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Thread: Polar to rectangular

  1. #1
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    Polar to rectangular

    I am trying to help grand daughter express this is rectangular form:

    thetha = pi/3
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  2. #2
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    Re: Polar to rectangular

    If $\displaystyle \displaystyle \begin{align*} \theta = \frac{\pi}{3} \end{align*}$ then you have a ray starting at the origin extending out infinitely long making this angle with the positive x-axis.

    The Cartesian (rectangular) form for the equation of a line is $\displaystyle \displaystyle y = mx + c$ where m is the gradient and c is the y intercept. c is clearly 0. As for m, it can be calculated using $\displaystyle \displaystyle \begin{align*} m = \tan{(\theta)} = \tan{\left( \frac{\pi}{3} \right)} = \sqrt{3} \end{align*}$.

    So the rectangular form of this equation is $\displaystyle \displaystyle \begin{align*} y = \sqrt{3}\,x \end{align*}$ where $\displaystyle \displaystyle \begin{align*} x \geq 0 \end{align*}$.
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    Re: Polar to rectangular

    Thank you for your help.
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    Re: Polar to rectangular

    Hello, dgable!

    $\displaystyle \text{Express in rectangular form: }\:\theta \,=\,\tfrac{\pi}{3}$

    There are a few identities for converting polar to rectangular:

    . . $\displaystyle \begin{array}{ccccccccc}r^2 &=&x^2+y^2 &\quad& r\cos\theta &=& x \\ \tan\theta &=&\dfrac{y}{x} && r\sin\theta &=& y \end{array}$


    $\displaystyle \text{We are given: }\qquad\;\;\theta \:=\:\tfrac{\pi}{3}$

    $\displaystyle \text{Take tangents: }\,\tan\theta \:=\:\tan\left(\tfrac{\pi}{3}\right)$

    $\displaystyle \text{Substitute: }\qquad\quad\,\frac{y}{x} \:=\:\sqrt{3}$

    Therefore: . . . . . $\displaystyle y \:=\:\sqrt{3}\,x$
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