Answer to Question #256607 in Calculus for Anuj

Question

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? 

Answer 1

$$V=pqr$$

$$\dfrac{dV}{dt}=qr\dfrac{dp}{dt}+pr\dfrac{dq}{dt}+pq\dfrac{dr}{dt}$$

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.