Rate measure

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A balloon in the form of a right cone surmiunted by a hemispere,having a diameter equal to height of the cone,is being inflated.how fast is its volume changing w.r.t its total height 'h' when h

  • Rate measure -

    I assume that h is the height of the cone from the floor to the base of the hemisphere. Thus the diameter of the cone at the floor is 2 h and the height of the cone if it went to the tip would be 2 h


    volume of cone of base diameter 2 h and height 2 h:
    (1/3)(2h)(pi/4)(4 h^2)= (2/3)pi h^3
    volume of cut off tip of cone:
    (1/3)(h)(pi/4)h^2 = (1/12) pi h^3
    so
    volume of cone base = (7/12)pi h^3
    now
    volume of hemisphere = (1/2)(4/3)pi(h/2)^3 = pi h^3/12
    so
    total balloon volume = (2/3) pi h^3
    dV/dh = 2 pi h^2
    dV/dt = dV/dh * dh/dt
    so
    dV/dt = 2 pi h^2 dh/dt

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