Question

The polar coordinates of a certain are $(r=4.30cm,\theta ={214}^{\circ })$. a) Find its Cartesian coordinate x and y. (a) Find its Cartesian coordinate of x and y. Find the polar coordinates of the points with Cartesian coordinates (b) $(-x,y)$(c) $(-2x,-2y)$and (d) (3x, -3y)

Short answer

$$\begin{gathered} a\left)\right(x,y)=(-3.56\hspace{0.33em}cm,-2.40\hspace{0.33em}cm\left) \\ b\right)(r,\theta )=(4.30\hspace{0.33em}cm,326.0^0) \\ c\left)\right(r,\theta )=(8.60\hspace{0.33em}cm,34.0^0\left) \\ d\right)(r,\theta )=(12.9\hspace{0.33em}cm,{146}^0) \end{gathered}$$

Step-by-step solution

Define vector

Step 2: (a)Calculate Cartesian coordinate of x and y

For polar coordinates, the Cartesian coordinates are

Hence, .

(a) Calculate Cartesian coordinate of x and y

For polar coordinates (r,$\theta$), the Cartesian coordinates are

$\begin{array}{l}(x,y)=(r\cos \theta ,rsin\theta )\\ (x,y)=(4.3cos{214}^{\circ },4.3sin{214}^o)\\ (x,y)=(-3.56,-2.40)cm\end{array}$

Hence, (x,y) = (-3,56,-2.40)cm

(b) Find the polar coordinates of the points with Cartesian coordinates (-x, y)

If the angle is measured relative to positive x axis$r=\surd (x^2+y^2),\theta ={\tan }^{-1}\frac{y}{x}$

Then

$$\begin{gathered} r=\sqrt{{(-x)}^2+y^2} \\ =\sqrt{{(-3.56)}^2+2.4^2} \\ =4.3cm \end{gathered}$$

$$\begin{gathered} \theta ={\tan }^{-1}\frac{y}{x} \\ ={\tan }^{-1}\frac{-2.4}{-(-3.56)} \\ =-{34}^{\circ } \end{gathered}$$

$\theta =360-34=326°$

$Hence,(r,\theta )=(4.30\hspace{0.33em}cm,326.0^0)$

(c) Find the polar coordinates of the points with Cartesian coordinates (-2x, -2y)

$$\begin{gathered} r=\sqrt{{(-2x)}^2+{(-2y)}^2} \\ =\sqrt{{(-2\times (-3.56\left)\right)}^2+(-2\times {(2.4)}^2} \\ =8.6cm \end{gathered}$$

$$\begin{gathered} \theta ={\tan }^{-1}\frac{-2y}{-2x} \\ ={\tan }^{-1}\frac{-2\times (-2.4)}{-2\times (-3.56)} \\ ={34}^{\circ } \end{gathered}$$

$Hence,(r,\theta )=(8.60\hspace{0.33em}cm,34.0^0)$

(d) Find the polar coordinates of the points with Cartesian coordinates (3x, -3y)

$$\begin{gathered} r=\sqrt{{\left(3x\right)}^2+{(-3y)}^2} \\ =\sqrt{{(3\times (-3.56\left)\right)}^2+(-3\times {(2.4)}^2} \\ =12.9cm \end{gathered}$$

$$\begin{gathered} \theta ={\tan }^{-1}\frac{-3y}{3x} \\ ={\tan }^{-1}\frac{-3\times (-2.4)}{3\times (-3.56)} \\ =-{34}^{\circ } \end{gathered}$$

$$\begin{gathered} \theta =180-34=146° \\ Hence,(r,\theta )=(12.9\hspace{0.33em}cm,{146}^0). \end{gathered}$$