Answer:
The answer is below
Step-by-step explanation:
a) dV/dt = 5 cm³/s
The volume (V) of the bubble is:
[tex]V=\frac{4}{3} \pi r^3\\\\Differentiating\ with\ respect\ to \ time(t)\\\\\frac{d}{dt}(V)=4\pi r^2 \frac{dr}{dt} \\\\Substituting\ \frac{d}{dt}(V)=5\ cm^3/s \ and\ r = 10\ cm\\\\5 = 4\pi *10^2*\frac{dr}{dt}\\\\5=400\pi *\frac{dr}{dt}\\\\\frac{dr}{dt}=0.004\ cm/s[/tex]
b) The surface area of the sphere (A) is:
[tex]A=4\pi r^2\\\\Differentiating\ w.r.t.\ t:\\\\\frac{dA}{dt}= 8\pi r\frac{dr}{dt} \\\\r=10, dr/dt=0.004\\\\\frac{dA}{dt}= 8\pi *10*0.004\\\\\frac{dA}{dt}=1\ cm^2/s[/tex]
The rate at which the radius of the bubble is changing at that time is [tex]0.0125 \pi cm^2/s[/tex]
The rate at which the surface area of the bubble is changing at that time is [tex]\pi^2 cm^2/s[/tex]
The formula for calculating the volume of a sphere is expressed as;
[tex]V =\frac{4}{3} \pi r^3[/tex]
r is the radius of the sphere
Differentiating the volume with respect to time;
[tex]\frac{dV}{dt} = \frac{dV}{dr} \times \frac{dr}{dt} \\\frac{dV}{dt} = 4 \pi r^2\frac{dr}{dt}[/tex]
Given the following parameters
dV/dt = 5 cm^3/sec.
r = 10 cm
Substitute into the formula to get dr/dt
[tex]\frac{dV}{dt} = 4 \pi (10)^2 \times \frac{dr}{dt}\\5 = 400 \pi \times \frac{dr}{dt}\\ \frac{dr}{dt}=0.0125 \pi cm/s\\[/tex]
The surface area of the bubble is expressed as:
[tex]S = 4 \pi r^2\\\frac{dS}{dt} = \frac{dS}{dr}\times \frac{dr}{dt}\\ \frac{dS}{dt} =8 \pi r \times 0.0125 \pi\\ \frac{dS}{dt} =\pi ^2 cm^2/s[/tex]
Hence the rate at which the surface area of the bubble is changing at that time is [tex]\pi^2 cm^2/s[/tex]
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