The edge of a cube is increasing at the rate of 2 cm/s. Find the rate of increase of the volume when each side is 8 cm
I can't think of a pre-calc method for this one
Let $\displaystyle s$ be the side of the cube and let $\displaystyle V$ be the volume.
From the volume of a cube $\displaystyle V = s^3$
it is known that $\displaystyle \frac{ds}{dt} = 2$
We want to know $\displaystyle \frac{dV}{dt}$
To get this use the chain rule:
$\displaystyle \frac{dV}{dt} = \frac{dV}{ds} \cdot \frac{ds}{dt}$
To find $\displaystyle \frac{dV}{ds}$ differentiate the first equation
$\displaystyle \frac{dV}{ds} = 3s^2$
$\displaystyle \frac{dV}{dt} = 3s^2 \times 2 = 6s^2$
To find the overall answer sub in $\displaystyle s=8$ in $\displaystyle \frac{dV}{dt}$
I get an answer of $\displaystyle 384 \text{cm}^3 \text{ s}^{-1}$