Answer to Question #256054 in Calculus for Latha

Question #256054

‘e lengths p, q, and r of the edges of a rectangular box are changing with time. At the

instant p = 2m, q = 3m, r = 4m,

dp

dt =

dq

dt = 1 m/sec and dr

dt = −2 m/sec. At what rate is the

box’s volume V changing at that instant?



1
Expert's answer
2021-10-29T02:53:36-0400

Answer:-


"V=pqr""\\dfrac{dV}{dt}=qr\\dfrac{dp}{dt}+pr\\dfrac{dq}{dt}+pq\\dfrac{dr}{dt}"

Given "p=2m,q=3m, r=4m"

"\\dfrac{dp}{dt}=\\dfrac{dq}{dt}=1\\ m\/sec, \\dfrac{dr}{dt}=-2\\ m\/sec"



"\\dfrac{dV}{dt}=(3m)(4m)(1m\/sec)+(2m)(4m)(1m\/sec)""+(2m)(3m)(-2m\/sec)=8m^3\/sec"

The box’s volume "V" is increasing at rate "8m^3\/sec" at that instant.



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