Question

A cube with edges of length x centimeters has volume v(x)=x^3 at the instant when x=2, what is the rate of change of x, in centimeters per minute, with respect to time?

Answer 1

Answer:

The rate of change of x, in centimeters per minute, with respect to time is found to be :

$\frac{2}{3}$

Step-by-step explanation:

Edge length of the cube is given to be x centimeters

$V(x) = x^3 ...........(1)\\\\\text{At any instant when x = 2}\\\\V(2) = 2^3= 8\:\:cm^3\\\\\implies\frac{dv}{dt}=8\\\\\text{Now, differentiating (1) w.r.t. t}\\\\\frac{dv}{dt}=3\cdot x^2\cdot\frac{dx}{dt}\\\\\implies 8=3\cdot x^2\cdot \frac{dx}{dt}\\\\\text{When x = 2}\\\\\implies\frac{dx}{dt}=\frac{8}{3\times 4}\\\\\implies\frac{dx}{dt}=\frac{2}{3}$

Hence, the rate of change of x, in centimeters per minute, with respect to time is found to be :

$\frac{2}{3}$

Answer 2

Dv/dt=40    dv/dt=3x^2dx/dt
40=3x^2dx/dt
40=12dx/dt
10/3=dx/dt