Respuesta :
Answer:
This is the rate at which the radius of the balloon is changing when the volume is 300 [tex]ft^3[/tex] [tex]\frac{dr}{dt}=-\frac{3}{225^{\frac{2}{3}}\pi ^{\frac{1}{3}}} \:\frac{ft}{h} \approx -0.05537 \:\frac{ft}{h}[/tex]
Step-by-step explanation:
Let [tex]r[/tex] be the radius and [tex]V[/tex] the volume.
We know that the gas is escaping from a spherical balloon at the rate of [tex]\frac{dV}{dt}=-12\:\frac{ft^3}{h}[/tex] because the volume is decreasing, and we want to find [tex]\frac{dr}{dt}[/tex]
The two variables are related by the equation
[tex]V=\frac{4}{3}\pi r^3[/tex]
taking the derivative of the equation, we get
[tex]\frac{d}{dt}V=\frac{d}{dt}(\frac{4}{3}\pi r^3)\\\\\frac{dV}{dt}=\frac{4}{3}\pi (3r^2)\frac{dr}{dt} \\\\\frac{dV}{dt}=4\pi r^2 \frac{dr}{dt}[/tex]
With the help of the formula for the volume of a sphere and the information given, we find [tex]r[/tex]
[tex]V=\frac{4}{3}\pi r^3\\\\300=\frac{4}{3}\pi r^3\\\\r^3=\frac{225}{\pi }\\\\r=\sqrt[3]{\frac{225}{\pi }}[/tex]
Substitute the values we know and solve for [tex]\frac{dr}{dt}[/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{12}{4\pi (\sqrt[3]{\frac{225}{\pi }})^2} \\\\\frac{dr}{dt}=-\frac{3}{\pi \left(\sqrt[3]{\frac{225}{\pi }}\right)^2}\\\\\frac{dr}{dt}=-\frac{3}{\pi \frac{225^{\frac{2}{3}}}{\pi ^{\frac{2}{3}}}}\\\\\frac{dr}{dt}=-\frac{3}{225^{\frac{2}{3}}\pi ^{\frac{1}{3}}} \approx -0.05537 \:\frac{ft}{h}[/tex]
Using implicit differentiation, it is found that the radius is changing at a rate of -0.0133 ft/hr.
The volume of a sphere of radius r is given by:
[tex]V = \frac{4\pi r^3}{3}[/tex]
In this problem, volume of 300 cubic feet, thus, this is used to find the radius.
[tex]300 = \frac{4\pi r^3}{3}[/tex]
[tex]4\pi r^3 = 900[/tex]
[tex]r = \sqrt[3]{\frac{900}{4\pi}}[/tex]
[tex]r = 4.153[/tex]
The rate of change is found, using implicit differentiation:
[tex]\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}[/tex]
Gas is escaping from a spherical balloon at the rate of 12 ft3/hr, thus:
[tex]\frac{dV}{dt} = -12[/tex]
The rate of change of the radius is now find:
[tex]\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}[/tex]
[tex]-12 = 4\pi(4.153)^2\frac{dr}{dt}[/tex]
[tex]\frac{dr}{dt} = -\frac{12}{4\pi(4.153)^2}[/tex]
[tex]\frac{dr}{dt} = -0.0133[/tex]
The radius is changing at a rate of -0.0133 ft/hr.
A similar problem is given at https://brainly.com/question/24158553
