How do you convert (-4, 3) into polar coordinates?

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How do you convert (-4, 3) into polar coordinates?

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Answer 1 · 2016-01-09T17:14:11

If $(a, b)$ $(a,b)$ is a are the coordinates of a point in Cartesian Plane, $u$ $u$ is its magnitude and $α$ $alpha$ is its angle then $(a, b)$ $(a,b)$ in Polar Form is written as $(u, α)$ $(u,alpha)$ .
Magnitude of a cartesian coordinates $(a, b)$ $(a,b)$ is given by $\sqrt{a^{2} + b^{2}}$ $sqrt(a^2+b^2)$ and its angle is given by ${tan}^{- 1} (\frac{b}{a})$ $tan^-1(b/a)$

Let $r$ $r$ be the magnitude of $(- 4, 3)$ $(-4,3)$ and $θ$ $theta$ be its angle.
Magnitude of $(- 4, 3) = {\sqrt{- 4}}^{2} + 3^{2}) = \sqrt{16 + 9} = \sqrt{25} = 5 = r$ $(-4,3)=sqrt(-4)^2+3^2)=sqrt(16+9)=sqrt25=5=r$
Angle of $(- 4, 3) = {tan}^{- 1} (\frac{3}{- 4}) = {tan}^{- 1} (- \frac{3}{4}) = - 36.869$ $(-4,3)=Tan^-1(3/-4)=Tan^-1(-3/4)=-36.869$ degree

$\Rightarrow$ $implies$ Angle of $(- 4, 3) = - 36.869$ $(-4,3)=-36.869$ degree

But since the point is in second quadrant so we have to add $180$ $180$ degree which will give us the angle.

$\Rightarrow$ $implies$ Angle of $(- 4, 3) = - 36.869 + 180 = 143.131$ $(-4,3)=-36.869+180=143.131$

$\Rightarrow$ $implies$ Angle of $(- 4, 3) = 143.131 = θ$ $(-4,3)=143.131=theta$

$\Rightarrow (- 4, 3) = (r, θ) = (5, 143.131)$ $implies (-4,3)=(r,theta)=(5,143.131)$
$\Rightarrow (- 4, 3) = (5, 143.131)$ $implies (-4,3)=(5,143.131)$
Note that the angle is given in degree measure.

Answer 2 · 2016-01-09T18:09:10

Given that a point $P \to (x, y) \to (- 4, 3) Cartesian$ $color(brown)(P-> (x,y)->(-4,3) " Cartesian")$

Then $P \to (5, {143.13}^{o}) Polar$ $color(blue)(P ->(5,143.13^o) " Polar ")$ to 2 decimal places

Explanation:

This is not a polar graph!!!
Tony B

$Where it all comes from$ $color(blue)("Where it all comes from")$

We are give the coordinates of (-4,3)
Suppose we viewed this in the context of Cartesian form and use $y = m x + c$ $y=mx+c$

Then $c = 0$ $c=0$ and $m = \frac{y}{x} = - \frac{3}{4}$ $m=y/x=-3/4$

So we would have $y = - \frac{3}{4} x$ $y=-3/4x$

Suppose the graph was only plotted over the range $x \to (0, - 4)$ $x->(0,-4)$

Then the above graph would not be continuous but be a line from
$(0, 0) \to (3, - 4)$ $(0,0) to (3,-4)$

All we need now is the angle that that line makes to the x-axis and the length of that line.
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$Finding the Polar r value$ $color(blue)("Finding the Polar r value")$

$Length = \sqrt{x^{2} + y^{2}} = \sqrt{3^{2} + 4^{2}} = 5$ $"Length "= sqrt(x^2+y^2) =sqrt(3^2+4^2) =5$

$So the polar r = 5$ $color(blue)("So the polar "r=5)$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$Finding the Polar θ value$ $color(blue)("Finding the Polar "theta" value")$
The Polar angle $θ$ $theta$ is measured from the positive x-axis counterclockwise.

Let the angle from the line to the negative x-axis be $ϕ$ $phi$

Then $ϕ = {tan}^{- 1} (\frac{y}{x}) = {tan}^{- 1} (\frac{3}{4})$ $phi = tan^(-1)(y/x) = tan^(-1)(3/4)$

But $. . ϕ + θ = 180$ $color(white)(..)phi+theta=180$

so $. . θ = 180 - {tan}^{- 1} (\frac{3}{4})$ $color(white)(..)theta =180 - tan^(-1)(3/4)$

$θ = 143.13$ $color(blue)( theta = 143.13$ to 2 decimal places
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$Putting at all together$ $color(blue)("Putting at all together")$

Given that a point $P \to (x, y) \to (- 4, 3) Cartesian$ $P-> (x,y)->(-4,3) "Cartesian"$

Then $P \to (5, {143.13}^{o}) Polar$ $P ->(5,143.13^o) " Polar "$ to 2 decimal places

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How do you convert (-4, 3) into polar coordinates?

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