Math, asked by nagoseamita, 3 months ago

Find the value of 'n so that the equation v=
r^n (3 cos^2
0 - 1) satisfies the relation​

Answers

Answered by mithun890
12

Given:

  •  Vr^{n} (3cos^{2}θ-1)
  • so that,

                (\frac{dv}{dr})_{r} =nr^{n-1 }(3cos^{2}θ-1)

              r^{2} (\frac{dv}{dr} )=nr^{n+1} (3cos^{2}θ-1)

           \frac{d}{dr} (r^{2} \frac{dv}{dr} )=n(n+1)r^{n} (3cos^{2}θ-1)

                          = n(n+1)v

   \frac{dv}{d}θ(sinθ  \frac{dv}{d}θ) =-6r^{n}(-sinθ sin^{2}θ+ cosθ .2sinθcosθ)

                      =-6r^{n} sinθ(2cos^{2}θ-sin^{2}θ)

  • Implying   \frac{1}{sin}θ {[{\frac{d}{d}θ(sinθ \frac{dv}{d}θ)]=-6r^{n} (2cos^{2}θ-sin^{2}θ)

                            =-6r^{n} (2cos^{2}θ+cos^{2}θ-1)

                            =-6r^{n} (3cos^{2}θ-1)

                            =-6V  (using given realation)

  • on ading expressions (2) and(5),we get

                       n(n+1)v-6v=0

  • Which implies ,

                   (n^{2} +n-6)=0

          ∴n^{2}+3n+2n-6=0

                (n+3) (n-2)=0

                     n=2,-3

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