Respuesta :
Using the concepts of Application of derivative, we got that 1657.92inches³/sec rate of the volume of the cone changing when the radius is 40 inches and the height is 40 inches for a right circular cone.
We know that volume of right circular cone is given by =(1/3πr²h)
We are given rate of change of radius(dr/dt) and rate of change of the height(dh/dt)
Therefore ,differentiating the volume with respect to time and applying the chain rule.
dV/dt = [(1/3)πr²×(dh/dt)+ (1/3)π·2·r·(dr/dt)·h]
=>dV/dt=[(1/3)× π×r × [(r×dh/dt)+2h×(dr/dt)]]
We are given that dr/dt=2inc/sec and dh/dt= -5inc/sec
So, on putting the values, we get
=>dV/dt=[ ( (1/3)×3.14×40)×[40×(-5)+2×40×2]]
=>dV/dt=[0.33×3.14×40×[-200+160]
=>dV/dt=[0.33×3.14×40×(-40)]
=>dV/dt= -1657.92inches³/sec(negative sign denote volume is decreasing)
Hence, if the radius of a right circular cone is increasing at the rate of 2 inches per second and its height is decreasing at the rate of 5 inches per second. the rate of the volume of the cone changing when the radius is 40 inches and the height is 40 inches is 1657.92inches³/sec.
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