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

 Aug 25, 2019
 #1
avatar+103674 
0

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

 

 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

.
 Aug 25, 2019
edited by Melody  Aug 25, 2019
 #2
avatar+23041 
+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 \(\mathbf{x = -1}\).

 

Source: https://www.youtube.com/watch?v=7Ae3VHPyMMQ

 

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

 

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

 

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) :
\(x = r \times \cos( \theta ) \\ y = r \times \sin( \theta )\)

 

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

 

laugh

 Aug 25, 2019
 #3
avatar+23041 
+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 \(\mathbf{x = -1}\).

source: https://www.youtube.com/watch?v=7Ae3VHPyMMQ

 

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

 

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

 

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: \((x^2+y^2) (1+x) = 2x^2 \quad | \quad r^2 = x^2+y^2,\ x = r \times \cos(\theta) ,\ y = r \times \sin(\theta) \)

 

laugh

 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
avatar+103674 
+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
avatar+23041 
+3

Thank you, Melody !

 

laugh

heureka  Aug 25, 2019

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