Respuesta :
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.
