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Sagot :
The question is incomplete. The complete question is :
The volume of a sphere is increasing at a rate of 7 cm^3/sec. Find the rate of change of its surface area when its volume is 4 pi / 3 cm^3. (Do not round your answer.)
Solution :
Volume of the sphere, [tex]$V=\frac{4}{3} \pi r^3$[/tex]
Given that [tex]$\frac{dV}{dt}= 7 \ cm^3/s $[/tex]
V = [tex]$\frac{4 \pi}{3} = \frac{4 \pi r^3}{3}$[/tex]
∴ r = 1
[tex]$\frac{dV}{dt} =\frac{4}{3} \pi .3.r^2 \ \frac{dr}{dt}$[/tex]
[tex]$7 = \frac{4}{3} ( \pi )(3) (1)^2 \frac{dr}{dt}$[/tex]
[tex]$\frac{dr}{dt}=\frac{7}{4 \pi}$[/tex] cm/s
Surface area, [tex]$A = 4 \pi r^2$[/tex]
[tex]$\frac{dA}{dt} = 8 \pi r \ \frac{dr}{dt}$[/tex]
[tex]$\frac{dA}{dt} = 8 \pi \times \frac{7}{4 \pi} \times 1$[/tex]
[tex]$=14 \ cm^2/s $[/tex]
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