If $V=4\pi r^3/3,$ at what rate in cubic units/ sec is V increasing when r= 10, and dr/dt=0.01 ?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $4\pi$
$\frac{dv}{dt}=4πr^2\frac{dr}{dt}$
$⇒4π(10)^2×0.01$
$=4π$
If $V=4\pi r^3/3,$ at what rate in cubic units/ sec is V increasing when r= 10, and dr/dt=0.01 ?
Correct answer: option 2
The correct answer is Option (2) → $4\pi$
$\frac{dv}{dt}=4πr^2\frac{dr}{dt}$
$⇒4π(10)^2×0.01$
$=4π$