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marin [14]
3 years ago
10

Carlos is filling a spherical balloon with water. if he increases the volume of the balloon from 4,188.79 cubic centimeters to 1

4,137.167 cubic centimeters in 12 seconds, then what is the average rate at which he has increased the balloon's surface area?
Mathematics
1 answer:
Harrizon [31]3 years ago
7 0

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.

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