Question

The edges of a cube increase at a rate of 2 cm divided by s. How fast is the volume changing when the length of each edge is 40 ​cm? Write an equation relating the volume of a​ cube, V, and an edge of the​ cube, a. nothing Differentiate both sides of the equation with respect to t. StartFraction dV Over dt EndFraction equals(nothing )StartFraction da Over dt EndFraction ​(Type an expression using a as the​ variable.) The rate of change of the volume is nothing cm cubed divided by sec. ​(Simplify your​ answer.)

Answer 1

Answer:

Step-by-step explanation:

Let the edge of the cube be a .

Given

$\frac{da}{dt} = 2 cm/s$

Volume V = a³

$\frac{dV}{dt} = 3a^ 2\frac{da}{dt}$

= 3a² x 2

= 6a²

If a = 40 cm

$\frac{dV}{dt} = 6 \times 40\times40$

= 9600 cm³/s .