How fast is the surface area of a spherical balloon increasing when the radius is 10 cm and the volume is increasing at 15 cm3 /sec?
Solution:
dVdt=15cm3/sec, r=10cm,dAdt=?\dfrac{dV}{dt}=15 cm^3/sec,\ r=10 cm, \dfrac{dA}{dt}=?dtdV=15cm3/sec, r=10cm,dtdA=?
V=43πr3V=\dfrac43\pi r^3V=34πr3
On differentiating w.r.t rrr,
dVdt=4πr2drdt⇒15=4π(10)2drdt⇒drdt=380πcm/sec\dfrac{dV}{dt}=4\pi r^2 \dfrac{dr}{dt} \\ \Rightarrow 15=4\pi (10)^2 \dfrac{dr}{dt} \\ \Rightarrow \dfrac{dr}{dt}=\dfrac{3}{80\pi} cm/secdtdV=4πr2dtdr⇒15=4π(10)2dtdr⇒dtdr=80π3cm/sec
Now, A=4πr2A=4\pi r^2A=4πr2
dAdt=8πrdrdt=8π(10)(380π)=3cm2/sec\dfrac{dA}{dt}=8\pi r \dfrac{dr}{dt}=8\pi(10)(\dfrac{3}{80\pi})=3 cm^2/secdtdA=8πrdtdr=8π(10)(80π3)=3cm2/sec
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