Respuesta :
Answer:
[tex]\frac{dr}{dt}\Bigr|_{\substack{r=3\:ft}}=4 \:\frac{ft}{min}[/tex]
[tex]\frac{dS}{dt}\Bigr|_{\substack{r=3\:ft}}=96\pi \:\frac{ft^2}{min}[/tex]
Step-by-step explanation:
Let [tex]r[/tex] be the radius, [tex]S[/tex] the surface area, and [tex]V[/tex] the volume of the spherical balloon.
Given:
[tex]\frac{dV}{dt}= 144\pi \:\frac{ft^3}{min}[/tex]
We want to find [tex]\frac{dr}{dt}\Bigr|_{\substack{r=3\:ft}}[/tex] and [tex]\frac{dS}{dt}\Bigr|_{\substack{r=3\:ft}}[/tex]
The volume [tex]V[/tex] of the sphere is given by [tex]V=\frac{4}{3}\pi r^3[/tex] while the surface area [tex]S[/tex] is [tex]S=4\pi r^2[/tex].
To find how fast is the balloon's radius increasing at the instant the radius is 3 ft, you need to:
Differentiate the volume formula with respect to time to 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]
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{144\pi }{4\pi (3)^2 } \\\\\frac{dr}{dt}=4 \:\frac{ft}{min}[/tex]
To find how fast is the surface area increasing, you need to:
[tex]r=(\frac{3V}{4\pi })^{1/3} \Rightarrow S=4\pi (\frac{3V}{4\pi })^{2/3}=(36\pi )^{1/3}V^{2/3}[/tex]
When [tex]r=3[/tex], we have
[tex]V=\frac{4}{3}\pi (3)^3=36\pi[/tex]
Differentiate the surface area formula with respect to time to get
[tex]\frac{d}{dt}S= \frac{d}{dt}((36\pi )^{1/3}V^{2/3})\\\\\frac{dS}{dt}=(36\pi )^{1/3}\frac{2}{3V^{1/3}}\frac{dV}{dt}[/tex]
Substitute the values we know
[tex]\frac{dS}{dt}=(36\pi )^{1/3}\frac{2}{3(36\pi )^{1/3}}(144\pi )\\\\\frac{dS}{dt}=\frac{2\left(36\pi \right)^{\frac{1}{3}}\cdot \:144\pi }{3\left(36\pi \right)^{\frac{1}{3}}}\\\\\frac{dS}{dt}=\frac{2\cdot \:144\pi }{3}\\\\\frac{dS}{dt}=\frac{288\pi }{3}\\\\\frac{dS}{dt}=96\pi \:\frac{ft^2}{min}[/tex]
