Answer to Question #179841 in Calculus for chan

Question #179841

Air is escaping from a spherical balloon at the rate of 2π‘π‘š3 per minute. How fast is the radius shrinking when the volume is 36πœ‹ π‘π‘š3 ?


Find the rate of change of the area 𝐴, of the circle with respect to its circumference C, 𝑖. 𝑒 𝑑𝐴 𝑑�


1
Expert's answer
2021-04-25T17:41:20-0400

Volume=43Ο€r3Volume={4 \over 3} \pi r^3

Since volume is 36πœ‹ π‘π‘š3

Then, 36Ο€=43Ο€r336\pi= {4 \over 3}\pi r^3

r3=27r^3=27

r=3cm


Since, V=43Ο€r3V={4 \over 3} \pi r^3 and dvdt={dv \over dt}= 2cm3 per mins

dvdt=4Ο€r2drdt{dv \over dt}=4\pi r^2{dr \over dt}


2=4Ο€(32)drdt2=4\pi(3^2) {dr \over dt}


drdt=118Ο€cm/mins{dr \over dt}={1 \over 18\pi}cm/mins




Area(A)=Ο€r2Area(A)=\pi r^2


Circumference(C)=2Ο€rCircumference (C)= 2\pi r

r=C2Ο€r={C \over 2\pi}


∴A=Ο€(C2Ο€)2\therefore A= \pi({C \over 2\pi})^2


A=C24Ο€A= {C^2 \over 4\pi}


dAdC=C2Ο€units{dA \over dC}={C \over 2\pi}units












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