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This is one of our many quizzes on Polar coordinates. Click on this to access more such competitive quizzes on 2D Coordinate Geometry.

Polar coordinates Quiz 2

mathematics

 Prateek has mapped polar coordinates as $\left( 2,\dfrac{\pi }{7} \right)$ . If the range is set to be $-2\pi \le \theta \le 2\pi $ and he has to find out three different sets of coordinates for the same point all of which have angles different from $\dfrac{\pi }{7}$ . What is his answer?

 Find the change in angle for the given range. The radius is kept constant and only angle is changed to map the coordinates.

The negative angle in the range specified can be found simply as $\dfrac{\pi }{7}-2\pi =-\dfrac{13\pi }{7}$ If the third quadrant is considered then the positive angle is found by adding the angles such that they lie in the given range $\dfrac{\pi }{7}+\pi =\dfrac{8\pi }{7}$ Likewise the negative angle in fourth quadrant is $\dfrac{\pi }{7}-\pi =-\dfrac{6\pi }{7}$ So the three points with the angles are $A\equiv \left( 2,\dfrac{-13\pi }{7} \right)\text{ B}\equiv \left( -2,\dfrac{-6\pi }{7} \right)\text{ C}\equiv \left( -2,\dfrac{-8\pi }{7} \right)$

 Frank has Cartesian coordinate of point as $(2,-6)$. He finds the polar coordinates of the point, and helps him in finding the polar coordinates.

 Find the value of r as $r=\sqrt{{{x}^{2}}+{{y}^{2}}}$ and the angle as $\theta ={{\tan }^{-1}}\left( \dfrac{y}{x} \right)$ . There is a possibility of 2 angles present for the value of r

Finding the value of r using the formula $r=\sqrt{{{x}^{2}}+{{y}^{2}}}$ &nbsp;$\Rightarrow r=\sqrt{{{2}^{2}}+{{(-6)}^{2}}}=\sqrt{40}=2\sqrt{10}$ Next, finding the value of the angle for the possible value of r There are two such values present, hence using the formula $\theta ={{\tan }^{-1}}\left( \dfrac{y}{x} \right)$ ${{\theta }_{1}}={{\tan }^{-1}}\left( \dfrac{-6}{2} \right)=-1.24$ ${{\theta }_{2}}={{\theta }_{1}}+\pi =1.86$ Here it is seen that $\dfrac{-\pi }{2}=-1.57<{{\theta }_{1}}=-1.24<0$ So ${{\theta }_{1}}$ lies in the fourth quadrant, likewise, ${{\theta }_{2}}$ lies in the second quadrant. So ${{\theta }_{1}}$ is the correct value as the point also lies in second quadrant Hence the polar coordinates are $(2\sqrt{10},-1.24)$

 Sudhir is a mechanical engineer who wishes to map the component $(-8,1)$ to its polar coordinate. What does his answer come out to be? Given ${{\tan }^{-1}}\left( \dfrac{-1}{8} \right)=-0.12$ 

 Find the value of r as $r=\sqrt{{{x}^{2}}+{{y}^{2}}}$ and the angle as $\theta ={{\tan }^{-1}}\left( \dfrac{y}{x} \right)$ . there is a possibility of 2 angles present for the value of r

Finding the value of r using the formula $r=\sqrt{{{x}^{2}}+{{y}^{2}}}$ &nbsp;$\Rightarrow r=\sqrt{{{1}^{2}}+{{(-8)}^{2}}}=\sqrt{65}$ Next, finding the value of the angle for the possible value of r There are two such values present, hence using the formula $\theta ={{\tan }^{-1}}\left( \dfrac{y}{x} \right)$ ${{\theta }_{1}}={{\tan }^{-1}}\left( \dfrac{-1}{8} \right)=-0.12$ ${{\theta }_{2}}={{\theta }_{1}}+\pi =3.01$ Here it is seen that $\dfrac{-\pi }{2}=-1.57<{{\theta }_{1}}=-0.12<0$ So ${{\theta }_{1}}$lies in the fourth quadrant, likewise,${{\theta }_{2}}$lies in the second quadrant. So ${{\theta }_{2}}$ is the correct value as the point also lies in second quadrant Hence the polar coordinates are $(\sqrt{65},3.01)$

 A vinyl disc is made such that the head of the pointer used to listen music always lies on the polar coordinate $(-5,0.23)$ . An innovator wishes to make changes in the device, helping her in jotting down the coordinate in the Cartesian plane.

 If the point is given in the form of polar coordinates then $x=r\cos \theta $  and $y=r\sin \theta $ . substitute the values to find the answer.

The point is given in form of polar coordinates as $(r,\theta )$ when the point is given in the form of polar coordinates then $x=r\cos \theta $ and $y=r\sin \theta $ Substituting values $\Rightarrow x=(-5)\cos (0.23)=-4.86$ And $\Rightarrow y=(-5)\sin (0.23)=-1.13$ So the coordinates are $(-4.86,-1.13)$

 A student wishes to find the equation $\dfrac{4x}{3{{x}^{2}}+3{{y}^{2}}}=6-xy$  as polar coordinates. What is his answer?

 Convert the $x$ and $y$ in terms of $r,\theta $ by using the formula $r=\sqrt{{{x}^{2}}+{{y}^{2}}}$ and $x=r\cos \theta $  and $y=r\sin \theta $ 

Converting the $x$ and $y$ in terms of $r,\theta $ Using the following formula $r=\sqrt{{{x}^{2}}+{{y}^{2}}}$ and $x=r\cos \theta $ and $y=r\sin \theta $ Substituting values $\Rightarrow \dfrac{4(r\cos \theta )}{3{{r}^{2}}}=6-(r\cos )(r\sin \theta )$ And $\Rightarrow \dfrac{4\cos \theta }{3{{r}^{2}}}=6-{{r}^{2}}\cos \theta sin\theta $