# What is the slope of the tangent line of r=-8sin(theta/4)+4cos(theta/2) at theta=(2pi)/3?

Jun 3, 2016

the tangent line is $y = - 1.73205 - 0.57735 \left(x - 1\right)$

#### Explanation:

Given $r \left(\theta\right)$ we have

{ (x(theta)=r(theta)cos(theta)), (y(theta)=r(theta)sin(theta)) :}

 { (dx/(d theta) = -r(theta)sin(theta)+cos(theta)((dr)/(d theta))), (dy/(d theta) = r(theta)cos(theta)+sin(theta)((dr)/(d theta))) :}

but

$r \left(\theta\right) = - 8 S \in \left(\frac{\theta}{4}\right) + 4 C o s \left(\frac{\theta}{2}\right)$

then

 { (dx/(d theta) = 3 Cos((3 theta)/4) - 5 Cos((5 theta)/4) - Sin(theta/2) - 3 Sin((3 theta)/2)), (dy/(d theta)=Cos(theta/2) + 3 (Cos((3 theta)/2) + Sin((3 theta)/4)) - 5 Sin((5 theta)/4)) :}

but

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(\frac{\mathrm{dy}}{d \theta}\right)}{\left(\frac{\mathrm{dx}}{d \theta}\right)}$

at point ${\theta}_{0} = 2 \frac{\pi}{3}$ we have

${p}_{0} = \left\{r \left({\theta}_{0}\right) \cos \left({\theta}_{0}\right) , r \left({\theta}_{0}\right) \sin \left({\theta}_{0}\right)\right\} = \left\{1.0 , - 1.73205\right\}$

and

${m}_{0} = {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{{\theta}_{0}} = - 0.57735$

$y = - 1.73205 - 0.57735 \left(x - 1\right)$