac{π}{3}) - 2 θ cos θ$ $f(theta)=2sin(3theta+pi/3) - 2thetacostheta$ at $θ = \frac{π}{4}$ $theta = pi/4$ ?

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Answer 1 · 2018-07-27T18:45:05

$y = \frac{- 8 - 8 \sqrt{3}}{- 12 - 4 \sqrt{3} + 2 π} (x - (\frac{2 - 2 \sqrt{3} - π}{4})) + (\frac{2 - 2 \sqrt{3} - π}{4})$ $y = (-8-8sqrt3)/(-12-4sqrt3+2pi)(x-((2-2sqrt3-pi)/4)) +((2-2sqrt3-pi)/4)$

Explanation:

To find the equation of the line, we need a $p o i n t$ $color(red)(poi nt)$ , and we need a $s l o p e$ $color(blue)(slope)$ .

First, let's find the $p o i n t$ $color(red)(poi nt)$ .

Plug in $\frac{π}{4}$ $pi/4$ into $f (θ)$ $f(theta)$ .

$f (\frac{π}{4}) = 2 sin (3 (\frac{π}{4}) + \frac{π}{3}) - 2 (\frac{π}{4}) (cos (\frac{π}{4}))$ $f(pi/4)=2sin(3(pi/4)+pi/3)-2(pi/4)(cos(pi/4))$
$f (\frac{π}{4}) = \frac{\sqrt{2} (2 - 2 \sqrt{3} - π)}{4}$ $f(pi/4)=(sqrt2(2-2sqrt3-pi))/4$

We have the polar coordinate $⎛ ⎜ ⎝ \frac{π}{4}, \frac{\sqrt{2} (2 - 2 \sqrt{3} - π)}{4} ⎞ ⎟ ⎠$ $(pi/4, (sqrt2(2-2sqrt3-pi))/4)$ .
We have to convert this to Cartesian form. Using the formulas:
$x = r cos θ$ $x = rcostheta$
$y = r cos θ$ $y = rcostheta$

We find that the Cartesian coordinate is $(\frac{2 - 2 \sqrt{3} - π}{4}, \frac{2 - 2 \sqrt{3} - π}{4})$ $color(red)(((2-2sqrt3-pi)/4,(2-2sqrt3-pi)/4)$ .

Good! We have found the $p o i n t$ $color(red)(poi nt)$ .

Now, we need to find the $s l o p e$ $color(blue)(slope)$ .

The formula for the tangent line of a polar function is:

$\frac{d y}{d x} = \frac{\frac{d r}{d θ} sin θ + r cos θ}{\frac{d r}{d θ} cos θ - r sin θ}$ $dy/dx = ((dr)/(d theta)sintheta+rcostheta)/((dr)/(d theta)costheta-rsintheta)$

If you notice, $sin θ$ $sintheta$ and $cos θ$ $costheta$ will be the same in this case, because $sin (\frac{π}{4})$ $sin(pi/4)$ and $cos (\frac{π}{4})$ $cos(pi/4)$ are the same, so for this case, our $\frac{d y}{d x}$ $dy/dx$ is simply:

$\frac{d y}{d x} = \frac{\frac{d r}{d θ} + r}{\frac{d r}{d θ} - r}$ $dy/dx = ((dr)/(d theta)+r)/((dr)/(d theta)-r)$

We solved $r$ $r$ already previously, which was $\frac{\sqrt{2} (2 - 2 \sqrt{3} - π)}{4}$ $(sqrt2(2-2sqrt3-pi))/4$ .

Now, we need to find $\frac{d r}{d θ}$ $(dr)/(d theta)$ .

$f'(theta)=(dr)/(d theta)=6cos(3theta+pi/3)+2thetasintheta-2costheta$

Now, we plug in $pi/4$ and get:

$f'(pi/4)=(dr)/(d theta)=(sqrt2(-10-6sqrt3+pi))/4$

Going back to the tangent line formula:

$dy/dx = ((sqrt2(-10-6sqrt3+pi))/4+(sqrt2(2-2sqrt3-pi))/4)/((sqrt2(-10-6sqrt3+pi))/4-(sqrt2(2-2sqrt3-pi))/4)$

After some tedious simplification, we find this reduces to $color(blue)(dy/dx = (-8-8sqrt3)/(-12-4sqrt3+2pi)$

We have our $color(red)(poi nt)$ :
$color(red)(((2-2sqrt3-pi)/4,(2-2sqrt3-pi)/4)$

We have our $color(blue)(slope)$ :
$color(blue)(dy/dx = (-8-8sqrt3)/(-12-4sqrt3+2pi)$

All that is left is to plug in these values into the point slope formula:
$(y - color(red)(y_1)) = color(blue)(m)(x - color(red)(x_1))$

We get:

$y - ((2-2sqrt3-pi)/4) = (-8-8sqrt3)/(-12-4sqrt3+2pi)(x-((2-2sqrt3-pi)/4))$

So the equation is:

$y = (-8-8sqrt3)/(-12-4sqrt3+2pi)(x-((2-2sqrt3-pi)/4)) +((2-2sqrt3-pi)/4)$

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What is the equation of the line that is normal to the polar curve $f (θ) = 2 sin (3 θ + \frac{π}{3}) - 2 θ cos θ$ $f(theta)=2sin(3theta+pi/3) - 2thetacostheta$ at $θ = \frac{π}{4}$ $theta = pi/4$ ?

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What is the equation of the line that is normal to the polar curve f(theta)=2sin(3theta+pi/3) - 2thetacostheta f(θ)=2sin(3θ+π3)−2θcosθf(theta)=2sin(3theta+pi/3) - 2thetacostheta at theta = pi/4θ=π4theta = pi/4?

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What is the equation of the line that is normal to the polar curve $f (θ) = 2 sin (3 θ + \frac{π}{3}) - 2 θ cos θ$ $f(theta)=2sin(3theta+pi/3) - 2thetacostheta$ at $θ = \frac{π}{4}$ $theta = pi/4$ ?