Question

Hi there! I'm not quite sure on how to solve this....

Answer 1

Answer:

Explanation:

$V=x^3\\\\\frac{dV}{dt}=3x^2\frac{dx}{dt}\\\\30\frac{cm^3}{s}=3x^2\frac{dx}{dt}\\\\\frac{dx}{dt}=\frac{30\frac{cm^3}{s}}{3x^2}~at~x=2cm,~\frac{dx}{dt}=\frac{30\frac{cm^3}{s}}{3*(2cm)^2}=\frac{5}{2}\frac{cm}{s}$

Answer 2

$\frac{dx}{dt} = 2.5 \: \frac{cm}{sec}$

Explanation:

The volume of a cube is V = x^3. Taking the time derivative of this expression, we get

$\frac{dV}{dt} = 3 {x}^{2} \frac{dx}{dt}$

or

$\frac{dx}{dt} = \frac{1}{3 {x}^{2}} \frac{dV}{dt}$

We know that dV/dt = 30 cm^3/sec so the value of dx/dt when x = 2 cm is

$\frac{dx}{dt} = \frac{1}{3 {(2 \: cm)}^{2}}(30 \: \frac{ {cm}^{3} }{sec} ) = 2.5 \: \frac{cm}{sec}$