Finding formula for parabolic cone with constant ratio of volume to surface area of each circular cross sectionI'm trying to express an equation for a parabolic cone (paraboloid?) where the ratio of volume to surface area of the circle through covering every cross section throughout the height of the 'cone' is constant. Presumably there is some parabola equation that creates a line that if spun around to create a parabolic cone shape would give this outcome.

Question

Finding formula for parabolic cone with constant ratio of volume to surface area of each circular cross sectionI&#039;m trying to express an equation for a parabolic cone (paraboloid?) where the ratio of volume to surface area of the circle through covering every cross section throughout the height of the &#039;cone&#039; is constant. Presumably there is some parabola equation that creates a line that if spun around to create a parabolic cone shape would give this outcome.

Seromaniaru · Accepted Answer

Step 1The ratio is a constant   c      /    2  =  k  ,    c  =  2  k, say                    (1a)                                                        ∫              π                              r                2                            d              y                                      π                              r                2                                                    =        k                                (1b)                                                        π                              r                2                                                                                      d                  (                  π                                      r                    2                                    )                                                  d                  y                                                                    =        k            Cancel   π and differentiate with respect to y using Chain Rule as above   (  u      /    v      )    ′    =  0  →      (    u          /        v    →          u      ′              /              v      ′        )                      r        2                    2        r                                            d              r                                      d              y                                            =  k                    (2)                                                        d              y                                      2              k                                      =                                            d              r                        r                          →                                            d              r                                      d              y                                      =                              r                          2              k                                          Step 2Integrate with BC   y  =  0  ,  r  =  a                      (3)                                          y                          2              k                                      =        log        ⁡                              r            a                              Or                     (4)                    r        =        a                  e                      y                          2              k                                          So it is an exponential curve rotated around Y axis.To verify the obtained result plug (2) into (1) and simplify                    (5)                                          Volume            Cross-Section Area                          =                                            π                              a                2                            ∫                              e                                  y                                      /                                    k                                            d              y                                      π              (                              a                2                                            e                                  y                                      /                                    (                  2                  k                  )                                                            )                2                                                    =                                                            e                                  y                                      /                                    k                                            ⋅              k                                      (                              e                                  y                                      /                                    (                  2                  k                  )                                                            )                2                                                    =        k             which is a constant.

Lilliana Livingston · Accepted Answer

Step 1Such a solid of revolution cannot exist. Let the solid be generated by the rotation around x-axis of the curve with equation   y  =  f  (  x  ). Then its volume, from   x  =  0 to   x  =  X is  V  (  X  )  =      ∫    0    X    π      f    2    (  x  )    d  xwhile the area of the cross section at   x  =  X is   A  (  X  )  =  π      f    2    (  X  ). If   A  (  X  )  =  k  V  (  X  ) for some constant k and any   X  ≥  0, then       A    ′    (  X  )  =  k      V    ′    (  X  ), that is:  2  π  f  (  X  )      f    ′    (  X  )  =  k  π      f    2    (  X  )  .Step 2Hence, either   f  (  X  )  =  0 or       f    ′    (  X  )  =  (  k      /    2  )  f  (  X  ), that is:   f  (  X  )  =  A      e          k      X              /            2      . But that solution doesn&#039;t work, because it gives:  V  (  X  )  =                    π                  A          2                    k                  (            e          k      X        −  1            )          and    A  (  X  )  =  π      A    2        e          k      X        .

Answered question

Answer & Explanation

New Questions in High school geometry