# What is the slope of the polar curve f(theta) = theta^2 - csctheta  at theta = (3pi)/4?

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#### Explanation:

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Feb 24, 2018

Slope of the curve at
$\theta = \frac{3 \pi}{4}$is -1

#### Explanation:

Given:
$f \left(\theta\right) = {\theta}^{2} - \csc \theta$

Let
$r = f \left(\theta\right)$

Then,
$r = {\theta}^{2} - \csc \theta$

At $\theta = \frac{3 \pi}{4}$

$r = {\left(\frac{3 \pi}{4}\right)}^{2} - \csc \left(\frac{3 \pi}{4}\right)$

${\left(\frac{3 \pi}{4}\right)}^{2} = \frac{9 {\pi}^{2}}{16}$

$\csc \left(\frac{3 \pi}{4}\right) = \sqrt{2}$

$r = \frac{9 {\pi}^{2}}{16} - \sqrt{2}$

$\frac{\mathrm{dr}}{d \left(\theta\right)} = 2 \theta - \left(- \csc \theta \cot \theta\right)$

$\frac{\mathrm{dr}}{d \left(\theta\right)} = 2 \theta + \csc \theta \cot \theta$

$\frac{\mathrm{dr}}{d \left(\theta\right)} = 2 \left(\frac{3 \pi}{4}\right) + \csc \left(\frac{3 \pi}{4}\right) \cot \left(\frac{3 \pi}{4}\right)$

$2 \left(\frac{3 \pi}{4}\right) = \frac{3 \pi}{2}$

$\csc \left(\frac{3 \pi}{4}\right) = \sqrt{2}$

$\cot \left(\frac{3 \pi}{4}\right) = - 1$

$\frac{\mathrm{dr}}{d \left(\theta\right)} = \frac{3 \pi}{2} + \sqrt{2} \left(- 1\right)$

$\frac{\mathrm{dr}}{d \left(\theta\right)} = \frac{3 \pi}{2} - \sqrt{2}$

To find

Slope of the curve at
$\theta = \frac{3 \pi}{4}$

$x = r \cos \theta$

$y = r \sin \theta$

$\frac{\mathrm{dx}}{d \left(\theta\right)} = r \left(- \sin \theta\right) + \cos \theta \frac{\mathrm{dr}}{d \left(\theta\right)}$

$\frac{\mathrm{dx}}{d \left(\theta\right)} = - r \left(\sin \theta\right) + \cos \theta \frac{\mathrm{dr}}{d \left(\theta\right)}$

$r = \frac{9 {\pi}^{2}}{16} - \frac{1}{\sqrt{2}}$

$\sin \theta = \sin \left(\frac{3 \pi}{4}\right)$

$\sin \left(\frac{3 \pi}{4}\right) = \frac{1}{\sqrt{2}}$

$\cos \theta = \cos \left(\frac{3 \pi}{4}\right)$

$\cos \left(\frac{3 \pi}{4}\right) = - \frac{1}{\sqrt{2}}$

$\cos \theta = - \frac{1}{\sqrt{2}}$

$\frac{\mathrm{dr}}{d \left(\theta\right)} = \frac{3 \pi}{2} - \sqrt{2}$

$\frac{\mathrm{dx}}{d \left(\theta\right)} = - \left(\frac{9 {\pi}^{2}}{16} - \frac{1}{\sqrt{2}}\right) \left(\frac{1}{\sqrt{2}}\right) + \left(- \frac{1}{\sqrt{2}}\right) \left(\frac{3 \pi}{2} - \sqrt{2}\right)$

$\frac{\mathrm{dx}}{d \left(\theta\right)} = \frac{1}{\sqrt{2}} \left(\frac{9 {\pi}^{2}}{16} - \frac{1}{\sqrt{2}}\right) - \frac{1}{\sqrt{2}} \left(\frac{3 \pi}{2} - \sqrt{2}\right)$

$\frac{\mathrm{dx}}{d \left(\theta\right)} = \frac{1}{\sqrt{2}} \left(\left(\frac{9 {\pi}^{2}}{16} - \frac{1}{\sqrt{2}}\right) - \left(\frac{3 \pi}{2} - \sqrt{2}\right)\right)$

$\frac{\mathrm{dy}}{d \left(\theta\right)} = r \cos \theta + \sin \theta \frac{\mathrm{dr}}{d \left(\theta\right)}$

$r = \frac{9 {\pi}^{2}}{16} - \frac{1}{\sqrt{2}}$

$\sin \theta = \sin \left(\frac{3 \pi}{4}\right)$

$\sin \left(\frac{3 \pi}{4}\right) = \frac{1}{\sqrt{2}}$

$\cos \theta = \cos \left(\frac{3 \pi}{4}\right)$

$\cos \left(\frac{3 \pi}{4}\right) = - \frac{1}{\sqrt{2}}$

$\cos \theta = - \frac{1}{\sqrt{2}}$

$\frac{\mathrm{dr}}{d \left(\theta\right)} = \frac{3 \pi}{2} - \sqrt{2}$

$\frac{\mathrm{dy}}{d \left(\theta\right)} = \left(\frac{9 {\pi}^{2}}{16} - \frac{1}{\sqrt{2}}\right) \left(- \frac{1}{\sqrt{2}}\right) + \left(\frac{1}{\sqrt{2}}\right) \left(\frac{3 \pi}{2} - \sqrt{2}\right)$

$\frac{\mathrm{dy}}{d \left(\theta\right)} = - \frac{1}{\sqrt{2}} \left(\frac{9 {\pi}^{2}}{16} - \frac{1}{\sqrt{2}}\right) + \left(\frac{1}{\sqrt{2}}\right) \left(\frac{3 \pi}{2} - \sqrt{2}\right)$

$\frac{\mathrm{dy}}{d \left(\theta\right)} = - \frac{1}{\sqrt{2}} \left(\left(\frac{9 {\pi}^{2}}{16} - \frac{1}{\sqrt{2}}\right) - \left(\frac{3 \pi}{2} - \sqrt{2}\right)\right)$

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

$= \frac{- \frac{1}{\sqrt{2}} \left(\left(\frac{9 {\pi}^{2}}{16} - \frac{1}{\sqrt{2}}\right) - \left(\frac{3 \pi}{2} - \sqrt{2}\right)\right)}{\frac{1}{\sqrt{2}} \left(\left(\frac{9 {\pi}^{2}}{16} - \frac{1}{\sqrt{2}}\right) - \left(\frac{3 \pi}{2} - \sqrt{2}\right)\right)}$

Cancelling

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 1$

Slope of the curve at
$\theta = \frac{3 \pi}{4}$is -1

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