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Graphing polar equations Quiz 2

mathematics

 What is the polar form of the equation $xy = 8$

 Substitute  $x = r\cos \theta $ and $y = r\sin \theta $to make the equation in terms of polar coordinates.

 Given: $ xy = 8$ $ \\  x = r\cos \theta ,y = r\sin \theta  \\   \Rightarrow r\cos \theta sin\theta  = 8 \\   \Rightarrow {r^2}\cos \theta sin\theta  = 8 \\   \Rightarrow {r^2} = \dfrac{8}{{\cos \theta \sin \theta }} \\  $ Multiply and divide by 2 $ \\   \Rightarrow {r^2} = \dfrac{{8 \times 2}}{{2\cos \theta \sin \theta }} \\   \Rightarrow {r^2} = \dfrac{{16}}{{\sin 2\theta }} = 16.\dfrac{1}{{\sin 2\theta }} \\   \Rightarrow {r^2} = 16\operatorname{cosec} 2\theta  \\  $

 The polar coordinates of a point are $\left( { - 2,\dfrac{\pi }{4}} \right)$. What are its Cartesian coordinates?

 Apply the formula for polar to Cartesian conversion.

 Given $r =  - 2$ and $\theta  = \dfrac{\pi }{4}$. $ \\  x = r\cos \theta  =  - 2\cos \dfrac{\pi }{4} =  - 2 \times \dfrac{1}{{\sqrt 2 }} =  - \sqrt 2  \\  y = r\sin \theta  =  - 2 \times \sin \dfrac{\pi }{4} \Rightarrow  - 2 \times \dfrac{1}{{\sqrt 2 }} =  - \sqrt 2  \\  $

 The Cartesian form of the polar equation  $\dfrac{4}{r} = \sec \theta  - \dfrac{1}{{\sin \theta }}$ is:

 Convert $\sec \theta $ to $\dfrac{1}{{\cos \theta }}$ and apply $r\cos \theta  = x,r\sin \theta  = y$ and ${r^2} = {x^2} + {y^2}$.

 Given, $ \dfrac{4}{r} = \sec \theta  - \dfrac{1}{{\sin x}} = \dfrac{1}{{\cos \theta }} - \dfrac{1}{{\sin \theta }}$ $ \\ \Rightarrow \dfrac{4}{r} = \dfrac{{\sin \theta  - \cos \theta }}{{\sin \theta  - \cos \theta }} \\   \Rightarrow {r^2}\left( {4\sin \theta .\cos \theta } \right) = \left( {r\sin \theta  - r\cos \theta } \right){r^2} \\   \Rightarrow 4r\sin \theta .r\cos \theta  = \left( {r\sin \theta  - r\cos \theta } \right){r^2} \\  y = r\sin \theta  \\  x = r\cos \theta  \\  {r^2} = {x^2} + {y^2} \\   \Rightarrow 4xy = \left( {y - x} \right)\left( {{x^2} + {y^2}} \right) \\  $

 Determine the value of $r$for the polar form of the Cartesian equation ${x^2} + {y^2} = 16$.

 Replace $x$ and $y$ by $x = r\cos \theta $ and $y = r\sin \theta $ and change in to polar form.

 ${x^2} + {y^2} = 16 \\  x = r\cos \theta ,{\text{ }}y = r\sin \theta  \\   \Rightarrow {r^2}{\cos ^2}\theta  + {r^2}{\sin ^2}\theta  = 16 \\   \Rightarrow {r^2}\left( {{{\cos }^2}\theta  + {{\sin }^2}\theta } \right) = 16 \\   \Rightarrow {r^2} = 16 \\   \Rightarrow r = \sqrt {16}  \\   \Rightarrow r = 4 \\  $

 Determine the correct expression that describes the following graph: $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/4a977879-8aca-4d07-9646-476723b1608b4777396081439521620.png"/>

 Observe the graph and determine some points and check the options to determine whether they satisfy the points or not.

 To solve this question, let us observe the graph and make a table of relevant points $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/209b06c4-4cc5-4e5d-a7b4-c68c1c45ccb65560084954630529746.png"/>$ \\ $  Now, if we see the options, the equation in the option a  nearly satisfies these points. $ \\ $ Thus, the answer is $r = 5\sin \theta $.