Question #160267

Find the cylindrical coordinates of the points where the Cartesian coordinates are

i.(6,6,8)

ii.(√2,1,1)


1
Expert's answer
2021-02-04T08:09:30-0500

Solution (i):

Given, x=6, y=6,z=8x=6,\ y=6, z=8

In rectangular coordinates,

x=rcosθ, y=rsinθ, z=zx=r\cos\theta,\ y=r\sin\theta,\ z=z

So, 6=rcosθ, 6=rsinθ, z=86=r\cos\theta,\ 6=r\sin\theta,\ z=8

Also, r2=x2+y2=62+62=36+36=72r^2=x^2+y^2=6^2+6^2=36+36=72

r=72=62\Rightarrow r=\sqrt{72}=6\sqrt2 (Taking positive value only)

And, tanθ=yx=66=1\tan\theta=\dfrac yx=\dfrac 66=1

θ=tan11=π4\Rightarrow \theta=\tan^{-1}1=\dfrac{\pi}4

So, cylindrical coordinates are (r,θ,z)=(62,π4,8)(r,\theta,z)=(6\sqrt2,\dfrac {\pi}4,8)

Solution (ii):

Given, x=2, y=1,z=1x=\sqrt2,\ y=1, z=1

In rectangular coordinates,

x=rcosθ, y=rsinθ, z=zx=r\cos\theta,\ y=r\sin\theta,\ z=z

So, 2=rcosθ, 1=rsinθ, z=1\sqrt2=r\cos\theta,\ 1=r\sin\theta,\ z=1

Also, r2=x2+y2=(2)2+12=2+1=3r^2=x^2+y^2=(\sqrt2)^2+1^2=2+1=3

r=3\Rightarrow r=\sqrt3 (Taking positive value only)

And, tanθ=yx=12\tan\theta=\dfrac yx=\dfrac 1{\sqrt2}

θ=tan1(12)=0.6154\Rightarrow \theta=\tan^{-1}(\dfrac 1{\sqrt2})=0.6154 rad

So, cylindrical coordinates are (r,θ,z)=(3,0.6154,1)(r,\theta,z)=(\sqrt3,0.6154,1)


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