Use differentials to approximate the change in volume of a cube if the length of a side changes from 2 to 2.01cm.
I have no idea how to even start this.
∆v/∆x ≈ dv/dx
∆v ≈ v' ∆x
so
v(2+∆x) ≈ f(2) + dv/dx * ∆x = 8 + 3*2^2 * 0.01 = 8.12
2.01^3 = 8.120601
First, recall that the equation for getting the volume of a cube is V=s³
So, we can now get the differential of the equation V=s³ to approximate the change in volume:
dV/ds = 3s²
dV = 3s²ds
Plug in the given values to the differential, (s = 2, ds = 2.01; since 2cm was the initial length of the sides and 2.01cm was after the change)
dV = 3(2)² (2.01)
dV = 24.12
I hope this helps 🙂