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# help

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Find the equation of the asymptote of the graph of  $$r = \cos 2 \theta \sec \theta.$$

Aug 25, 2019

#1
+105468
+1

$$r=cos2\theta sec\theta\\ r=\frac{cos^2\theta-sin^2\theta}{cos\theta}\qquad \theta \ne\frac{\pi }{2}\pm n\pi$$

It seems that the asymptotes are    $$\theta=\frac{\pi}{2}+ n\pi \quad where \quad n\in Z \quad (\text{n is an integer})$$

https://www.desmos.com/calculator/uvjgzlf5mr

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Aug 25, 2019
edited by Melody  Aug 25, 2019
#2
+23273
+2

Find the equation of the asymptote of the graph of  $$r = \cos 2 \theta \sec \theta$$.

I assume the function is a polar function:

polar plot see: https://www.wolframalpha.com/input/?i=polar+plot+r%3Dcos%282x%29sec%28x%29

The asymptote of the graph is $$\mathbf{x = -1}$$.

$$\begin{array}{|rcll|} \hline \mathbf{r} &=& \mathbf{\cos (2\theta) \sec( \theta)} \\\\ r &=& \dfrac{\cos (2\theta)} {\cos( \theta)} \quad | \quad \theta=\dfrac{\pi}{2} \Rightarrow r\to \infty \\\\ && \text{Set \alpha = \dfrac{\pi}{2} } \\\\ \dfrac{dr}{d\theta } &=& \dfrac{-2\sin(2\theta)\cos(\theta)+\cos(2\theta)\sin(\theta) } {\cos^2(\theta)} \\\\ \dfrac{d\theta }{dr} &=& \dfrac{\cos^2(\theta)} {-2\sin(2\theta)\cos(\theta)+\cos(2\theta)\sin(\theta) } \\\\ p &=& \lim \limits_{\theta\to \dfrac{\pi}{2}} r^2\dfrac{d\theta}{dr}\\ \\ p &=& \lim \limits_{\theta\to \dfrac{\pi}{2}} \dfrac{\cos^2(2\theta)} {\cos^2( \theta)} \left(\dfrac{\cos^2(\theta)} {-2\sin(2\theta)\cos(\theta)+\cos(2\theta)\sin(\theta) }\right) \\\\ p &=& \lim \limits_{\theta\to \dfrac{\pi}{2}} \dfrac{\cos^2(2\theta)} {-2\sin(2\theta)\cos(\theta)+\cos(2\theta)\sin(\theta) } \\\\ p &=& \lim \limits_{\theta\to \dfrac{\pi}{2}} \dfrac{\cos^2(2\dfrac{\pi}{2})} {-2\sin(2\dfrac{\pi}{2})\cos(\dfrac{\pi}{2})+\cos(2\dfrac{\pi}{2})\sin(\dfrac{\pi}{2}) } \\\\ p &=& \lim \limits_{\theta\to \dfrac{\pi}{2}} \dfrac{\cos^2(\pi)} {-2\sin(\pi)\cos(\dfrac{\pi}{2})+\cos(\pi)\sin(\dfrac{\pi}{2}) } \\\\ p &=& \dfrac{(-1)^2} {-2\cdot 0\cdot 0+(-1)(1)} \\\\ p &=& \dfrac{1} {-1} \\\\ \mathbf{p} &=& \mathbf{-1} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \mathbf{p} &=& \mathbf{r \sin(\alpha-\theta)} \quad | \quad r=-1,\ \alpha = \dfrac{\pi}{2} \\\\ -1 &=& r \sin(\dfrac{\pi}{2} -\theta) \\\\ -1 &=& r \cos( \theta) \\\\ \mathbf{r} &=&\mathbf{ -\dfrac{1}{\cos( \theta)} \quad | \quad \text{the asymptote of the graph }} \\ \hline \end{array}$$

polar plot asymtote see: https://www.wolframalpha.com/input/?i=polar+plot+r%3D+-1%2Fcos%28x%29

Asymptote in Cartesian Coordinates:

To convert from Polar Coordinates (r,θ) to Cartesian Coordinates (x,y) :
$$x = r \times \cos( \theta ) \\ y = r \times \sin( \theta )$$

$$\begin{array}{|rcll|} \hline \mathbf{r} &=&\mathbf{ -\dfrac{1}{\cos( \theta)} } \quad | \quad \cos{\theta}=\dfrac{x}{r} \\\\ r &=& -\dfrac{1}{ \dfrac{x}{r} } \\ r &=& -\dfrac{r}{ x } \\ xr &=& -r \quad | \quad :r \\\\ \mathbf{x} &=& \mathbf{-1 } \\ && \mathbf{\text{the asymptote of the graph in Cartesian Coordinates}} \\ \hline \end{array}$$

Aug 25, 2019
#3
+23273
+2

Find the equation of the asymptote of the graph of  $$r = \cos ( 2\theta) \sec( \theta)$$.

I assume the function is a polar function:
https://www.wolframalpha.com/input/?i=polar+plot+r%3Dcos%282x%29sec%28x%29

The asymptote of the graph is $$\mathbf{x = -1}$$.

$$\begin{array}{|rcll|} \hline \mathbf{r} &=& \mathbf{\cos (2\theta) \sec( \theta)} \\\\ r &=& \dfrac{\cos (2\theta)} {\cos( \theta)} \quad | \quad \theta=\dfrac{\pi}{2} \Rightarrow r\to \infty \\\\ && \text{Set \alpha = \dfrac{\pi}{2} } \\\\ \dfrac{dr}{d\theta } &=& \dfrac{-2\sin(2\theta)\cos(\theta)+\cos(2\theta)\sin(\theta) } {\cos^2(\theta)} \\\\ \dfrac{d\theta }{dr} &=& \dfrac{\cos^2(\theta)} {-2\sin(2\theta)\cos(\theta)+\cos(2\theta)\sin(\theta) } \\\\ p &=& \lim \limits_{\theta\to \dfrac{\pi}{2}} r^2\dfrac{d\theta}{dr}\\ \\ p &=& \lim \limits_{\theta\to \dfrac{\pi}{2}} \dfrac{\cos^2(2\theta)} {\cos^2( \theta)} \left(\dfrac{\cos^2(\theta)} {-2\sin(2\theta)\cos(\theta)+\cos(2\theta)\sin(\theta) }\right) \\\\ p &=& \lim \limits_{\theta\to \dfrac{\pi}{2}} \dfrac{\cos^2(2\theta)} {-2\sin(2\theta)\cos(\theta)+\cos(2\theta)\sin(\theta) } \\\\ p &=& \lim \limits_{\theta\to \dfrac{\pi}{2}} \dfrac{\cos^2(2\dfrac{\pi}{2})} {-2\sin(2\dfrac{\pi}{2})\cos(\dfrac{\pi}{2})+\cos(2\dfrac{\pi}{2})\sin(\dfrac{\pi}{2}) } \\\\ p &=& \lim \limits_{\theta\to \dfrac{\pi}{2}} \dfrac{\cos^2(\pi)} {-2\sin(\pi)\cos(\dfrac{\pi}{2})+\cos(\pi)\sin(\dfrac{\pi}{2}) } \\\\ p &=& \dfrac{(-1)^2} {-2\cdot 0\cdot 0+(-1)(1)} \\\\ p &=& \dfrac{1} {-1} \\\\ \mathbf{p} &=& \mathbf{-1} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \mathbf{p} &=& \mathbf{r \sin(\alpha-\theta)} \quad | \quad r=-1,\ \alpha = \dfrac{\pi}{2} \\\\ -1 &=& r \sin(\dfrac{\pi}{2} -\theta) \\\\ -1 &=& r \cos( \theta) \\\\ \mathbf{r} &=&\mathbf{ -\dfrac{1}{\cos( \theta)} \quad | \quad \text{the asymptote of the graph }} \\ \hline \end{array}$$

polar plot asymptote see: https://www.wolframalpha.com/input/?i=polar+plot+r%3D+-1%2Fcos%28x%29

Asymptote in Cartesian Coordinates:
To convert from Polar Coordinates (r,$$\theta$$) to Cartesian Coordinates (x,y) :
$$x = r \times \cos(\theta) \\ y = r \times \sin(\theta)$$

$$\begin{array}{|rcll|} \hline \mathbf{r} &=&\mathbf{ -\dfrac{1}{\cos( \theta)} } \quad | \quad \cos{\theta}=\dfrac{x}{r} \\\\ r &=& -\dfrac{1}{ \dfrac{x}{r} } \\ r &=& -\dfrac{r}{ x } \\ xr &=& -r \quad | \quad :r \\\\ \mathbf{x} &=& \mathbf{-1 } \\ && \mathbf{\text{the asymptote of the graph in Cartesian Coordinates}} \\ \hline \end{array}$$

hint: The function $$r = \cos ( 2\theta) \sec( \theta)$$ in cartesian coordinates: $$(x^2+y^2) (1+x) = 2x^2 \quad | \quad r^2 = x^2+y^2,\ x = r \times \cos(\theta) ,\ y = r \times \sin(\theta)$$

Aug 25, 2019
edited by heureka  Aug 25, 2019
edited by heureka  Aug 25, 2019
edited by heureka  Aug 25, 2019
edited by heureka  Aug 25, 2019
#4
+105468
+2

Hi Heueka and guest,

Heureka is right, it should be looked at in polar form.

Aug 25, 2019
edited by Melody  Aug 25, 2019
#5
+23273
+3

Thank you, Melody !

heureka  Aug 25, 2019