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Question

Question: How do you find the polar coordinates of the point?...

How do you find the polar coordinates of the point?

Explanation

Solution

in this question, we would convert the Cartesian point into polar coordinates of the point. The final answer would be in the form of (r,θ)\left( {r,\theta } \right). This question also involves the operation of addition/ subtraction/ multiplication/ division. We need to know Pythagoras#39;s theorem
for finding the formula to find the value of rr from Cartesian points.

Complete step by step solution:
In this question, we need to convert the Cartesian points (x,y)\left( {x,y} \right) into polar
points (r,θ)\left( {r,\theta } \right). For that, we assume the following Cartesian point,
(x,y)=(0,2)\left( {x,y} \right) = \left( {0,2} \right)
First, we have to find the value of rrin(r,θ)\left( {r,\theta } \right).
We know that,
r=x2+y2r = \sqrt {{x^2} + {y^2}} (By using Pythagoras theorem)
Here, x=0x = 0andy=2y = 2

r=(02)+(22) r=4 r = \sqrt {\left( {{0^2}} \right) + \left( {{2^2}} \right)} \\\ r = \sqrt 4 \\\

We know that, the square root value of 44is22
So, we get
r=2r = 2
Next, we have to find the value of θ\theta in(r,θ)\left( {r,\theta } \right)
We know that,
tanθ=(yx)\tan \theta = \left( {\dfrac{y}{x}} \right)
Here, x=0x = 0andy=2y = 2
So, we get
tanθ=(20)\tan \theta = \left( {\dfrac{2}{0}} \right)
We know that anything divided by zero is equal to infinity.
So, we get

tanθ=  \tan \theta = \infty \\\ \\\

θ=arctan()\theta = arc\tan (\infty )
(When tan\tan it moves from left side to right side of the equation it converts intoarctan\arctan .)
θ=(90)\theta = \left( {{{90}^ \circ }} \right)
The above equation can also be written as,
θ=(π2)\theta = \left( {\dfrac{\pi }{2}} \right)
So, the final answer is, (r,θ)=(2,π2)\left( {r,\theta } \right) = \left( {2,\dfrac{\pi }{2}} \right)

Note: In this type of question remember that the polar coordinates are (r,θ)\left( {r,\theta } \right).
Note that, the formula for finding the values of rrandθ\theta (r=x2+y2r = \sqrt {{x^2} + {y^2}} ,tanθ=(yx)\tan \theta = \left( {\dfrac{y}{x}} \right)). So, the final answer would be in the form of(r,θ)\left( {r,\theta } \right). Also, remember the trigonometric table values for finding the value of θ\theta fromtanθ\tan \theta . Note that when tan\tan it moves from the left side to the right side of the equation it converts into arctan\arctan . Also, remember that anything divided by zero becomes infinity.