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Answer:
[tex]105m^3/min[/tex]
Step-by-step explanation:
We are given that
Base side of square pyramid, a=3 m
Height of square pyramid, h=9m
[tex]\frac{da}{dt}=6m/min[/tex]
[tex]\frac{dh}{dt}=-1/min[/tex]
We have to find the rate of change of the volume of the pyramid at that instant.
Volume of square pyramid, V=[tex]\frac{1}{3}a^2h[/tex]
Differentiate w.r.t t
[tex]\frac{dV}{dt}=\frac{1}{3}(2ah\frac{da}{dt}+a^2\frac{dh}{dt})[/tex]
Substitute the values
[tex]\frac{dV}{dt}=\frac{1}{3}(2(3)(9)(6)+(3^2)(-1)[/tex]
[tex]\frac{dV}{dt}=105m^3/min[/tex]
Hence, the rate of change of the volume of the pyramid at that instant=[tex]105m^3/min[/tex]