What are the components of the vector between the origin and the polar coordinate (3, (19pi)/12)(3,19π12)?

1 Answer
sankarankalyanam
Feb 23, 2018

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Explanation:

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Given r = 3, theta = (19pi)/12Givenr=3,θ=19π12

To find components of the vector between origin and the given point (3,(19pi)/12#

theta =( (19pi)/12)^c = 285^@θ=(19π12)c=285∘

x = r cos theta = 3 * cos ((19pi)/12) = 0.7765x=rcosθ=3⋅cos(19π12)=0.7765

y = r sin theta = 3 * sin ((19pi)/12) = -2.8978y=rsinθ=3⋅sin(19π12)=−2.8978

tan theta = tan ((19pi)/12) = y / x = - 2.8978 / 0.7765 = -3.7321tanθ=tan(19π12)=yx=−2.89780.7765=−3.7321

In rectangular form (x,y) = (0.7765, -2.8978)(x,y)=(0.7765,−2.8978)

In polar form (r, theta) = (3, (19pi)/12)(r,θ)=(3,19π12)

Slope of the vector m = tan theta = - 3.7321#

Vector is in the IV quadrant.

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