What is the equation of the tangent line of r=cos^2(theta-pi) +sin^2(theta-3pi)-theta at theta=(-13pi)/4?

Mar 10, 2016

$y = \frac{8 + 13 \pi}{13 \pi} x + \frac{{\left(4 + 13 \pi\right)}^{2}}{26 \sqrt{2} \pi}$

$\approx \left(1.196\right) x + 17.406$

Explanation:

First, simplify $r$.

$r = {\cos}^{2} \left(\theta - \pi\right) + {\sin}^{2} \left(\theta - 3 \pi\right) - \theta$

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

$= 1 - \theta$

When $\theta = - \frac{13 \pi}{4}$, $r = 1 + \frac{13 \pi}{4}$.

In cartesian coordinates, that would be

$x = r \cos \theta$

$= \left(1 + \frac{13 \pi}{4}\right) \cdot \left(- \frac{1}{\sqrt{2}}\right)$
$\approx - 7.927$

$y = r \sin \theta$

$= \left(1 + \frac{13 \pi}{4}\right) \cdot \left(\frac{1}{\sqrt{2}}\right)$
$\approx 7.927$

The tangent line has to pass through the point $\left(- 7.927 , 7.927\right)$.

The gradient of the tangent line is equal to frac{"d"y}{"d"x} at $\theta = - \frac{13 \pi}{4}$. To find frac{"d"y}{"d"x}, we use the chain rule and the product rule.

frac{"d"y}{"d"x} = frac{frac{"d"y}{"d" theta}}{frac{"d"x}{"d" theta}}

= frac{frac{"d"}{"d" theta}(rsintheta)}{frac{"d"}{"d" theta}(rcostheta)}

= frac{sinthetafrac{"d"}{"d" theta}(r)+rfrac{"d"}{"d" theta}(sintheta)}{costhetafrac{"d"}{"d" theta}(r)+rfrac{"d"}{"d" theta}(costheta)}

= frac{sinthetafrac{"d"r}{"d" theta}+rcostheta}{costhetafrac{"d"r}{"d" theta}-rsintheta}

Next, we find frac{"d"r}{"d"theta}.

$\frac{\text{d"r}{"d"theta} = frac{"d"}{"d} \theta}{1 - \theta}$

$= - 1$

Therefore,

$\frac{\text{d"y}{"d} x}{=} \frac{\sin \theta \left(- 1\right) + \left(1 - \theta\right) \cos \theta}{\cos \theta \left(- 1\right) - \left(1 - \theta\right) \sin \theta}$

$= \frac{- \sin \theta + \cos \theta - \theta \cos \theta}{- \cos \theta - \sin \theta + \theta \sin \theta}$.

So, now we can find frac{"d"y}{"d"x} at $\theta = - \frac{13 \pi}{4}$.

${\frac{\text{d"y}{"d} x}{|}}_{\theta = - \frac{13 \pi}{4}}$

$= \frac{- \sin \left(- \frac{13 \pi}{4}\right) + \cos \left(- \frac{13 \pi}{4}\right) - \left(- \frac{13 \pi}{4}\right) \cos \left(- \frac{13 \pi}{4}\right)}{- \cos \left(- \frac{13 \pi}{4}\right) - \sin \left(- \frac{13 \pi}{4}\right) + \left(- \frac{13 \pi}{4}\right) \sin \left(- \frac{13 \pi}{4}\right)}$

$= \frac{\sin \left(\frac{13 \pi}{4}\right) + \cos \left(\frac{13 \pi}{4}\right) + \left(\frac{13 \pi}{4}\right) \cos \left(\frac{13 \pi}{4}\right)}{- \cos \left(\frac{13 \pi}{4}\right) + \sin \left(\frac{13 \pi}{4}\right) + \left(\frac{13 \pi}{4}\right) \sin \left(\frac{13 \pi}{4}\right)}$

$= \frac{\left(- \frac{1}{\sqrt{2}}\right) + \left(- \frac{1}{\sqrt{2}}\right) + \left(\frac{13 \pi}{4}\right) \left(- \frac{1}{\sqrt{2}}\right)}{- \left(- \frac{1}{\sqrt{2}}\right) + \left(- \frac{1}{\sqrt{2}}\right) + \left(\frac{13 \pi}{4}\right) \left(- \frac{1}{\sqrt{2}}\right)}$

$= \frac{1 + 1 + \left(\frac{13 \pi}{4}\right) \left(1\right)}{- \left(1\right) + \left(1\right) + \left(\frac{13 \pi}{4}\right) \left(1\right)}$

$= \frac{2 + \frac{13 \pi}{4}}{\frac{13 \pi}{4}}$

$= \frac{8}{13 \pi} + 1$

$\approx = 1.196$

So now that we have the gradient of the line and one point that it passes through, we can write the equation of the line in point-slope form.

$y - \left(\frac{4 + 13 \pi}{4 \sqrt{2}}\right) = \frac{8 + 13 \pi}{13 \pi} \left(x - \left(- \frac{4 + 13 \pi}{4 \sqrt{2}}\right)\right)$

Now we rearrange the terms to get the slope-intercept form.

$y = \frac{8 + 13 \pi}{13 \pi} x + \left(\frac{8 + 13 \pi}{13 \pi} + 1\right) \frac{4 + 13 \pi}{4 \sqrt{2}}$

$= \frac{8 + 13 \pi}{13 \pi} x + \frac{{\left(4 + 13 \pi\right)}^{2}}{26 \sqrt{2} \pi}$

$\approx \left(1.196\right) x + 17.406$

Below is a graph for additional reference

$- 4 \pi \le \theta < 4 \pi$