56. A spherical balloon is being inflated and the radius of the balloon is increasing at a rate of 2 cmys. (a) Express the radius r of the balloon as a function of the time t (in seconds). (b) If V is the volume of the balloon as a function of the radius, find V 8 r and interpret it.

Answer: a) [tex]r=2t + r_0[/tex]b) [tex]V=\frac{4\pi }{3}(2t+r_0)^3[/tex]Step-by-step explanation:a) The rate means that the initial radius [tex]r_0[/tex] is changing at that speed, therefore for each unit of time (for example one second, if the rate is 2 cm per second), the radius is increased by 2 cm. Hence, after 1 second the ballon will have a radius of [tex]r_0+2[/tex], after 2 seconds [tex]r_0+4[/tex], and so on. b) The volume of a sphere is, in general, [tex]V=\frac{4\pi }{3}(r)^3[/tex]. If you replace r by what we found in the previuos answer it gives you the specific equation for this problem.You can interprete this result by noticing that even though the radius is only increased by 2 cm each second,  the volume is increased much more in the same time, which can be explained by the cubic relation between volume and radius. For instance, every time the radius is doubled, the volume is multiplied by 8.