Preview Activity 3.5.1. A spherical balloon is being inflated at a constant rate of 20 cubic inches per second. How fast is the radius of the balloon changing at the instant the balloon's diameter is 12 inches? Is the radius changing more rapidly when d =12 or When d =16? Why?

Answer:The radius is increasing at a rate of approximately 0.044 in/s when the diameter is 12 inches.Because [tex]\frac{dr}{dt}=\frac{5}{36\pi }>\frac{dr}{dt}=\frac{5}{64\pi }[/tex] the radius is changing more rapidly when the diameter is 12 inches.Step-by-step explanation:Let [tex]r[/tex] be the radius, [tex]d[/tex] the diameter, and [tex]V[/tex] the volume of the spherical balloon.We know [tex]\frac{dV}{dt}=20 \:{in^3/s}[/tex] and we want to find [tex]\frac{dr}{dt}[/tex]The volume of a spherical balloon is given by[tex]V=\frac{4}{3} \pi r^3[/tex]Taking the derivative with respect of time of both sides gives[tex]\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}[/tex]We now substitute the values we know and we solve for [tex]\frac{dr}{dt}[/tex]:[tex]d=2r\\\\r=\frac{d}{2}[/tex][tex]r=\frac{12}{2}=6[/tex][tex]\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}\\\\\frac{dr}{dt}=\frac{\frac{dV}{dt}}{4\pi r^2} \\\\\frac{dr}{dt}=\frac{20}{4\pi(6)^2 } =\frac{5}{36\pi }\approx 0.044[/tex]The radius is increasing at a rate of approximately 0.044 in/s when the diameter is 12 inches.When d = 16, r = 8 and [tex]\frac{dr}{dt}[/tex] is:[tex]\frac{dr}{dt}=\frac{20}{4\pi(8)^2}=\frac{5}{64\pi }\approx 0.025[/tex]The radius is increasing at a rate of approximately 0.025 in/s when the diameter is 16 inches.Because [tex]\frac{dr}{dt}=\frac{5}{36\pi }>\frac{dr}{dt}=\frac{5}{64\pi }[/tex] the radius is changing more rapidly when the diameter is 12 inches.