Question

The volume of a cube increases at a rate of 4 m^3 /sec . Find the rate at which the side of the cube changes when its length is 6 m. Submit an exact answer in fractional form. Provide your answer below:

Answer 1

1/27 m/s

Explanation:

To address the problem of determining the rate at which the side of a cube changes given that its volume increases at a rate of 4 m³/s, we need to apply the concept of related rates in calculus. This involves using derivatives to relate the rates of change of different quantities.

Given:

• dV/dt = 4 m³/s
• L = 6 m

We know the volume of a cube is:

`V = L^3`

Where 'L' is the length of one side of the cube. Take the derivative of this function with respect to time, 't'.

`\Longrightarrow (d)/(dt)[V] = (d)/(dt)[L^3]\\\\\\\\\Longrightarrow (dV)/(dt)= 3L^2\cdot(dL)/(dt)`

Solve for dL/dt:

`\therefore (dL)/(dt)= (\left((dV)/(dt)\right))/(3L^2)`

Plug in our given values:

`\Longrightarrow (dL)/(dt)= \frac{\left(4 \text{ m$^3$/s}\right)}{3(6 \text{ m})^2}\\\\\\\\\Longrightarrow (dL)/(dt)= \frac{4 \text{ m$^3$/s}}{3(36 \text{ m$^2$})}\\\\\\\\\Longrightarrow (dL)/(dt)= \frac{4 \text{ m$^3$/s}}{108 \text{ m$^2$}}\\\\\\\\\therefore (dL)/(dt) = \boxed{(1)/(27) \text{ m/s}}`

Therefore, the rate at which the side of the cube changes when its length is 6 m is 1/27 m/s.