[Math] the rate of change of the diagonal, volume, and surface of a box with dimensions a, b, and c

calculusmultivariable-calculus

I am working through some problems in my calculus workbook, and this one has me a bit stuck.
If someone wouldnt mind lending a hand, I would much appreciate it.

This is the problem:

A rectangular box has dimensions a, b, and c which are changing with time. At $t = t_0$:

$a = 1 m$, $b = 2 m$, and $c = 3 m$ (m is meters).

$da/dt = db/dt = 1 m/sec$ and $dc/dt = -3 m/sec$
What is the rate of change of the volume, surface area, and the diagonals of the box at $t = t_0$.

If someone wouldnt mind lending me a hand as to the way you would solve this, I would greatly appreciate it.

Thanks
Corey

Best Answer

Notice, the dimensions of the box $a$, $b$ & $c$ are changing w.r.t. time $t$ hence, we have

  1. Volume, $$V=abc$$$$\implies \frac{d V}{dt}=bc\frac{da}{dt}+ac\frac{d b}{dt}+ab\frac{dc}{dt}$$ substitute all the corresponding values, the rate of change of volume is $$\color{red}{\frac{dV}{dt}}=(2\cdot 3)(1)+(1\cdot 3)(1)+(1\cdot 2)(-3)=\color{red}{3\ m^3/s}$$
  2. Surface area, $$S=2(ab+bc+ac)$$$$\implies \frac{d S}{dt}=2\left(b\frac{da}{dt}+c\frac{d b}{dt}+a\frac{dc}{dt}\right)$$ substitute all the corresponding values, the rate of change of surface is $$\color{red}{\frac{dS}{dt}}=2(2\cdot 1+3\cdot 1+1\cdot (-3))=\color{red}{6\ m^2/s}$$
  3. Diagonal, $$D=\sqrt{a^2+b^2+c^2}\implies D^2=a^2+b^2+c^2$$ $$\implies 2D\frac{d D}{dt}=\left(2a\frac{da}{dt}+2b\frac{d b}{dt}+2c\frac{dc}{dt}\right)$$ $$\implies \frac{d D}{dt}=\frac{1}{D}\left(a\frac{da}{dt}+b\frac{d b}{dt}+c\frac{dc}{dt}\right)=\frac{1}{\sqrt{a^2+b^2+c^2}}\left(a\frac{da}{dt}+b\frac{d b}{dt}+c\frac{dc}{dt}\right)$$

substitute all the corresponding values, the rate of change of diagonal is $$\color{red}{\frac{dD}{dt}}=\frac{1}{\sqrt{1^2+2^2+3^2}}\left(1\cdot 1+2\cdot 1+3\cdot (-3)\right)=\color{red}{\frac{-6}{\sqrt {14}}\ m/s}$$

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