The radius r of the base of right circular cone is decreasing at the rate of 2cm/min . and its height h is increasing at the rate of 3cm/min . when r = 3.5cm and h = 6 cm ,find the rate of change of the volume of the cone (use π = 22/7)
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The radius r of the base of right circular cone is decreasing at the rate of 2cm/min
dr/dt=-2
its height h is increasing at the rate of 3cm/min
dh/dt=3
Volume (V) is =1/3pi*r2*h
Take derivative w.r.t t
dV/dt=1/3*pi*(r2*dh/dt+h*2*r*dr/dt)
=1/3*pi*(3.52*3+6*2*3.5*(-2))
=1/3*pi*(36.75-84)
=1/3*pi*(-47.25)
=-49.4801.
HOPE , IT HELPS ... ✌
____________________________
The radius r of the base of right circular cone is decreasing at the rate of 2cm/min
dr/dt=-2
its height h is increasing at the rate of 3cm/min
dh/dt=3
Volume (V) is =1/3pi*r2*h
Take derivative w.r.t t
dV/dt=1/3*pi*(r2*dh/dt+h*2*r*dr/dt)
=1/3*pi*(3.52*3+6*2*3.5*(-2))
=1/3*pi*(36.75-84)
=1/3*pi*(-47.25)
=-49.4801.
HOPE , IT HELPS ... ✌
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The answer for this question is 48.4901. Please mark as brilliant.
murugaiah61:
The answer for this question is 49.4801
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