Answer:
Please see attachment
Step-by-step explanation:
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In this exercise we have to use our knowledge of volume, in this way it will be possible to calculate the spherical volume from the differential equations, so we will have to:
a)[tex]-6^{2\beta}k\pi ^{\beta}V^{2\beta}[/tex]
b) [tex]-K[/tex]
c) [tex]-4K\pi ^{1/2} S^{1/2}[/tex]
Knowing that the volume and sufrace area of the spherical raindrop of radius r is given by:
[tex]V=\frac{4}{3} \pi r^3\\S=4\pi r^2[/tex]
a) Given that the ratechange of the volume exactly equals -K times the surface area, so we have:
[tex]\frac{dV}{dT}=-KS\\=-K(4\pi r^2)\\=-4K\pi r^2\\=-4K\pi(r^3)^{2\beta}\\=-4K\pi(\frac{3V}{4\pi})^{2\beta}[/tex]
Knowing that:
[tex]V=\frac{4}{3}\pi r^3\\r^3= \frac{3V}{4\pi}[/tex]
Applying in the calculation before, that is:
[tex]=-4K\pi \frac{3^{2\beta}V^{2\beta}}{4^{2\beta}\pi ^{2\beta}} \\=4^{\beta}\pi ^{\beta}3^{2\beta}KV^{2\beta}\\= -2^{2\beta}3^{2\beta}K\pi^{\beta}V^{2\beta}\\=-6^{2\beta}K\pi^{\beta}V^{2\beta}[/tex]
b) The volume of the spherical raindrop of radius r is given by;
[tex]V=\frac{4}{3}\pi r^3 \\\frac{dV}{dt}= 4\pi r^2 \frac{dr}{dt}[/tex]
From part (a), we have:
[tex]\frac{dV}{dt}=-4K\pi r^2 \\4\pi r^2\frac{dr}{dt}=-4K\pi r^2 \\\frac{dr}{dt}=\frac{-4K\pi r^2}{4\pi r^2} = -K[/tex]
c) The volume of the spherical raindrop of radius r is given by:
[tex]V=\frac{4}{3}\pi r^3\\=\frac{4}{3}\pi (r^2)^{3/2}\\=\frac{4}{3}\pi \frac{S}{4\pi} ^{3/2}\\=\frac{\pi^{-1/2}S^{1/2}}{4}[/tex]
From the part (a):
[tex]\frac{\pi^{-1/2}S^{1/2}}{4}=-KS\\\frac{dS}{dt}=\frac{-4KS}{\pi^{-1/2}S^{1/2}}\\=-4K\pi^{1/2}S^{1/2}[/tex]
See more about volume at brainly.com/question/1578538
