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

Jan 14, 2017

Slope at $\theta = \frac{3 \pi}{8}$ is (-2cos((3pi)/8)sin(pi/16)+sin((3pi)/8)cos(pi/16))/(2sin((3pi)/8)sin(pi/16)+cos((3pi)/8)cos(pi/16)

Explanation:

$r = f \left(\theta\right) = 3 \sin \left(\frac{\theta}{2} - \frac{\pi}{4}\right)$

Relating this to Cartesian coordinates, we know that

$x = r \cos \theta = 3 \cos \theta \sin \left(\frac{\theta}{2} - \frac{\pi}{4}\right)$

Hence $\frac{\mathrm{dx}}{d \theta} = - 3 \sin \theta \sin \left(\frac{\theta}{2} - \frac{\pi}{4}\right) + \frac{3}{2} \cos \theta \cos \left(\frac{\theta}{2} - \frac{\pi}{4}\right)$

$y = r \sin \theta = 3 \sin \theta \sin \left(\frac{\theta}{2} - \frac{\pi}{4}\right)$

Hence $\frac{\mathrm{dy}}{d \theta} = 3 \cos \theta \sin \left(\frac{\theta}{2} - \frac{\pi}{4}\right) + \frac{3}{2} \sin \theta \cos \left(\frac{\theta}{2} - \frac{\pi}{4}\right)$

and hence $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3 \cos \theta \sin \left(\frac{\theta}{2} - \frac{\pi}{4}\right) + \frac{3}{2} \sin \theta \cos \left(\frac{\theta}{2} - \frac{\pi}{4}\right)}{- 3 \sin \theta \sin \left(\frac{\theta}{2} - \frac{\pi}{4}\right) + \frac{3}{2} \cos \theta \cos \left(\frac{\theta}{2} - \frac{\pi}{4}\right)}$

= $\frac{2 \cos \theta \sin \left(\frac{\theta}{2} - \frac{\pi}{4}\right) + \sin \theta \cos \left(\frac{\theta}{2} - \frac{\pi}{4}\right)}{- 2 \sin \theta \sin \left(\frac{\theta}{2} - \frac{\pi}{4}\right) + \cos \theta \cos \left(\frac{\theta}{2} - \frac{\pi}{4}\right)}$

and slope at $\theta = \frac{3 \pi}{8}$ is

$\frac{2 \cos \left(\frac{3 \pi}{8}\right) \sin \left(\frac{3 \pi}{16} - \frac{\pi}{4}\right) + \sin \left(\frac{3 \pi}{8}\right) \cos \left(\frac{3 \pi}{16} - \frac{\pi}{4}\right)}{- 2 \sin \left(\frac{3 \pi}{8}\right) \sin \left(\frac{3 \pi}{16} - \frac{\pi}{4}\right) + \cos \left(\frac{3 \pi}{8}\right) \cos \left(\frac{3 \pi}{8} - \frac{\pi}{4}\right)}$

= (-2cos((3pi)/8)sin(pi/16)+sin((3pi)/8)cos(pi/16))/(2sin((3pi)/8)sin(pi/16)+cos((3pi)/8)cos(pi/16)