The radius of a sphere shrinks from 10 cm to 9.8 cm. Find approximate

1.	The radius of a sphere shrinks from 10 cm to 9.8 cm. Find approximate decrease in its volume.
Answer» Given Radius of sphere = 10 cm ∆r = radius shrinks = 9.8 – 10 = – 0.2 cm Volume of sphere, V = 4/3πr³ Diff. w.r.t. r, dV/dx = 4/3 π 3r² = 4 πr² ∵ Approximation error in calculation of volume of sphere, dV = dV/dx × (∆r) dV = 4πr² × (∆r) ⇒ dV = 4π × (10)² × (- 0.2) ⇒ dV = – 400 × 0.2 π cm³ = -80 π cm³ Hence approximation error in volume is 80 π cm³

The radius of a sphere shrinks from 10 cm to 9.8 cm. Find approximate decrease in its volume.

Answer»

Given

Radius of sphere = 10 cm

∆r = radius shrinks

= 9.8 – 10

= – 0.2 cm

Volume of sphere, V = 4/3πr³

Diff. w.r.t. r,

dV/dx = 4/3 π 3r² = 4 πr²

∵ Approximation error in calculation of volume of sphere,

dV = dV/dx × (∆r)

dV = 4πr² × (∆r)

⇒ dV = 4π × (10)² × (- 0.2)

⇒ dV = – 400 × 0.2 π cm³

= -80 π cm³

Hence approximation error in volume is 80 π cm³