Question

The volume v of a right circular cone of radius r and height h is given by v = 1/3 pi r^2 h. suppose that the height decreases from 20 in to 19.95 in and the radius increases from 4 in to 4.05 in. compare the change in volume of the cone with an approximation of this change using a total differential.

Answer 1

V = (π/3)*r^2*h
dV = (π/3)*(2r*h*dr +r^2*dh)

The approximate change in volume is
.. ∆V ≈ (π/3)*(2*4*20*0.05 +4^2*(-.05))
.. = (π/3)*(8 - 0.8) = 7.2(π/3) . . . . . in^3

The actual change in volume is
.. (π/3)*(4.05^2*19.95 -4^2*20) = 7.229875(π/3) . . . . . in^3

The actual change in volume is slightly more than the linear approximation would suggest.