Answer :
Answer:
[tex]\displaystyle \frac{dr}{dt}\approx 0,0112\ cm/sec[/tex]
Step-by-step explanation:
Rates of Change as Derivatives
If some variable V is a function of another variable r, we can compute the rate of change of one with respect to the other as the first derivative of V, or
[tex]\displaystyle V'=\frac{dV}{dr}[/tex]
The volume of a sphere of radius r is
[tex]\displaystyle V=\frac{4}{3}\pi r^3[/tex]
The volume of the balloon is growing at a rate of [tex]9\ cm^3/sec[/tex]. This can be written as
[tex]\displaystyle \frac{dV}{dt}=9[/tex]
We need to compute the rate of change of the radius. Note that both the volume and the radius are functions of time, so we need to use the chain rule. Differentiating the volume with respect to t, we get
[tex]\displaystyle \frac{dV}{dt}=\displaystyle \frac{dV}{dr}\displaystyle \frac{dr}{dt}[/tex]
[tex]\displaystyle \frac{dV}{dt}=4\pi r^2 \frac{dr}{dt}[/tex]
solving for [tex]\displaystyle \frac{dr}{dt}[/tex]
[tex]\displaystyle \frac{dr}{dt}=\frac{\frac{dV}{dt}}{4\pi r^2}[/tex]
We need to find the value of r, which can be obtained by using the condition that in that exact time
[tex]\displaystyle V=\frac{2048}{3}\pi\ cm^3[/tex]
[tex]\displaystyle \frac{2048}{3}\pi=\frac{4}{3}\pi r^3[/tex]
Simplifying and isolating r
[tex]\displaystyle r^3=512[/tex]
[tex]\displaystyle r=\sqrt[3]{512}=8\ cm[/tex]
Replacing in the rate of change
[tex]\displaystyle \frac{dr}{dt}=\frac{9}{4\pi 8^2}[/tex]
[tex]\displaystyle \frac{dr}{dt}=\frac{9}{256\pi }[/tex]
[tex]\displaystyle \frac{dr}{dt}\approx 0,0112\ cm/sec[/tex]