Given that `v=sqrt(sin u)` ,show that `4v^3*((d^2v)/(du^2)) +v^4+1 = 0`

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justaguide | College Teacher | (Level 2) Distinguished Educator

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It is given that `v = sqrt(sin u)` .

 `(dv)/(du) = (1/2)*(1/sqrt(sin u))*cos u`

= `cos u/(2*sqrt(sin u))`

 `(d^2v)/(du^2) = ((cos u)'*2*sqrt(sin u)-cos u*(2*sqrt(sin u))')/(2*sqrt(sin u))^2`

= `(-sin u*(2*sqrt(sin u))-cos u*(2*cos u/(2*sqrt(sin u))))/(4*sin u)`

= `(-2*sin u*sqrt(sin u) - (cos^2u)/sqrt(sin u))/(4*sin u)`

`4v^3*((d^2v)/(du^2)) +v^4+1`

= `4*(sqrt(sin u))^3*((-2*sin u*sqrt(sin u) - (cos^2u)/sqrt(sin u))/(4*sin u)) + (sqrt(sin u))^4 + 1`

= `4*sin u*sqrt(sin u)*((-2*sin u*sqrt(sin u) - (cos^2u)/sqrt(sin u))/(4*sin u)) + (sqrt(sin u))^4 + 1`

= `sqrt(sin u)*(-2*sin u*sqrt(sin u) - (cos^2u)/sqrt(sin u)) + (sqrt(sin u))^4 + 1`

= `-2*sin u*sin u - cos^2u + sin^2 u + 1`

= `-2*sin^2u - cos^2u + sin^2 + 1`

= `-sin^2u - cos^2u + 1`

= `-(sin^2u+cos^2u) + 1`

= -1 + 1

= 0

This proves that `4v^3*((d^2v)/(du^2)) +v^4+1 = 0`

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