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s2008m [1.1K]
3 years ago
9

Find parametric equations for the sphere centered at the origin and with radius 3. Use the parameters s and t in your answer.

Mathematics
2 answers:
sdas [7]3 years ago
6 0
Radius, r = 3

The equation of a sphere entered at the origin in cartesian coordinates is

x^2 + y^2 + z^2 = r^2 

That in spherical coordinates is:

x = rcos(theta)*sin(phi)
y= r sin(theta)*sin(phi)
z = rcos(phi)

where you can make u = r cos(phi) to obtain the parametrical equations

x = √[r^2 - u^2] cos(theta)
y = √[r^2 - u^2] sin (theta)
z = u

where theta goes from 0 to 2π and u goes from -r to r.

In our case r = 3, so the parametrical equations are:

Answer:
x = √[9 - u^2] cos(theta)
y = √[9 - u^2] sin (theta)
z = u
shepuryov [24]3 years ago
4 0

Answer:

The parametric equation are

x=3\cos (s)\sin (t)

y=3\sin (s)\cos (t)

z=3\cos (t)

where, 0\le s\le 2\pi and 0\le t\le \pi.

Step-by-step explanation:

The general equation of a sphere is

(x-h)^2+(y-k)^2+(z-l)^2=r^2

where, (h,k,l) is center of the sphere and r is radius.

It is given that the sphere centered at the origin and with radius 3. It means h=0, k=0, l=0, r=3.

(x-0)^2+(y-0)^2+(z-0)^2=3^2

x^2+y^2+z^2=9

The parametric equation of a sphere are

x=h+r\cos (s)\sin (t)

y=k+r\sin (s)\cos (t)

z=l+r\cos (t)

where, 0\le s\le 2\pi and 0\le t\le \pi.

Substitute h=0, k=0, l=0, r=3 in the above equations. So, the parametric equation are

x=0+3\cos (s)\sin (t)=3\cos (s)\sin (t)

y=0+3\sin (s)\cos (t)=3\sin (s)\cos (t)

z=0+3\cos (t)=3\cos (t)

where, 0\le s\le 2\pi and 0\le t\le \pi.

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