What is the Cartesian form of #( -9 , ( - 15pi)/2 ) #?

Question

What is the Cartesian form of #(  -9  , (  - 15pi)/2  ) #?

Scarlett Allison · Accepted Answer

(0. -9)

We have the coordinate #(-9, (-15pi)/2)# in polar form. Coordinates in polar form have the standard form #(color(green)(r), color(purple)(Θ))#. To convert from polar form to Cartesian form, we use the following formulas: Now, let's plug stuff in. We know #color(green)(r) = -9# and #color(purple)(Θ) = (-15pi)/2# #color(red)(x) = (-9)*cos((-15pi)/2)# #color(red)(x) = (-9)*(0)# #color(red)(x) = 0# #color(blue)(y) = (-9)*sin((-15pi)/2)# #color(blue)(y) = (-9)*(1)# #color(blue)(y) = -9# We get #color(red)(x) = 0# and #color(blue)(y) = -9#, making our Cartesian coordinate #(0, -9).#

Axel Alvarez · Answer

undefined To convert the polar coordinates $ (-9, -\frac{15\pi}{2}) $ to Cartesian coordinates, you can use the following formulas:

$$ x = r \cos(\theta) $$
$$ y = r \sin(\theta) $$

Substitute $ r = -9 $ and $ \theta = -\frac{15\pi}{2} $ into these formulas:

$$ x = -9 \cdot \cos\left(-\frac{15\pi}{2}\right) $$
$$ y = -9 \cdot \sin\left(-\frac{15\pi}{2}\right) $$

Since $ \cos\left(-\frac{15\pi}{2}\right) = \cos\left(\frac{\pi}{2}\right) = 0 $ and $ \sin\left(-\frac{15\pi}{2}\right) = \sin\left(\frac{\pi}{2}\right) = 1 $, we have:

$$ x = -9 \cdot 0 = 0 $$
$$ y = -9 \cdot 1 = -9 $$

Therefore, the Cartesian coordinates corresponding to the polar coordinates $ (-9, -\frac{15\pi}{2}) $ are $ (0, -9) $.