What is the polar form of #( 4,9 )#?

Question

Aurora Alvarado · Accepted Answer

#(10, 66°)# in Polar Coordinates

To solve this question you must imagine a triangle that has a base distance of #4# and a height of #9#. We want to do this so that we can find out how far and at what angle the polar coordinates are and so that we can convert the current cartesian coordinates #(x,y)# to the polar coordinates #(r,θ)#. The triangle Base#=x = 4# Height#=y=9# Hypotenuse#=r# The angle between #r# and #x = θ# Use Pythagoras Theorem to find the hypotenuse (#r#) #r=sqrt(4^2+9^2)# #r=sqrt(97)=9.848857802# #r = 10# (1d.p) Use the Tangent Function to find the desired angle #tan(θ)=9/4# #θ = tan^-1(9/4)=66.03751103# #θ = 66°# (1d.p) #:.# the cartesian coordinates #(4,9)# are #(10, 66°)# in Polar Coordinates.

Elijah Abram · Answer

#(sqrt(97), arctan (4/9))#

The line joining the origin to the given point, that is the line joining #(0, 0)# to #(4, 9)#

is the hypotenuse of a right angled triangle with

the line along the x axis from #(0, 0)# to #(4, 0)#

and the line along the y axis from #(0, 0)# to #(0, 9)#

forming the other two sides that enclose the right angle.

The side on the x axis has length 4, and that on the y axis has length 9, so that, by the Pythagorean relationship, the length of the hypotenuse is

#sqrt(4^2 + 9^2) = sqrt(16 + 81) = sqrt (97)#

That is the modulus of the polar form, conventionally denoted by #r#.

The tangent of the angle (in radians, conventionally denoted by #theta#) between the hypotenuse and the positive x axis is #4/9# that is, #theta = arctan (4/9)#

so, the point has polar form #(r, theta)#

#(r, theta) = (sqrt(97), arctan (4/9))#

Charles Arocha · Answer

undefined To convert the Cartesian coordinates $(4, 9)$ to polar coordinates, we use the formulas:

$$ r = \sqrt{x^2 + y^2} $$
$$ \theta = \arctan\left(\frac{y}{x}\right) $$

Substitute $x = 4$ and $y = 9$ into these formulas:

$$ r = \sqrt{4^2 + 9^2} = \sqrt{16 + 81} = \sqrt{97} $$
$$ \theta = \arctan\left(\frac{9}{4}\right) $$

Now, to find $\theta$, you can use a calculator to find the arctan or inverse tangent of $\frac{9}{4}$.

So, the polar form of the point $(4, 9)$ is $(\sqrt{97}, \arctan\left(\frac{9}{4}\right))$.