Question

Carlos is filling a spherical balloon with water. if he increases the volume of the balloon from 4,188.79 cubic centimeters to 14,137.167 cubic centimeters in 12 seconds, then what is the average rate at which he has increased the balloon's surface area?

Answer 1

We know that the volume of a sphere of radius r is given by

$V=\frac{4}{3}\pi r^3$

Now, we have been given the initial volume was 4,188.79 cubic centimeters. Let us find the radius at this time.

Substituting the value of volume in the above equation, we get

$4,188.79=\frac{4}{3}\pi r^3\\ \\ 4\pi r^3=4,188.79 \times 3\\ \\ r^3=\frac{4188.79 \times 3}{4\pi}\\ \\ r^3=1000\\ \\ r=10$

Now, the final volume is given by 14,137.167 cubic centimeters. Thus, we have

$14137.167 =\frac{4}{3}\pi r^3\\ \\ 4\pi r^3=14137.167 \times 3\\ \\ r^3=\frac{14137.167\times 3}{4\pi}\\ \\ r^3=3375\\ \\ r=15$

Now, let us find the surface areas for these two radii.

For $r=10$

Surface area is given by

$S=4\pi r^2\\ \\ S=4\times 3.14 \times (10)^2\\ \\ S=1256 \text{ square centimeters}$

Similarly, for

$r=15\\ \\ S=4\times 3.14\times (15)^2\\ \\ S=2826 \text{ square centimeters}$

Therefore, the increase in the surface area is given by

$\Delta S= 2826-1256= 1570 \text{ square centimeters}$

Hence, the average rate at which the surface area is changing is given by

$\frac{1570}{12}=130.8$

Therefore, the average rate at which the surface area is changing is given by 130.8 square centimeters per second.