Let r ( t ) = ⟨ 1 + 3 cos ⁡ ( t ) , 4 cos ⁡ ( t ) , − 1 + 5 sin ⁡ ( t ) ⟩. Find an equation for a sphere that this curve lies on.As it says in the title, any help would be greatly appreciated.

Question

Let       r    (  t  )  =  ⟨  1  +  3  cos  ⁡  (  t  )  ,  4  cos  ⁡  (  t  )  ,  −  1  +  5  sin  ⁡  (  t  )  ⟩. Find an equation for a sphere that this curve lies on.As it says in the title, any help would be greatly appreciated.

Damarion Pierce · Accepted Answer

r        (        t        )            →        =  ⟨  1  ,  0  ,  −  1  ⟩  +  ⟨  3  cos  ⁡  t  ,  4  cos  ⁡  t  ,  5  sin  ⁡  t  ⟩And,      |    ⟨  3  cos  ⁡  t  ,  4  cos  ⁡  t  +  5  sin  ⁡  t  ⟩      |    =      (    3    cos    ⁡    t          )      2        +    (    4    cos    ⁡    t          )      2        +    (    5    sin    ⁡    t          )      2        =  5So we have that no matter what the value of t is the second term is at a distance of 5 from the point   ⟨  1  ,  0  ,  −  1  ⟩

Jayvion Caldwell · Accepted Answer

r(t) defines a curve in 3-dimensions as a functions of every real t from       R   onto             R        3  . We will have to get rid of its parameter t solving the following system of equations:  x  =  1  +  3  cos  ⁡  (  t  )  y  =  4  cos  ⁡  (  t  )  z  =  −  1  +  5  sin  ⁡  (  t  )Do squares over all the equations:  (  x  −  1      )    2    =  9      cos    2    ⁡  (  t  )      y    2    =  16      cos    2    ⁡  (  t  )  (  z  +  1      )    2    =  25      sin    2    ⁡  (  t  )Now add the three of them together:  (  x  −  1      )    2    +      y    2    +  (  z  +  1      )    2    =  9      cos    2    ⁡  (  t  )  +  16      cos    2    ⁡  (  t  )  +  25      sin    2    ⁡  (  t  )Applying       cos    2    ⁡  (  u  )  +      sin    2    ⁡  (  u  )  =  1 for every real number.  (  x  −  1      )    2    +      y    2    +  (  z  +  1      )    2    =  25The last equation defines a sphere of radius 5 and center (1,0,−1) in 3D space.

Let r(t)=⟨1+3 cos (t),4 cos (t),−1+5 sin (t)⟩. Find an equation for a sphere that this curve lies on.

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