Question

(Urgent!!) A spherical balloon is leaking air at 2 cubic inches per hour. How fast is the balloon’s radius changing when the radius is 3 inches?

Answer 1

initial volume of balloon =

$\frac{4}{3} \times \frac{22}{7} \times r^{3} \\ \frac{4}{3} \times \frac{22}{7} \times {3}^{3} \\ 113.14$

so intial volume is 113.14 cubic inches

volume after one hour will be 113.14 - 2 = 111.14 inches

new radius

$111.14 = \frac{4}{3} \times \frac{22}{7} \times {x}^{3} \\ \frac{2333.94}{88} = {x}^{3} \\ 26.52 = {x}^{3} \\ x = 2.98$

Rate change of radius is

$\frac{2.98}{3} \times 100 \\ \frac{298}{3} \\ 99.93\%$

Rate change is 0.7% per hour the radius is decreasing