Question

A beach ball is inflated so its diameter increases from 12 inches to 12.5 inches. Given that the formula for the volume of a sphere is V=4/3π,r^3 the change in the volume of the beach ball using differentials is approximately.
a. 367π cubic inches.
b. 727π cubic inches.
c. 1447π cubic inches.
d. 288.7π cubic inches.

Answer 1

Answer:

V = 36π cubic inches.

Step-by-step explanation:

Given that:

The volume of the sphere is $V = \dfrac{4}{3} \pi r^3$

If the diameter is represented by P.

Then P = 2r

and r = P/2

∴

$V = \dfrac{4}{3} \pi ( \dfrac{P}{2})^3$

By differentiating both sides;

$dV = \dfrac{4}{3} \pi \times 3 ( \dfrac{P}{2})^2 \times \dfrac{1}{2} \times dP$

$dV = \dfrac{1}{2} \pi P^2 \ dP$

where;

P = 12

dP = 12.5 - 12

dP = 0.5

Then;

$\dfrac{1}{2} \pi ( 12)^2 \times 0.5$

= 36π

Thus, the volume is V = 36π cubic inches.