Question

Two flies sit exactly opposite each other on the surface of a spherical balloon. If the balloon's volume doubles, by what factor does the distance between the flies change?

Short answer

Answer: The distance between the two flies changes by a factor of (\frac{2}{1})^{1/3}.

Step-by-step solution

Express the volume of a sphere

The volume (V) of a sphere with radius (r) can be written as: V = \frac{4}{3} \pi r^3 Since the volume doubles, we can express the new volume (V') as: V' = 2V

Determine the relationship between the new radius and the original radius

Using the formula for the volume of a sphere, we can write the new volume (V') in terms of the new radius (r'): V' = \frac{4}{3} \pi r'^3 Since V' = 2V, we can write: \frac{4}{3} \pi r'^3 = 2(\frac{4}{3} \pi r^3) Now we can solve for the ratio \frac{r'}{r}: \frac{r'^3}{r^3} = 2 \Rightarrow \frac{r'}{r} = (\frac{2}{1})^{1/3}

Apply the relationship to the surface distance

Let's denote the initial distance between the flies as d, and the new distance between them as d'. Note that the flies are exactly on opposite sides of the sphere, so the distance between them is equivalent to the spherical distance between the two antipodal points. The initial spherical distance between the flies can be represented as d = \pi r, since it is half the circumference of the sphere. Similarly, the new spherical distance can be represented as d' = \pi r'.

Determine the factor by which the distance between the flies changes

Now, we can write the ratio \frac{d'}{d} in terms of the relationship between r' and r: \frac{d'}{d} =\frac{\pi r'}{ \pi r} = \frac{r'}{r} = (\frac{2}{1})^{1/3} Thus, the factor by which the distance between the two flies changes is (\frac{2}{1})^{1/3}.